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Logarithmic Integrals, Polylogarithms, and associated functions

DreamWeaver

Well-known member
Sep 16, 2013
337
This is not so much a tutorial, but rather a collection of useful results and techniques. Some of the proofs will be quite long, since as much as possible, I'll aim to prove most results and functional relations used herein, rather than just present another's identity as fact. There will be a few exceptions - to avoid clutter - but not many (and they'll be along more basic lines, like Leibniz's rule for differentiation of a product).


-------------------


A wide variety of logarithmic integrals can be obtained by evaluating a simpler parametric integral, and then differentiating that w.r.t the parameter in question. Here's the simplest example of all...

Let \(\displaystyle 0 < z \le 1\,\) and \(\displaystyle q > 0\,\) be real numbers, then


\(\displaystyle \mathcal{I}(z,q)=\int_0^zx^{q-1}\,dx=\frac{z^q}{q}\)


Repeated differentiation of this integral w.r.t. the parameter \(\displaystyle q\,\) then gives:


\(\displaystyle \mathcal{I}^{(m)}(z,q)=\frac{d^m}{dq^m}\,\int_0^zx^{q-1}\,dx=\int_0^z\left[\frac{d^m}{dq^m}\,x^{q-1}\right]\,dx=\)


\(\displaystyle \int_0^zx^{q-1}(\log x)^m=\frac{d^m}{dq^m}\,\frac{z^q}{q}=\frac{(-1)^mm!\,z^q}{q^{m+1}}\)


This simple result can be used to help evaluate vast numbers of considerably more complicated logarithmic integrals. The general idea is the same, however. Let's say we have a continuous function \(\displaystyle \phi(x)\,\), such that the following integral is meaningful (and at worst has singularities near \(\displaystyle x=0\,\), or \(\displaystyle x=z\,\), but nowhere in between):

\(\displaystyle \mathcal{P} (q,z)=\int_0^zx^{q-1}\,\phi(x)\,dx\)


If we can evaluate this integral, the differentiation of that result will give us a closed form for integrals of the type


\(\displaystyle \mathcal{P}^{(m)}(q,z)=\int_0^zx^{q-1}(\log x)^m\,\phi(x)\,dx\)


and


\(\displaystyle \mathcal{P}^{(m)}(1,z)=\lim_{q\to 1}\int_0^zx^{q-1}(\log x)^m\,\phi(x)\,dx=\int_0^1(\log x)^m\,\phi(x)\,dx\)


-------------------



Here's an example. Let \(\displaystyle \phi(x)=\log(1-x^p)\,\), for \(\displaystyle p\in \mathbb{R}^+\,\). Using the series definition for the logarithm

\(\displaystyle \log(1-x)=-\sum_{k=1}^{\infty}\frac{x^k}{k}\)


We get

\(\displaystyle \int_0^zx^{q-1}\log(1-x^p)\,dx=-\sum_{k=1}^{\infty}\frac{1}{k}\int_0^zx^{kp+q-1}\,dx=\)


\(\displaystyle -\sum_{k=1}^{\infty}\frac{1}{k}\left[\frac{x^{kp+q}}{(kp+q)}\right]_0^z=-\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k(kp+q)}=

-\sum_{k=1}^{\infty}\frac{\frac{1}{q}[(kp+q)-kp]}{k(kp+q)}z^{kp+q}=\)


\(\displaystyle -\frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k}+ \frac{p}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{(kp+q)}=

-\frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{k}+ \frac{1}{q}\sum_{k=1}^{\infty}\frac{z^{kp+q}}{(k+q/p)}\)


Now, if we let \(\displaystyle z=1\,\), then those last two series are expressible in terms of the polygamma function. The polygamma function has the series representation


\(\displaystyle \psi_0(z)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)-\gamma\)


Or, rearranging the terms


\(\displaystyle \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)=\gamma+\psi_0(z)\)


Where \(\displaystyle \gamma\,\) is the Euler-Mascheroni constant. Using this definition, we can re-write our series - in the case \(\displaystyle z=1\,\) - as:


\(\displaystyle \int_0^1x^{q-1}\log(1-x^p)\,dx=-\frac{1}{q}\,\left[\sum_{k=1}^{\infty}\frac{1}{k}- \sum_{k=1}^{\infty}\frac{1}{(k+q/p)}\right]=\)


\(\displaystyle -\frac{1}{q}\,\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k-1+\left(\frac{p+q}{p}\right)}\right)=\)


\(\displaystyle -\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)


-------------------


Before we move on, a quick word about the polygamma function. By definition, the polygamma function is the logarithmic derivative of the gamma function:


\(\displaystyle \psi_0(z)=\frac{d}{dz}\, \log\Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}\)


Similarly, higher order polygamma functions are defined as the higher order derivatives of the loggamma function \(\displaystyle \log\Gamma(x)\,\), so for \(\displaystyle m\ge 1\,\)


\(\displaystyle \psi_m(z)=\frac{d^{m+1}}{dz^{m+1}}\log\Gamma(z)= \frac{d^m}{dz^m} \psi_0(z)=\)


\(\displaystyle \frac{d^m}{dz^m} \left[\psi_0(z)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n-1}\right)-\gamma\right]=\)


\(\displaystyle \frac{d^m}{dz^m}\sum_{n=1}^{\infty}\frac{1}{z+n-1}=(-1)^mm!\,\sum_{n=1}^{\infty}\frac{1}{(z+n-1)^{m+1}}\)


We now have a series expansion for the higher order polygamma functions:


\(\displaystyle \psi_{m \ge 1}(z)=(-1)^mm!\,\sum_{n=1}^{\infty}\frac{1}{(z+n-1)^{m+1}}\)


-------------------



Returning to our logarithmic integral


\(\displaystyle \int_0^1x^{q-1}\log(1-x^p)\,dx=-\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)


We can now differentiate both sides of this equation m-times with respect to q, to obtain


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=-\frac{d^m}{dq^m}\,\frac{1}{q}\left[\gamma+\psi_0 {\left(\frac{p+q}{p}\right)}\right]\)


To evaluate the RHS, we use Leibniz's rule for differentiation of a product:


\(\displaystyle \frac{d^m}{dx^m} [f(x)\, g(x)]=\sum_{j=0}^m\binom{m}{j}\left[\frac{d^j}{dx^j}\, f(x)\right]\, \left[\frac{d^{m-j}}{dx^{m-j}}\, g(x)\right]\)



\(\displaystyle \Rightarrow\)


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=\)


\(\displaystyle \left(-\gamma\, \frac{d^m}{dq^m}\,\frac{1}{q}\right)-\sum_{j=0}^m\binom{m}{j} \left(\frac{d^{m-j}}{dx^{m-j}}\,\frac{1}{q}\right)\left[\frac{d^j}{dx^j}\, \psi_0 {\left(\frac{p+q}{p}\right)} \right]=\)


\(\displaystyle \frac{(-1)^{m+1}m!\,\gamma}{q^{m+1}}-\sum_{j=0}^m\binom{m}{j}\left[\frac{(-1)^{m-j}(m-j!)}{q^{m-j+1}}\right]\left[\frac{1}{p^j}\, \psi_j {\left(\frac{p+q}{p}\right)}\right]=\)


\(\displaystyle \frac{(-1)^{m+1}m!\,\gamma}{q^{m+1}}-\sum_{j=0}^m\frac{m!}{j!\,(m-j)!}\left[\frac{(-1)^{m-j}(m-j!)}{q^{m-j+1}}\right]\left[\frac{1}{p^j}\, \psi_j {\left(\frac{p+q}{p}\right)}\right]=\)


\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\left[\gamma+\sum_{j=0}^m\frac{1}{j!}\left(-\frac{q}{p}\right)^j \psi_j {\left(\frac{p+q}{p}\right)} \right]\)




-------------------


Final Result:



\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=\)



\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\left[\gamma+\sum_{j=0}^m\frac{1}{j!}\left(-\frac{q}{p}\right)^j \psi_j {\left(\frac{p+q}{p}\right)} \right]\)

Comments and/or questions should be posted here:

http://mathhelpboards.com/commentar...polylogarithms-associated-functions-6524.html
 
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337
Continuing on from the previous result, further evaluations are easily obtained via nothing more than basic manipulation of logarithms.

For the first example, note that

\(\displaystyle \log(1-x^{2p})=\log(1+x^p)+\log(1-x^p)\)


And so


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1+x^p)\,dx=\)


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^{2p})\,dx-\int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx\)


Using the formula

\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^p)\,dx=\)



\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\left[\gamma+\sum_{j=0}^m\frac{1}{j!}\left(-\frac{q}{p}\right)^j \psi_j {\left(\frac{p+q}{p}\right)} \right]\)

we obtain


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1+x^p)\,dx=\)


\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\sum_{j=0}^m\frac{1}{j!} \left(-\frac{q}{2p}\right)^j \psi_j {\left(\frac{2p+q}{2p}\right)}-
\)

\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\sum_{j=0}^m\frac{1}{j!} \left(-\frac{q}{p}\right)^j \psi_j {\left(\frac{p+q}{p}\right)}=\)


\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\sum_{j=0}^m\frac{1}{j!} \left(-\frac{q}{p}\right)^j \left[ \frac{1}{2^j} \psi_j {\left(\frac{2p+q}{2p}\right)}- \psi_j {\left(\frac{p+q}{p}\right)} \right]\)



------------------------


For the second example, we use the simple factorizations


\(\displaystyle (1-x^3)=(1-x)(1+x+x^2)\)
\(\displaystyle (1-x^4)=(1-x)(1+x+x^2+x^3)\)
.. .. .. ..
.. .. .. ..
.. .. .. ..
\(\displaystyle (1-x^{n+1})=(1-x)(1+x+\, \cdots \, +x^n)\)


to obtain


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1+x+\, \cdots \, +x^p)\,dx=\)


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x^{p+1})\,dx-\int_0^1x^{q-1}(\log x)^m\log(1-x)\,dx=\)


\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\sum_{j=0}^m\frac{(-q)^j}{j!} \left[ \frac{1}{(p+1)^j} \psi_j {\left(\frac{p+q+1}{p+1}\right)}- \psi_j {\left(q+1\right)} \right]\)



------------------------


For the third example, we use the similar factorizations


\(\displaystyle (1+x^3)=(1-x)(1-x+x^2)\)
\(\displaystyle (1+x^4)=(1-x)(1-x+x^2-x^3)\)
.. .. .. ..
.. .. .. ..
.. .. .. ..
\(\displaystyle (1+x^{n+1})=(1-x)(1-x+x^2- \, \cdots \, +(-1)^nx^n)\)


to obtain


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x+x^2- \, \cdots \, +(-1)^px^p)\,dx=\)


\(\displaystyle \frac{(-1)^{m+1}m!} {q^{m+1}}\,\sum_{j=0}^m\frac{1}{j!} \left(-\frac{q}{p+1}\right)^j \left[ \frac{1}{2^j} \psi_j {\left(\frac{2p+q+2}{2p+2}\right)}- \psi_j {\left(\frac{p+q+1}{p+1}\right)} \right] \)

\(\displaystyle -\frac{(-1)^{m+1}m!} {q^{m+1}}\,\sum_{j=0}^m\frac{(-q)^j}{j!}
\, \psi_j(q+1) - \frac{(-1)^{m+1}m!} {q^{m+1}}\, \gamma\)



------------------------
------------------------



Result 1:


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1+x^p)\,dx=\)


\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\sum_{j=0}^m\frac{1}{j!} \left(-\frac{q}{p}\right)^j \left[ \frac{1}{2^j} \psi_j {\left(\frac{2p+q}{2p}\right)}- \psi_j {\left(\frac{p+q}{p}\right)} \right]\)




Result 2:


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1+x+\, \cdots \, +x^p)\,dx=\)


\(\displaystyle \frac{(-1)^{m+1}m!}{q^{m+1}}\,\sum_{j=0}^m\frac{(-q)^j}{j!} \left[ \frac{1}{(p+1)^j} \psi_j {\left(\frac{p+q+1}{p+1}\right)}- \psi_j {\left(q+1\right)} \right]\)




Result 3:


\(\displaystyle \int_0^1x^{q-1}(\log x)^m\log(1-x+x^2- \, \cdots \, +(-1)^px^p)\,dx=\)


\(\displaystyle \frac{(-1)^{m+1}m!} {q^{m+1}}\,\sum_{j=0}^m\frac{1}{j!} \left(-\frac{q}{p+1}\right)^j \left[ \frac{1}{2^j} \psi_j {\left(\frac{2p+q+2}{2p+2}\right)}- \psi_j {\left(\frac{p+q+1}{p+1}\right)} \right] \)

\(\displaystyle -\frac{(-1)^{m+1}m!} {q^{m+1}}\,\sum_{j=0}^m\frac{(-q)^j}{j!}
\, \psi_j(q+1) - \frac{(-1)^{m+1}m!} {q^{m+1}}\, \gamma\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
The Gamma Function, as well as the closely-related Beta and Polygamma functions, are all essential tools in the evaluation of all sorts of logarithmic integrals. So too, are the derivatives of the Gamma function - especially at the special value \(\displaystyle x=1\,\). For this reason, the results presented below might at first appear to be a bit of a detour into the land of off-topic. Later posts, however, prove just how relevant these constants are.

Define


\(\displaystyle \Gamma^{(n)}(x)\,\Big|_{x=1}=\frac{d^n}{dx^n}\, \Gamma(x)\,\Big|_{x=1} =\gamma_n\)


where \(\displaystyle \gamma_n\,\), are called the the Stieltjes constants. To find these constants, we could simply start out by investigating the (earlier) series definition of the polygamma function, but let's take it back one step further. The following expression for the gamma function is due to Weierstrauss:


\(\displaystyle \frac{1}{\Gamma(x)}=xe^{\gamma x}\prod_{k=1}^{\infty}\left[e^{-x/k}\left(1+\frac{x}{k}\right)\right]\)


Where the Euler-Mascheroni constant, \(\displaystyle \gamma\, \) is defined by the limit:


\(\displaystyle \gamma = \lim_{k\to\infty}\,\left(1+\frac{1}{2}+\frac{1}{3}+\, \cdots \, +\frac{1}{k}-\log k\right) \approx 0.57721566\cdots\)


Taking the natural logarithm on both sides of the Weierstrauss product form gives


\(\displaystyle \log\Gamma(x)=-\Bigg\{ \log x +\gamma x+\sum_{k=1}^{\infty}\left[
\log\left(1+\frac{x}{k}\right)-\frac{x}{k}\right] \Bigg\}\)


Let's check that differentiation of both sides of this relation does indeed give the series definition for the polygamma function that we used earlier:


\(\displaystyle \frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}\equiv \psi_0(x)= \)


\(\displaystyle -\frac{d}{dx}\,\Bigg\{ \log x +\gamma x+\sum_{k=1}^{\infty}\left[
\log\left(1+\frac{x}{k}\right)-\frac{x}{k}\right] \Bigg\}=\)


\(\displaystyle -\Bigg\{ \frac{1}{x} +\gamma +\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{1+x/k}\right)-
\sum_{k=1}^{\infty}\frac{1}{k} \Bigg\}=\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}-\Bigg\{ \sum_{k=1}^{\infty}\left(\frac{1}{k+x}\right)+ \frac{1}{x} \Bigg\} - \gamma=\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}-\Bigg\{ \sum_{k=1}^{\infty}\left(\frac{1}{k-1+x}\right) \Bigg\} - \gamma=\)


\(\displaystyle \sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k-1+x}\right)-\gamma = \psi_0(x)\, . \, \Box\)


Incidentally, it is occasionally convenient to express this series in terms of the Zeta function \(\displaystyle \zeta(s)\,\)


\(\displaystyle \zeta(s)=\sum_{k=1}^{\infty}\frac{1}{k^s}\)


In particular, since

\(\displaystyle \zeta(1)=\sum_{k=1}^{\infty}\frac{1}{k}\)


We can re-write the polygamma function series as


\(\displaystyle \psi_0(x)=\zeta(1)-\gamma-\sum_{k=1}^{\infty}\frac{1}{k-1+x}=\zeta(1)-\gamma-\sum_{k=0}^{\infty}\frac{1}{k+x}
\)


Note that the zeta function is divergent at 1 - where it has a simple pole, and similarly, the infinite series in x that's subtracted from it also diverges, but at a different rate. The end result is that, for finite x not equal to zero or a negative integer, the difference of these two series is a finite limit.


--------------------


Now let's calculate some of the Stieltjes constants. Recall that


\(\displaystyle \psi_0(x)=\frac{\Gamma'(x)}{\Gamma(x)}\)


So


\(\displaystyle \Gamma'(x)=\Gamma(x)\psi_0(x)\)


Hence


\(\displaystyle \Gamma^{(n+1)}(x)=\frac{d^n}{dx^n}\, \Gamma(x)\psi_0(x)=\sum_{j=0}^n\binom{n}{j}\Gamma^{(n-j)}(x)\,\psi_{(j)}(x)
\)


[By Leibniz's rule for the repeated differentiation of a product of two functions].

When \(\displaystyle x=1\,\), we recover the Stieltjes constants:


\(\displaystyle \gamma_1=\Gamma'(1)=\Gamma(1)\psi_0(1)\)


\(\displaystyle \gamma_{n+1}=\Gamma^{(n+1)}(1)=\sum_{j=0}^n\binom{n}{j}\Gamma^{(n-j)}(1)\,\psi_{(j)}(1)\)


From the integral definition of the Gamma function


\(\displaystyle \Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}\,dt\)


So


\(\displaystyle \Gamma(1)=\int_0^{\infty}e^{-t}\,dt=-e^{-t}\,\Bigg|_0^{\infty}=1\)


Hence


\(\displaystyle \gamma_1=\psi_0(1)=\lim_{x\to 1}\left[\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k-1+x}\right)-\gamma\right]=-\gamma\)


To evaluate higher order Stieltjes constants, we will first need to evaluate higher order polygamma functions at \(\displaystyle x=1\,\).


\(\displaystyle \psi_0(x)=\zeta(1)-\gamma-\sum_{k=0}^{\infty}\frac{1}{k+x}
\)


\(\displaystyle \Rightarrow\)


\(\displaystyle \psi_{m\ge 1}(x)=\frac{d^m}{dx^m} \, \Bigg\{\zeta(1)-\gamma-\sum_{k=0}^{\infty}\frac{1}{k+x} \, \Bigg\} =
\)


\(\displaystyle (-1)^{m+1}m!\, \sum_{k=0}^{\infty}\frac{1}{(k+x)^{m+1}}\)


Setting \(\displaystyle x=1\,\) into this series gives


\(\displaystyle \psi_{m\ge 1}(1)=(-1)^{m+1}m!\,\sum_{k=0}^{\infty}\frac{1}{(k+1)^{m+1}}=\)


\(\displaystyle (-1)^{m+1}m!\,\sum_{k=1}^{\infty}\frac{1}{k^{m+1}}=(-1)^{m+1}m!\,\zeta(m+1)\)


A little later on in this thread I'll present a (relatively) simple method for determining the exact value of lower order zeta functions of the type \(\displaystyle \zeta(2m)\,\) where \(\displaystyle m\in\mathbb{Z}^+\,\) (these are rational powers of \(\displaystyle \pi\,\) ), but for now I'll just state a few lower order values. Incidentally, for odd, nonnegative integer values, the zeta functions have no know closed-form evaluation in terms of simpler functions. And one final point, the integral to be used later on to evaluate various zeta values is itself an important logarithmic integral.


\(\displaystyle \zeta(2)=\frac{\pi^2}{6}\)

\(\displaystyle \zeta(4)=\frac{\pi^4}{90}\)

\(\displaystyle \zeta(6)=\frac{\pi^6}{945}\)

\(\displaystyle \zeta(8)=\frac{\pi^8}{9450}\)

And more generally...

\(\displaystyle \zeta(2m)=(-1)^{m+1}\frac{(2\pi)^{2m}B_{2m}}{2\, (2m)!}\)


Now we can work out the precise values of a few particular higher order polygamma functions at \(\displaystyle x=1\,\)


\(\displaystyle \psi_{m\ge 1}(1)=(-1)^{m+1} m!\,\zeta(m+1)\)


\(\displaystyle \Rightarrow\)


\(\displaystyle \psi_1(1)=(-1)^21!\, \zeta(2)=\frac{\pi^2}{6}\)


\(\displaystyle \psi_2(1)=(-1)^32!\, \zeta(3)=-2\zeta(3)\)


\(\displaystyle \psi_3(1)=(-1)^43!\, \zeta(4)=\frac{\pi^4}{15}\)


\(\displaystyle \psi_4(1)=(-1)^54!\, \zeta(5)=-24\zeta(5)\)


\(\displaystyle \psi_5(1)=(-1)^65!\, \zeta(6)=\frac{8\pi^6}{63}\)


\(\displaystyle \psi_6(1)=(-1)^76!\, \zeta(7)=-720\zeta(7)\)



Part 2 to follow very shortly...
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Apologies, folks, but that last post was getting too long, so here's the rest...

Here's a brief summary of the key results from the previous post:



\(\displaystyle \Gamma(1)=1\)

\(\displaystyle \gamma_1\equiv \psi_0(1)\equiv \Gamma^{(1)}(1)=-\gamma\)

\(\displaystyle \gamma_{n+1}=\Gamma^{(n+1)}(1)=\sum_{j=0}^n\binom{n}{j}\Gamma^{(n-j)}(1)\,\psi_{(j)}(1)\)

-----------------------

\(\displaystyle \psi_1(1)=(-1)^21!\, \zeta(2)=\frac{\pi^2}{6}\)


\(\displaystyle \psi_2(1)=(-1)^32!\, \zeta(3)=-2\zeta(3)\)


\(\displaystyle \psi_3(1)=(-1)^43!\, \zeta(4)=\frac{\pi^4}{15}\)


\(\displaystyle \psi_4(1)=(-1)^54!\, \zeta(5)=-24\zeta(5)\)


\(\displaystyle \psi_5(1)=(-1)^65!\, \zeta(6)=\frac{8\pi^6}{63}\)


\(\displaystyle \psi_6(1)=(-1)^76!\, \zeta(7)=-720\zeta(7)\)


Using the finite series expansion above, for \(\displaystyle \gamma_k\,\), we get


\(\displaystyle \gamma_2=\Gamma^{(1)}(1)\psi_0(1)+\Gamma(1)\psi_1(1)=(-\gamma)^2+\frac{\pi^2}{6}=\gamma^2+\frac{\pi^2}{6}\)


-----------------------


\(\displaystyle \gamma_3=\Gamma^{(2)}(1)\psi_0(1)+2\Gamma^{(1)}(1)\psi_1(1)+\Gamma(1)\psi_2(1)=\)


\(\displaystyle \left(\gamma^2+\frac{\pi^2}{6}\right)(-\gamma)+2(-\gamma)\frac{\pi^2}{6}+[-2\zeta(3)]=\)


\(\displaystyle -\gamma^3-\frac{\pi^2\gamma}{2}-2\zeta(3)\)


-----------------------


\(\displaystyle \gamma_4=\Gamma^{(3)}(1)\psi_0(1)+3\Gamma^{(2)}(1)\psi_1(1)+3\Gamma^{(1)}(1)\psi_2(1)+\Gamma(1)\psi_3(1)=\)


\(\displaystyle \left(-\gamma^3-\frac{\pi^2\gamma}{2}-2\zeta(3)\right)(-\gamma)+3\left(\gamma^2+\frac{\pi^2}{6}\right) \frac{\pi^2}{6}+3(-\gamma)[-2\zeta(3)]+\frac{\pi^4}{15}=\)


\(\displaystyle \gamma^4+\pi^2\gamma^2+8\gamma\zeta(3)+\frac{3\pi^4}{20}\)


-----------------------


\(\displaystyle \gamma_5=\Gamma^{(4)}(1)\psi_0(1)+4\Gamma^{(3)}(1)\psi_1(1)+6\Gamma^{(2)}(1)\psi_2(1)+4\Gamma^{(1)}(1)\psi_3(1)+\Gamma(1)\psi_4(1)=\)


\(\displaystyle \left(\gamma^4+\pi^2\gamma^2+8\gamma\zeta(3)+ \frac{3\pi^4}{20}\right)(-\gamma)+4\left(-\gamma^3-\frac{\pi^2\gamma}{2}-2\zeta(3)\right) \frac{\pi^2}{6}+\)


\(\displaystyle 6\left(\gamma^2+\frac{\pi^2}{6}\right)[-2\zeta(3)]+ 4(-\gamma)\frac{\pi^4}{15}+[-24\zeta(5)]=\)


\(\displaystyle -\gamma^5-\frac{5\pi^2\gamma^3}{3}-\frac{3\pi^4\gamma}{4}-10\left(\frac{\pi^2}{3}+2\gamma^2\right)\zeta(3)-24\zeta(5)\)


-----------------------


Although we've only evaluated the first five Euler-Stieltjes constants, it's worth nothing that - at least symbolically - we are now able to produce a Taylor series for the Gamma function, centred around \(\displaystyle x_0=1\,\). By Taylor's Theorem:


\(\displaystyle f(x)=f(x_0)+\sum_{k=1}^{\infty}\frac{(x-x_0)^k}{k!}f^{(k)}(x_0)=\)


\(\displaystyle f(x_0)+(x-x_0)f^{(1)}(x_0)+\frac{(x-x_0)^2}{2!}f^{(2)}(x_0)+\frac{(x-x_0)^3}{3!}f^{(3)}(x_0)+\, \cdots \, \)


Accordingly, the Taylor series for the Gamma function is:


\(\displaystyle \Gamma(x)=\Gamma(1)+\sum_{k=1}^{\infty}\frac{(x-1)^k\Gamma^{(1)}(1)}{k!}=1+\sum_{k=1}^{\infty} \frac{(x-1)^k\gamma_k}{k!}\)


Alternatively, by applying the reflection transform \(\displaystyle x \to (1-x)\,\) to both sides, we get


\(\displaystyle \Gamma(1-x)=1+\sum_{k=1}^{\infty}(-1)^k\frac{\gamma_k\,x^k}{k!}\)



Results:


\(\displaystyle \gamma_1=-\gamma\)


\(\displaystyle \gamma_2=\gamma^2+\frac{\pi^2}{6}\)


\(\displaystyle \gamma_3=-\gamma^3-\frac{\pi^2\gamma}{2}-2\zeta(3)\)


\(\displaystyle \gamma_4=\gamma^4+\pi^2\gamma^2+8\gamma\zeta(3)+ \frac{3\pi^4}{20}\)


\(\displaystyle \gamma_5=-\gamma^5-\frac{5\pi^2\gamma^3}{3}-\frac{3\pi^4\gamma}{4}-10\left(\frac{\pi^2}{3}+2\gamma^2\right)\zeta(3)-24\zeta(5)\)


\(\displaystyle \Gamma(x)=1+\sum_{k=1}^{\infty} \frac{(x-1)^k\gamma_k}{k!}\)


\(\displaystyle \Gamma(1-x)=1+\sum_{k=1}^{\infty}(-1)^k\frac{\gamma_k\,x^k}{k!}\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Returning to the Euler-integral definition of the Gamma function, it's worth mentioning that the uniform convergence of the integral makes differentiation of the integrand meaningful. Hence

\(\displaystyle \Gamma^{{m}}(x)=\frac{d^m}{dx^m}\int_0^{\infty} e^{-t}t^{x-1}\,dt\equiv\)


\(\displaystyle \int_0^{\infty} e^{-t}\left[\frac{d^m}{dx^m}t^{x-1}\right]\,dt=\int_0^{\infty} e^{-t}t^{x-1}(\log t)^m\,dt
\)


Setting \(\displaystyle x=1\,\) gives us the Euler-Stieltjes constants, since


\(\displaystyle \Gamma^{{m}}(1)=\gamma_m\)


So the results in the previous post give us the following (classic) integral evaluations:



\(\displaystyle \int_0^{\infty}e^{-t}\log t\,dt=-\gamma\)



\(\displaystyle \int_0^{\infty}e^{-t}(\log t)^2\,dt=\gamma^2+\frac{\pi^2}{6}\)



\(\displaystyle \int_0^{\infty}e^{-t}(\log t)^3\,dt=-\gamma^3-\frac{\pi^2\gamma}{2}-2\zeta(3)\)



\(\displaystyle \int_0^{\infty}e^{-t}(\log t)^4\,dt=\gamma^4+\pi^2\gamma^2+8\gamma\zeta(3)+ \frac{3\pi^4}{20}\)



\(\displaystyle \int_0^{\infty}e^{-t}(\log t)^5\,dt=-\gamma^5-\frac{5\pi^2\gamma^3}{3}-\frac{3\pi^4\gamma}{4}-10\left(\frac{\pi^2}{3}+2\gamma^2\right)\zeta(3)-24\zeta(5)\)



----------------------


As I said before, the Euler-Stieltjes constants will crop up in quite a few future post, but for now, let's apply the substitution \(\displaystyle y=e^{-t}\,\) on one of the integrals above, say, that final example.


\(\displaystyle y=e^{-t};\quad t=-\log y;\quad dt=-\frac{dy}{y}\quad \Rightarrow\)


\(\displaystyle \int_0^{\infty}e^{-t}(\log t)^5\,dt=\int_0^1\log^5(-\log y)\,dy=\int_0^1\log^5\left(\log \frac{1}{y}\right)\,dy\)


These nested-logarithm integrals are interesting enough in their own right, but form a small part of a much larger class of integrals having the general form


\(\displaystyle \int_0^1\frac{P(x)}{Q(x)}(\log x)^n\log^m(-\log x)\,dx\)


where \(\displaystyle P(x)\,\) and \(\displaystyle Q(x)\,\) are rational functions.

In order to make this not-tutorial self-contained, and therefore - hopefully - more accessible, we'll need to prove a few more functional relations for the Gamma and Polygamma functions first. The aim will be to break it up into stages, and thereby alternate between proving a few new identities, applying them to the evaluation of specific classes of logarithmic integrals, rinse, and repeat. To that end, the next goal is a closed form for integrals of the type



\(\displaystyle \mathcal{I}(m, n; q)=\int_0^1y^{q-1}(\log y)^n\log^m(-\log x)\,dx\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Firstly, it would be instructive to prove that the Gamma function does indeed generalize the factorial function (to real/complex numbers):

\(\displaystyle \Gamma(x+1)=x\Gamma(x)\)


Starting with the Euler-integral form of the Gamma function, and noting that \(\displaystyle e^{-t}\,\) is - essentially - it's own derivatve

\(\displaystyle (d/dt)e^{-t}=-e^{-t}\)

we can use this fact to repeatedly integrate by parts


\(\displaystyle \Gamma(x+1)=\int_0^{\infty}e^{-t}t^{(x+1)-1}\,dt=\int_0^{\infty}e^{-t}t^{x}\,dt=\)

\(\displaystyle -e^{-t}t^{x}\,\Bigg|_0^{\infty}+\int_0^{\infty}e^{-t}\Big\{xt^{x-1}\Big\}\,dt=\)

\(\displaystyle x\, \int_0^{\infty}e^{-t}t^{x-1}\,dt = x\Gamma(x) = \Gamma(x+1)\)

since

\(\displaystyle \text{limit}_{\, t\to\infty}\,(e^{-t})=0\)

and

\(\displaystyle \text{limit}_{\, t\to 0^+}\,(t^{x})=0\, ;\quad \quad (x \ne 0)\)


Next, let's suppose this is true for \(\displaystyle x+n, \quad n\in\mathbb{Z}^+\,\), that is to say,

\(\displaystyle (x+n)\Gamma(x+n)=\Gamma(x+n+1)\)

Then

\(\displaystyle \Gamma(x+n+1)=\int_0^{\infty}e^{-t}t^{(x+n+1)-1}\,dt=\int_0^{\infty}e^{-t}t^{x+n}\,dt\)

This time, however, let's integrate by parts the other way around, by expressing the above as

\(\displaystyle \Gamma(x+n+1)=\int_0^{\infty}e^{-t}t^{x+n}\,dt=\int_0^{\infty}e^{-t}\Bigg\{\frac{1}{x+n+1}\frac{d}{dt}t^{x+n+1}\Bigg\}\,dt=\)

\(\displaystyle \frac{e^{-t}t^{x+n+1}}{(x+n+1)}\,\Bigg|_0^{\infty}-\frac{1}{(x+n+1)}\int_0^{\infty}(-e^{-t})t^{(x+n+1)}\,dt=\)

\(\displaystyle \frac{1}{(x+n+1)}\int_0^{\infty}e^{-t}t^{(x+n+2)-1}\,dt=\frac{\Gamma(x+n+2)}{(x+n+1)}\)

After rearranging, this becomes:

\(\displaystyle \Gamma(x+n+2)=(x+n+1)\Gamma(x+n+1)\)


So, having assumed the result to be true for some arbitrary non-zero positive integer \(\displaystyle n\,\), we've shown that if it's true for some particular \(\displaystyle n\,\), then it is also true for \(\displaystyle (n+1)\,\). However, right back at the start we showed the result was true for \(\displaystyle n=1\,\):

\(\displaystyle \Gamma(x+1)=x\Gamma(x)\)

hence the general result has been proven by induction, and for \(\displaystyle n\in{\mathbb{Z}^+}\,\), we conclude that:

\(\displaystyle \Gamma(x+n+1)=(x+n)\Gamma(x+n)\, . \, \Box\)



---------------------------------


Next, we introduce the so-called rising factorial (rising Pochhammer symbol) for the finite product:

\(\displaystyle (x)_n=x(x+1)(x+2)\, \cdots\, (x+n-1)=\prod_{j=0}^{n-1}(x+j)
\)
This product has exactly \(\displaystyle n\,\)terms. In terms of the Gamma function,

\(\displaystyle \Gamma(x+1)=x\Gamma(x)\)

\(\displaystyle \Gamma(x+2)=(x+1)\Gamma(x+1)=x(x+1)\Gamma(x)\)

\(\displaystyle \Gamma(x+3)=(x+2)\Gamma(x+2)=x(x+1)(x+2)\Gamma(x)\)

and so on. Using the rising Pochhammer symbol, the general relation can be written as:

\(\displaystyle \Gamma(x+n)=x(x+1)(x+2)\, \cdots\, (x+n-1)\Gamma(x)=(x)_n\,\dot\,\Gamma(x)\)

\(\displaystyle \therefore\, (x)_n=\prod_{j=0}^{n-1}(x+j)=\frac{\Gamma(x+n)}{\Gamma(x)}\)


---------------------------------


Writing the previous relation in the form

\(\displaystyle \Gamma(x)\, \prod_{j=0}^{n-1}(x+j)=\Gamma(x+n)\)

And then taking the logarithm of both sides, the (finite) Pochhammer product then becomes a (finite) sum of logarithmic terms:

\(\displaystyle \log\Gamma(x)+\sum_{j=0}^{n-1}\log(x+j)=\log\Gamma(x+n)\)

If we differentiate this identity once, we get a relation for the digamma function, do it twice, and we get one for the trigamma function, etc. So, for \(\displaystyle m\ge 0\,\)

\(\displaystyle \frac{d^{m+1}}{dx^{m+1}}\log\Gamma(x+n)=\psi_m(x+n)=\)

\(\displaystyle \frac{d^{m+1}}{dx^{m+1}}\Bigg\{\log\Gamma(x)+\sum_{j=0}^{n-1}\log(x+j)\Bigg\}=\)

\(\displaystyle \psi_m(x)+\sum_{k=0}^{n-1} \Bigg\{ \frac{d^{m+1}}{dx^{m+1}}\frac{1}{(x+j)}\Bigg\}=\)

\(\displaystyle \psi_m(x)+(-1)^mm!\,\sum_{j=0}^{n-1}\frac{1}{(x+j)^{m+1}}\)


That last finite sum/function is called the generalized harmonic function. Various different notations are used, but I'll write it as

\(\displaystyle \mathcal{H}(n, p;\,x)=\sum_{j=0}^{n-1}\frac{1}{(x+j)^p}=\frac{1}{x^p}+\frac{1}{(x+1)^p}+\, \cdots\, \frac{1}{(x+n-1)^p}\)


Changing the summation index, by the relation \(\displaystyle k=j+1\,\) , gives the equivalent form:

\(\displaystyle \mathcal{H}(n, p;\,x)=\sum_{k=1}^{n}\frac{1}{(x+k-1)^p}=\frac{1}{x^p}+\frac{1}{(x+1)^p}+\, \cdots\, \frac{1}{(x+n-1)^p}\)


Setting \(\displaystyle x=1\,\) in the generalized harmonic function of order \(\displaystyle p\,\) gives the generalized Harmonic Number of order \(\displaystyle p\,\)

\(\displaystyle \mathcal{H}(n, p;\, 1)=\mathcal{H}_{n, p}=1+\frac{1}{2^p}+\frac{1}{3^p}+\, \cdots\, +\frac{1}{n^p}\)


Returning to the polygamma relation above, we can re-write it in the form:

\(\displaystyle \psi_m(x+n)=\psi_m(x)+(-1)^mm!\,\mathcal{H}(n, m+1;\, x)\)


Similarly, by setting \(\displaystyle x=1\,\), we obtain


\(\displaystyle \psi_m(n+1)=\psi_m(1)+(-1)^mm!\, \mathcal{H}_{n, m+1}\)


Or, equivalently,


\(\displaystyle \psi_m(n+1)=\psi_m(1)+(-1)^mm!\sum_{k=1}^{n}\frac{1}{k^{m+1}}\)



---------------------------------



This last relation, as well as a little familiarity with the rising factorial (Pochhammer symbol), sets the stage for the next step; a closed form evaluation for integrals of the type:



\(\displaystyle \mathcal{I}(m, n; q)=\int_0^1y^{q-1}(\log y)^n\log^m(-\log x)\,dx\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
To tackle integrals of the form


\(\displaystyle \mathcal{I}(m, n; q)=\int_0^1y^{q-1}(\log y)^n\log^m(-\log x)\,dx\)


We start off by considering the parametric integral


\(\displaystyle \mathcal{I}(q)=\int_0^{\infty}e^{-qt}t^{x-1}\,dt\)


Let \(\displaystyle q > 0 \in \mathbb{R}^+\, \), and then apply the substitution \(\displaystyle y=qt\)


\(\displaystyle \mathcal{I}(q)=\int_0^{\infty}e^{-qt}t^{x-1}\,dt=\frac{1}{q}\int_0^{\infty}e^{-y}\left(\frac{y}{q}\right)^{x-1}\,dy=\)


\(\displaystyle \frac{1}{q^x}\int_0^{\infty}e^{-y}y^{x-1}\,dy\)


That last integral is the Gamma function, so


\(\displaystyle \mathcal{I}(q)=q^{-x}\Gamma(x)\)


Next, we perform the exponential substitution \(\displaystyle y=e^{-t}\,\) on \(\displaystyle \mathcal{I}(q)\,\):


\(\displaystyle y=e^{-t}; \quad t=-\log y; \quad dt=-\frac{dy}{y}\)


\(\displaystyle \Rightarrow\)


\(\displaystyle \mathcal{I}(q)=\int_0^{\infty}e^{-qt}t^{x-1}\,dt=\int_0^{\infty}\left(e^{-t}\right)^qt^{x-1}\,dt=\)


\(\displaystyle -\int_1^0(y)^{q}(-\log y)^{x-1}\,\frac{dy}{y}=\int_0^1y^{q-1}(-\log y)^{x-1}\,dy=q^{-x}\Gamma(x)\)



----------------------------



That last result represents the preferred form of the integral:


\(\displaystyle \mathcal{I}(q)=\int_0^1y^{q-1}(-\log y)^{x-1}\,dy=q^{-x}\Gamma(x)\)


If we differentiate both side of this equation \(\displaystyle n\)-times with respect to the parameter \(\displaystyle q\,\) we get:


\(\displaystyle \frac{d^n}{dq^n}\,\int_0^1y^{q-1}(-\log y)^{x-1}\,dy=
\int_0^1\left[\frac{d^n}{dq^n}y^{q-1}\right](-\log y)^{x-1}\,dy=\)


\(\displaystyle \int_0^1y^{q-1}(\log y)^n(-\log y)^{x-1}\,dy=\frac{d^n}{dq^n}\,q^{-x}\,\Gamma(x)=\Gamma(x)\left[\frac{d^n}{dq^n}\,q^{-x}\right]=\)


\(\displaystyle (-1)^n\,\frac{x(x+1)(x+2)\, \cdots\, (x+n-1)}{q^{x+n}}\,\Gamma(x)=(-1)^n\,\frac{(x)_n}{q^{x+n}}\, \Gamma(x)\)


Note the Pochhammer symbol in that last result... And as we proved earlier, the Gamma function satisfies the relation:


\(\displaystyle \Gamma(x)\, \dot\, (x)_n=\Gamma(x+n)\)


So that last result can be simplified to:


\(\displaystyle \int_0^1y^{q-1}(\log y)^n(-\log y)^{x-1}\,dy=(-1)^n\, \frac{\Gamma(x+n)}{q^{x+n}}\)



----------------------------


Next, we take the new integral


\(\displaystyle \int_0^1y^{q-1}(\log y)^n(-\log y)^{x-1}\,dy=(-1)^n\, \frac{\Gamma(x+n)}{q^{x+n}}\)


and differentiate it \(\displaystyle m\)-times with respect to the parameter \(\displaystyle x\). This time we have a product of two functions of \(\displaystyle x\) to repeatedly differentiate, so we use Leibniz's rule to get:




\(\displaystyle \frac{d^m}{dx^m}\,\int_0^1y^{q-1}(\log y)^n(-\log y)^{x-1}\,dy=\)


\(\displaystyle \int_0^1y^{q-1}(\log y)^n\left[\frac{d^m}{dx^m}\,(-\log y)^{x-1}\right]\,dy=\)


\(\displaystyle \int_0^1y^{q-1}(\log y)^n(-\log y)^{x-1}\log^m(-\log y)\,dy=\)


\(\displaystyle
\frac{d^m}{dx^m}\,(-1)^n\, \frac{\Gamma(x+n)}{q^{x+n}}=\)


\(\displaystyle
(-1)^n\sum_{j=0}^m\binom{m}{j}\left[\frac{d^{m-j}}{dx^{m-j}}\,\Gamma(x+n)\right]\, \frac{d^j}{dx^j}\, \Bigg\{\frac{1}{q^{x+n}}\Bigg\}=
\)


\(\displaystyle (-1)^n\sum_{j=0}^m\binom{m}{j}\Gamma^{(m-j)}(x+n)\left[\frac{(-1)^j(x+n)(x+n+1)\, \cdots\, (x+n+j-1)}{q^{x+n+j}}\right]=\)


\(\displaystyle (-1)^n\sum_{j=0}^m\binom{m}{j}\Gamma^{(m-j)}(x+n)\,\frac{(-1)^j\, (x+n)_j}{q^{x+n+j}}\)


Now, using the earlier relation for the Pochhammer symbol:


\(\displaystyle (x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}\, \Rightarrow \, (x+n)_j=\frac{\Gamma(x+n+j)}{\Gamma(x+n)}\)


we can re-write that last result in the more convenient form


\(\displaystyle (-1)^n\sum_{j=0}^m\binom{m}{j}\Gamma^{(m-j)}(x+n)\,\frac{(-1)^j\, (x+n)_j}{q^{x+n+j}}=\)


\(\displaystyle \frac{(-1)^n}{q^{x+n}\,\Gamma(x+n)}\sum_{j=0}^m\binom{m}{j}\frac{\Gamma^{(m-j)}(x+n)\,\Gamma(x+n+j)}{(-q)^j}\)


----------------------------


We now have a (slightly inelegant) closed form for the integral


\(\displaystyle \int_0^1y^{q-1}(\log y)^n(-\log y)^{x-1}\log^m(-\log y)\,dy=\)


\(\displaystyle \frac{(-1)^n}{q^{x+n}\,\Gamma(x+n)}\sum_{j=0}^m\binom{m}{j}\frac{\Gamma^{(m-j)}(x+n)\,\Gamma(x+n+j)}{(-q)^j}\)


So we're very nearly there...! (Bandit)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
To find a more concise - and indeed workable - closed form for integrals of the form


\(\displaystyle \int_0^1y^{q-1}(\log y)^n\log^m(-\log y)\,dy\)


Let's have a quick re-cap of relevant results:


\(\displaystyle \int_0^1y^{q-1}(\log y)^n(-\log y)^{x-1}\log^m(-\log y)\,dy=\)


\(\displaystyle \frac{(-1)^n}{q^{x+n}\,\Gamma(x+n)}\sum_{j=0}^m\binom{m}{j}\frac{\Gamma^{(m-j)}(x+n)\,\Gamma(x+n+j)}{(-q)^j}\)


\(\displaystyle \psi_{m \ge 1}(z)=(-1)^mm!\,\sum_{n=1}^{\infty}\frac{1}{(z+n-1)^{m+1}}\)


\(\displaystyle \Gamma^{(n+1)}(x)=\frac{d^n}{dx^n}\, \Gamma(x)\psi_0(x)=\sum_{j=0}^n\binom{n}{j}\Gamma^{(n-j)}(x)\,\psi_{(j)}(x)
\)


\(\displaystyle \psi_0(1)=-\gamma\)


\(\displaystyle \psi_1(1)=(-1)^21!\, \zeta(2)=\frac{\pi^2}{6}\)


\(\displaystyle \psi_2(1)=(-1)^32!\, \zeta(3)=-2\zeta(3)\)


\(\displaystyle \psi_3(1)=(-1)^43!\, \zeta(4)=\frac{\pi^4}{15}\)


\(\displaystyle \psi_4(1)=(-1)^54!\, \zeta(5)=-24\zeta(5)\)


\(\displaystyle \psi_5(1)=(-1)^65!\, \zeta(6)=\frac{8\pi^6}{63}\)


\(\displaystyle \psi_6(1)=(-1)^76!\, \zeta(7)=-720\zeta(7)\)


\(\displaystyle \Gamma(1) = 1\)


\(\displaystyle \Gamma(x+1)=x\, \Gamma(x)\)


\(\displaystyle \gamma_1=-\gamma\)


\(\displaystyle \gamma_2=\gamma^2+\frac{\pi^2}{6}\)


\(\displaystyle \gamma_3=-\gamma^3-\frac{\pi^2\gamma}{2}-2\zeta(3)\)


\(\displaystyle \gamma_4=\gamma^4+\pi^2\gamma^2+8\gamma\zeta(3)+ \frac{3\pi^4}{20}\)


\(\displaystyle \gamma_5=-\gamma^5-\frac{5\pi^2\gamma^3}{3}-\frac{3\pi^4\gamma}{4}-10\left(\frac{\pi^2}{3}+2\gamma^2\right)\zeta(3)-24\zeta(5)\)


\(\displaystyle \psi_m(n+1)=\psi_m(1)+(-1)^mm!\, \mathcal{H}_{n, m+1}\)


Or, equivalently,


\(\displaystyle \psi_m(n+1)=\psi_m(1)+(-1)^mm!\sum_{k=1}^{n}\frac{1}{k^{m+1}}\)


Also, here's something that I essentially proved above, but never actually stated explicitly. For \(\displaystyle n\in \mathbb{Z}^+\):


\(\displaystyle \Gamma(n+1) = n!\)


Proof:


\(\displaystyle (x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}\)


as has been shown. Similarly,


\(\displaystyle (x)_n=x(x+1)(x+2)\, \cdots \, (x+n-1)\)


Let \(\displaystyle x\to 1\) in both of the above, and you get


\(\displaystyle \text{limit}_{\, x\to 1}\, (x)_n= \frac{\Gamma(1+n)}{\Gamma(1)}=\Gamma(1+n)\)


\(\displaystyle \text{limit}_{\, x\to 1}\, (x)_n=1(1+1)(1+2)\, \cdots \, (1+n-1)=\)


\(\displaystyle 1\, \dot\, 2\, \dot\, 3\, \dot\, \cdots\, (n-1) \, \dot\, (n)\equiv n!\)


Both expressions are equivalent, hence


\(\displaystyle \Gamma(n+1) = n!\, . \, \Box\)


---------------------------------


Applying that last result to the main result of the previous post, namely


\(\displaystyle \int_0^1y^{q-1}(\log y)^n(-\log y)^{x-1}\log^m(-\log y)\,dy=\)


\(\displaystyle \frac{(-1)^n}{q^{x+n}\,\Gamma(x+n)}\sum_{j=0}^m\binom{m}{j}\frac{\Gamma^{(m-j)}(x+n)\,\Gamma(x+n+j)}{(-q)^j}\)


by letting \(\displaystyle x\to 1\) then gives


\(\displaystyle \int_0^1y^{q-1}(\log y)^n\log^m(-\log y)\,dy=\)


\(\displaystyle \frac{(-1)^n}{q^{n+1}\,\Gamma(n+1)}\sum_{j=0}^m\binom{m}{j}\frac{\Gamma^{(m-j)}(n+1)\,\Gamma(n+j+1)}{(-q)^j}=\)


\(\displaystyle \frac{(-1)^n}{q^{n+1}\,n!}\sum_{j=0}^m\binom{m}{j}\frac{ (n+j)! \, \Gamma^{(m-j)}(n+1)}{(-q)^j}\)


For the next step, we want to remove that final term from the finite sum... In simplified notation, if \(\displaystyle T_j\) represents the \(\displaystyle j\)-th term in the series above, then we can re-write the sum as


\(\displaystyle \sum_{j=0}^mT_j=\sum_{j=0}^{m-1}T_j\, + T_{(j=m)}=T_m+\sum_{j=0}^{m-1}T_j\)


Why isolate it? Well, the next few steps will hopefully make it clear why it's convenient, but for now, suffice to say that all of the terms except that last one involve derivatives of the Gamma function. That final term, however, contains \(\displaystyle \Gamma^{(m-j)}(n+1)|_{j=m}=\Gamma(n+1)=n!\,\)

I hope that was both clear, and yet not patronizing. [I'm trying to write this thread so it's widely accessible, regardless of the reader's familiarity with the various functions herein]. :D


So, now that I've had my little paranoia moment, let's split off that last term from the sum to obtain the relation:


\(\displaystyle \int_0^1y^{q-1}(\log y)^n\log^m(-\log y)\,dy=\)


\(\displaystyle \frac{(-1)^n}{q^{n+1}\,n!}\sum_{j=0}^m\binom{m}{j}\frac{ (n+j)! \, \Gamma^{(m-j)}(n+1)}{(-q)^j}=\)


\(\displaystyle \frac{(-1)^n}{q^{n+1}\,n!}\binom{m}{m}\frac{(n+m)!\,\Gamma(n+1)}{(-q)^m}+

\frac{(-1)^n}{q^{n+1}\,n!}\, \sum_{j=0}^{m-1}\binom{m}{j}\frac{ (n+j)! \, \Gamma^{(m-j)}(n+1)}{(-q)^j}\)


And since


\(\displaystyle \binom{j}{k}=\frac{j!}{k!\, (j-k)!}\, \Rightarrow \binom {j}{j}\equiv 1\)


and


\(\displaystyle \Gamma(n+1)=n!\)


That separated (final) term can be simplified, giving the overall result:


\(\displaystyle \frac{(-1)^{n+m}(n+m)!}{q^{n+m+1}}+

\frac{(-1)^n}{q^{n+1}\,n!}\sum_{j=0}^{m-1}\binom{m}{j}\frac{ (n+j)! \, \Gamma^{(m-j)}(n+1)}{(-q)^j}\)



---------------------------------


Next step:



Noting that the higher order derivatives of the Gamma function can be expressed as a sum of lower order derivatives:


\(\displaystyle \Gamma^{(n+1)}(x)=\frac{d^n}{dx^n}\, \Gamma(x)\psi_0(x)=\sum_{j=0}^n\binom{n}{j}\Gamma^{(n-j)}(x)\,\psi_{(j)}(x)
\)


Which, in turn, can be expressed as Euler-Stieltjes constants plus generalized Harmonic numbers via


\(\displaystyle \psi_m(n+1)=\psi_m(1)+(-1)^mm!\sum_{k=1}^{n}\frac{1}{k^{m+1}}\)



Then, although somewhat laborious, it is nonetheless possible to evaluate integrals of the form:


\(\displaystyle \int_0^1y^{q-1}(\log y)^n\log^m(-\log y)\,dy=\)


\(\displaystyle \frac{(-1)^{n+m}(n+m)!}{q^{n+m+1}}+

\frac{(-1)^n}{q^{n+1}\,n!}\sum_{j=0}^{m-1}\binom{m}{j}\frac{ (n+j)! \, \Gamma^{(m-j)}(n+1)}{(-q)^j}\)




Apologies for not going all the way on this one, but without making the answer altogether too convoluted, the best closed form - imho - is the one given above...



Result:


\(\displaystyle \int_0^1y^{q-1}(\log y)^n\log^m(-\log y)\,dy=\)


\(\displaystyle \frac{(-1)^{n+m}(n+m)!}{q^{n+m+1}}+

\frac{(-1)^n}{q^{n+1}\,n!}\sum_{j=0}^{m-1}\binom{m}{j}\frac{ (n+j)! \, \Gamma^{(m-j)}(n+1)}{(-q)^j}\)




ps. This was a VERY long proof, so I will go back and double check it just to make sure I haven't made any unforced errors.... Might need some coffee first though. Or maybe something stronger...


(Beer)(Beer)(Beer)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
At a later date, I'll give a more systematic approach to integrals like that last one, involving nested logarithms and rational functions, but for now, here's another special class. Let's start with the integral


\(\displaystyle \mathcal{S}(p)=\int_0^{\infty}\frac{x^{p-1}}{\sinh x}\,dx\)



Expressing the hyperbolic sine in exponential form, and then expanding the denominator as an infinite series, gives


\(\displaystyle \mathcal{S}(p)=2\int_0^{\infty}\frac{x^{p-1}e^{-x}}{1-e^{-2x}}\,dx=2\,\sum_{k=0}^{\infty}\int_0^{\infty}x^{p-1}e^{-(2k+1)x}\,dx\)


Next, apply the substitution \(\displaystyle y=(2k+1)x\)


\(\displaystyle 2\,\sum_{k=0}^{\infty}\int_0^{\infty}x^{p-1}e^{-(2k+1)x}\,dx=2\,\sum_{k=0}^{\infty}\int_0^{\infty}\left(\frac{y}{2k+1}\right)^{p-1}e^{-y}\frac{dy}{(2k+1)}=\)


\(\displaystyle 2\,\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{p}}\int_0^{\infty}y^{p-1}e^{-y}\,dy=2\Gamma(p)\, \sum_{k=0}^{\infty}\frac{1}{(2k+1)^{p}}\)


That last series can be expressed in terms of the Zeta Function, since


\(\displaystyle \left(1+\frac{1}{3^p}+\frac{1}{5^p}+ \cdots \right)=\)


\(\displaystyle \left(1+\frac{1}{2^p}+\frac{1}{3^p}+ \cdots \right)-\frac{2}{2^p}\left(1+\frac{1}{2^p}+\frac{1}{3^p}+ \cdots \right)=\)


\(\displaystyle \left(1-\frac{1}{2^{p-1}}\right)\, \zeta(p)\)


So, for \(\displaystyle p>0 \in\mathbb{R}\), our hyperbolic integral has the following closed form:


\(\displaystyle \int_0^{\infty}\frac{x^{p-1}}{\sinh x}\,dx= \left(\frac{2^{p+1}-1}{2^p}\right)\, \zeta(p)\Gamma(p)\)


Differentiating both sides \(\displaystyle m\)-times with respect to the parameter \(\displaystyle p\) gives:


\(\displaystyle \int_0^{\infty}\frac{x^{p-1}(\log x)^m}{\sinh x}\,dx=\)


\(\displaystyle -\frac{1}{2^p}\sum_{j=0}^{m}\sum_{k=0}^{m-j}\binom{m}{j} \binom{m-j}{k}(-\log 2)^{m-j-k}\, \zeta^{(j)}(p)\,\Gamma^{(k)}(p)\)


\(\displaystyle +2\, \sum_{j=0}^{m}\zeta^{(j)}(p)\,\Gamma^{(m-j)}(p)
\)


On the other hand, expressing our (multiply-differentiated) hyperbolic integral in exponential terms - via the substitution \(\displaystyle y=e^{-x}\) gives:


\(\displaystyle \int_0^{\infty}\frac{x^{p-1}(\log x)^m}{\sinh x}\,dx=2\int_0^{\infty}\frac{x^{p-1}e^{-x}(\log x)}{(1-e^{-2x})}\,dx=\)


\(\displaystyle 2\, \int_0^1\frac{(-\log y)^{p-1}\log^m(-\log y)}{1-y^2}\,dy\)


Hence


\(\displaystyle \int_0^1\frac{(-\log y)^{p-1}\log^m(-\log y)}{1-y^2}\,dy=\)


\(\displaystyle -\frac{1}{2^{p+1}}\sum_{j=0}^{m}\sum_{k=0}^{m-j}\binom{m}{j} \binom{m-j}{k}(-\log 2)^{m-j-k}\, \zeta^{(j)}(p)\,\Gamma^{(k)}(p)\)


\(\displaystyle +\sum_{j=0}^{m}\zeta^{(j)}(p)\,\Gamma^{(m-j)}(p)
\)



----------------------------


The denominator of that last integral suggests that an inverse hyperbolic substitution might be a good idea, since \(\displaystyle d/dx\,\tanh^{-1}x=1/(1-x^2)\)...


Indeed, the substitution above gives:


\(\displaystyle \int_0^{\infty}(-\log\tanh x)^{p-1}\log^m(-\log\tanh x)\,dx=\)


\(\displaystyle \int_0^{\infty}(\log\coth x)^{p-1}\log^m(\log\coth x)\,dx\)



----------------------------



Result:


\(\displaystyle \int_0^1\frac{(-\log y)^{p-1}\log^m(-\log y)}{1-y^2}\,dy=\)


\(\displaystyle \int_0^{\infty}(\log\coth x)^{p-1}\log^m(\log\coth x)\,dx=\)


\(\displaystyle -\frac{1}{2^{p+1}}\sum_{j=0}^{m}\sum_{k=0}^{m-j}\binom{m}{j} \binom{m-j}{k}(-\log 2)^{m-j-k}\, \zeta^{(j)}(p)\,\Gamma^{(k)}(p)\)


\(\displaystyle +\sum_{j=0}^{m}\zeta^{(j)}(p)\,\Gamma^{(m-j)}(p)
\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
A little later on, I'll come back to a few more classes of integrals similar to all of the previous ones, but each time I see that thread title I can't help but feel I really should add some integrals that either contain or can be solved in terms of polylogarithms. So that's exactly what I'll do for a few posts or ten...

(Heidy)



-----------------------------------


Using a similar methodology to the first few entries on here, we will now evaluate integrals of the type:


\(\displaystyle \mathcal{I}=\int_0^zx^{q-1}(\log x)^m\text{Li}_n(x)\,dx\)


Where \(\displaystyle \text{Li}_m(x)\) is the \(\displaystyle m\)-th order polylogarithm. For \(\displaystyle -1 \le < x \le 1\), the \(\displaystyle m\)-th order polylogarithm has the series representation:


\(\displaystyle \text{Li}_m(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^m}\)


If you aren't particularly familiar with polylogarithms, and wonder where the name came from, it's worth comparing that series above to the following series representation for the logarithm:


\(\displaystyle \log(1-x)=-\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4} \cdots \, \right)=-\sum_{k=1}^{\infty}\frac{x^k}{k}\)


Notwithstanding the minus sign in the logarithmic series above, this is identical to that of the polylogarithm when \(\displaystyle m=1\).

Hence


\(\displaystyle \text{Li}_1(x)=\sum_{k=1}^{\infty}\frac{x^k}{k}=-\log(1-x)\)


The next question to ask and answer might be "what happens to this series if we divide it by \(\displaystyle x\), and then integrate term by term over the interval \(\displaystyle 0\) to \(\displaystyle z \, (\le 1)\)...???"


\(\displaystyle \int_0^z\frac{1}{x}\left(\sum_{k=1}^{\infty}\frac{x^k}{k}\right)\, dx=\sum_{k=1}^{\infty}\frac{1}{k}\int_0^zx^{k-1}\, dx\)


Sine that last integral contains powers of x no lower than zero, the result is simply:


\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}\int_0^zx^{k-1}\, dx=\sum_{k=1}^{\infty}\frac{1}{k}\left[\frac{x^k}{k}\right]_0^z=\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{z^k}{k^2}=\text{Li}_2(z)\)


... the Dilogarithm!!!

However, the series that we divided by \(\displaystyle x\) and then integrated was none other than \(\displaystyle f(x)=-\log(1-x)\), so we now have an integral representation for the Dilogarithm:


\(\displaystyle \text{Li}_2(z)=-\int_0^z\frac{\log(1-x)}{x}\,dx\equiv\int_0^z\frac{\text{Li}_1(x)}{x}\,dx\)


More generally, the same iterated (integral) approach can be used to demonstrate the recursive relation between polylogarithms that differ in order by one:


\(\displaystyle \text{Li}_{m+1}(z)=\int_0^z\frac{\text{Li}_m(x)}{x}\,dx\)


This is particularly useful when integrating Polylogarithmic Integrals, since - by the fundamental theorem of calculus - differentiating both sides of the above gives the relation:


\(\displaystyle \frac{d}{dz}\, \text{Li}_{m+1}(z)=\frac{\text{Li}_m(z)}{z}\)


And in particular


\(\displaystyle \frac{d}{dx}\, \text{Li}_2(x)=-\frac{\log(1-x)}{x}\)


We will use this last result quite frequently...



-----------------------------------


Now, back to the problem at hand. In light of the series representations presented above, which are valid inside and on the unit circle, we can now write:


\(\displaystyle \int_0^zx^{q-1}\text{Li}_n(x)\,dx=\int_0^1x^{q-1}\left(\sum_{k=1}^{\infty}\frac{x^k}{k^n}\right)\, dx=
\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^n}\int_0^zx^{k+q-1}\,dx=\sum_{k=1}^{\infty}\frac{1}{k^n}\left[\frac{z^{k+q}}{(k+q)}\right]=z^q\,\sum_{k=1}^{\infty}\frac{z^k}{k^n(k+q)}\)


That last series is - in a very loose sense - 'almost polylogarithmic'. To resolve the problem, and express it terms that are genuinely polylogarithmic, we introduce the factor:

\(\displaystyle \frac{(k+q)-k}{q}\equiv 1\)

into the series, to obtain:


\(\displaystyle \sum_{k=1}^{\infty}\frac{(k+q)-k}{q}\frac{z^k}{k^n(k+q)}=\frac{1}{q}\sum_{k=1}^{ \infty}\frac{z^k}{k^n}-\frac{1}{q}\sum_{k=1}^{\infty}\frac{z^k}{k^{n-1}(k+q)}\)


The first series on the right is a Polylogarithm of order \(\displaystyle n\). For the second series, we introduce the same factor again, to obtain:


\(\displaystyle \frac{1}{q}\sum_{k=1}^{ \infty}\frac{z^k}{k^n}-\frac{1}{q^2}\Bigg\{ \sum_{k=1}^{\infty}\frac{z^k}{k^{n-1}}-\sum_{k=1}^{\infty}\frac{z^k}{k^{n-2}(k+q)}\Bigg\}\)


Repeating the process \(\displaystyle n\)-times gives the general result:


\(\displaystyle \frac{1}{q}\sum_{k=1}^{ \infty}\frac{z^k}{k^n}-
\frac{1}{q^2}\sum_{k=1}^{ \infty}\frac{z^k}{k^{n-1}}+
\frac{1}{q^3}\sum_{k=1}^{ \infty}\frac{z^k}{k^{n-2}}+ \cdots \, \)


\(\displaystyle \cdots\, +(-1)^{n-1}\frac{1}{q^n}\sum_{k=1}^{ \infty}\frac{z^k}{k}+(-1)^n\frac{1}{q^n}\, \sum_{k=1}^{ \infty}\frac{z^k}{(k+q)}=\)


\(\displaystyle \sum_{j=0}^{n-1}\frac{(-1)^j}{q^{j+1}}\left(\sum_{k=1}^{ \infty}\frac{z^k}{k^{n-j}}\right)+(-1)^n\frac{1}{q^n}\, \sum_{k=1}^{ \infty}\frac{z^k}{(k+q)}\)


Which, by using the series definition of the Polylogarithm, can be simplified to:


\(\displaystyle \sum_{j=0}^{n-1}\frac{(-1)^j}{q^{j+1}}\text{Li}_{n-j}(z)+(-1)^n\frac{1}{q^n}\, \sum_{k=1}^{ \infty}\frac{z^k}{(k+q)}\)


Combining this with:


\(\displaystyle \int_0^zx^{q-1}\text{Li}_n(x)\,dx=z^q\,\sum_{k=1}^{\infty}
\frac{z^k}{k^n(k+q)}\)


gets us halfway toward a full solution - which will be in the next post...




Result:


\(\displaystyle \int_0^zx^{q-1}\text{Li}_n(x)\,dx=z^q\, \Bigg\{ \sum_{j=0}^{n-1}\frac{(-1)^j}{q^{j+1}}\text{Li}_{n-j}(z)+(-1)^n\frac{1}{q^n}\, \sum_{k=1}^{ \infty}\frac{z^k}{(k+q)} \Bigg\}\)





To be continued... (Heidy)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Before we proceed, a couple of remarks about that last (partial) result:


\(\displaystyle \int_0^zx^{q-1}\text{Li}_n(x)\,dx=z^q\, \Bigg\{ \sum_{j=0}^{n-1}\frac{(-1)^j}{q^{j+1}}\text{Li}_{n-j}(z)+(-1)^n\frac{1}{q^n}\, \sum_{k=1}^{ \infty}\frac{z^k}{(k+q)} \Bigg\}\)


That last series on the right isn't particularly easy to sum, although a number of important special cases can be evaluated. Furthermore, it's also a named function - called the Lerch Transcendent - and is closely related to the Polygamma, Polylogarithm, Zeta, and Hurwitz Zeta functions, to name but a few. The standard notation for the Lerch Transcendent is as follows (the 3 parameters are usually named \(\displaystyle z\), \(\displaystyle s\), and \(\displaystyle a\) respectively, as shown below):


\(\displaystyle \Phi(z,s,a)=\sum_{k=0}^{\infty}\frac{z^k}{(k+a)^s}\)


Notice that the Lerch Transcendent is typically summed over \(\displaystyle 0\) to \(\displaystyle \infty\), whereas our Lerch-like sum is from \(\displaystyle 1\) to \(\displaystyle \infty\). To address this, we change the summation variable in our series via \(\displaystyle k=j+1\):


\(\displaystyle \sum_{k=1}^{ \infty}\frac{z^k}{(k+q)}=\sum_{j=0}^{ \infty}\frac{z^{j+1}}{(j+1+q)}=z\, \Phi(z,1,q+1)\)


So, in this new notation, the main result from the last post can be written as:


\(\displaystyle \int_0^zx^{q-1}\text{Li}_n(x)\,dx=z^q\, \sum_{j=0}^{n-1}\frac{(-1)^j}{q^{j+1}}\text{Li}_{n-j}(z)+(-1)^nz^{q+1}\frac{ \Phi(z,1,1+q)}{q^n}\)


As with numerous previous integrals, to get the logarithmic form of that last one, we will need to differentiate both sides w.r.t. the parameter \(\displaystyle q\). So we'll need the m-th derivative of the Lerch Transcendent w.r.t. its third argument.

\(\displaystyle \frac{\partial^m}{\partial q^m}\, \Phi(z,1,1+q)=\Phi^{(m)}(z,1,1+q)=\frac{\partial^m}{\partial q^m}\,
\sum_{k=0}^{\infty}\frac{z^k}{(k+q+1)}=\)


\(\displaystyle (-1)^mm!\, \sum_{k=0}^{\infty}\frac{z^k}{(k+q+1)^{m+1}}=(-1)^mm!\,\Phi(z,m+1,1+q)\)


More about the Lerch Transcendent a little later...


------------------------------------



For the next step, we need the \(\displaystyle m\)-th derivative of both sides of:


\(\displaystyle \int_0^zx^{q-1}\text{Li}_n(x)\,dx=z^q\, \sum_{j=0}^{n-1}\frac{(-1)^j}{q^{j+1}}\text{Li}_{n-j}(z)+(-1)^nz^{q+1}\frac{ \Phi(z,1,1+q)}{q^n}\)


The LHS is simply:


\(\displaystyle \frac{\partial^m}{\partial q^m}\, \int_0^zx^{q-1}\text{Li}_n(x)\,dx=\int_0^zx^{q-1}(\log x)^m\text{Li}_n(x)\,dx\)


The RHS is slightly trickier, but nothing that a double application of Leibniz's rule for repeated differentiation of a product can't handle. The first half is given by:


\(\displaystyle \frac{\partial^m}{\partial q^m}\, z^q\, \sum_{j=0}^{n-1}\frac{(-1)^j}{q^{j+1}}\text{Li}_{n-j}(z)= \sum_{j=0}^{n-1}(-1)^j\text{Li}_{n-j}(z)\left[ \frac{\partial^m}{\partial q^m} \frac{z^q}{q^{j+1}}\right]=\)


\(\displaystyle \sum_{j=0}^{n-1}(-1)^j\text{Li}_{n-j}(z)\left[ \sum_{k=0}^m\binom{m}{k}z^q(\log z)^{m-k}\frac{(-1)^k\, (j+1)_k}{q^{k+j+1}}\right]\)


Recalling a few earlier results helps us simplify this double sum somewhat (although to be fair, not by much). We start by expressing the Pochhammer symbol in terms of the Gamma function:


\(\displaystyle (x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}\quad \Rightarrow\)


\(\displaystyle (j+1)_k=\frac{\Gamma(j+k+1)}{\Gamma(j+1)}\)


Next, we utilize the factorial property of the Gamma Function:


\(\displaystyle \Gamma(n+1)=n!\quad \text{if}\quad n\in \mathbb{Z}^+\quad \Rightarrow\)


\(\displaystyle (j+1)_k=\frac{\Gamma(j+k+1)}{\Gamma(j+1)}=\frac{(j+k)!}{j!}\)


So the first half of our differentiated form is:


\(\displaystyle z^q\, \sum_{j=0}^{n-1}\text{Li}_{n-j}(z)\, \sum_{k=0}^m(-1)^{j+k}\binom{m}{k}\frac{(\log z)^{m-k}(j+k)!}{j!\, q^{j+k+1}}\)


The second half involves evaluating:


\(\displaystyle \frac{\partial^m}{\partial q^m}\,(-1)^nz^{q+1}\frac{ \Phi(z,1,q+1)}{q^n}\)


And since we have a product of three functions of \(\displaystyle q\), we need to apply Leibniz's Rule twice (once in nested form):


\(\displaystyle \frac{\partial^m}{\partial q^m}\,(-1)^nz^{q+1}\frac{ \Phi(z,1,q+1)}{q^n}=\)


\(\displaystyle (-1)^n\, \sum_{j=0}^m\binom{m}{j}\left[\frac{\partial^j}{\partial q^j}\frac{1}{q^n}\right]\, \dot \, \left[\frac{\partial^{m-j}}{\partial q^{m-j}}z^{q+1}\, \Phi(z,1,q+1)\right]=\)


\(\displaystyle (-1)^n\, \sum_{j=0}^m\binom{m}{j}\left[\frac{(-1)^j\, (n)_j}{q^{n+j}}\right]\, \dot \, \left[\frac{\partial^{m-j}}{\partial q^{m-j}}z^{q+1}\, \Phi(z,1,q+1)\right]=\)


\(\displaystyle \sum_{j=0}^m\binom{m}{j}\frac{(-1)^{n+j}\, (n+j-1)!}{q^{n+j}(n-1)!}\, \dot \, \left[\frac{\partial^{m-j}}{\partial q^{m-j}}z^{q+1}\, \Phi(z,1,q+1)\right]=\)


\(\displaystyle \sum_{j=0}^m\binom{m}{j}\frac{(-1)^{n+j}\, (n+j-1)!}{q^{n+j}(n-1)!}\, \sum_{k=0}^{m-j}\binom{m-j}{k}z^{q+1}(\log z)^{m-j-k}\frac{\partial^k}{\partial q^k} \Phi(z,1,q+1)\)




To be continued... (Wait)