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A generalized Clausen Function, and associated loggamma integrals

DreamWeaver

Well-known member
Sep 16, 2013
337
I've recently been working on a number of integrals related to the loggamma function, so I thought I'd share my results here. I'll have to post as and when I have time, and there will be a fair bit of preliminary work before we get to the final results, but - loosely speaking - the main aim here is to find a closed form for parametric integrals of the form:


\(\displaystyle \Gamma i_m(q)=\int_0^qx^m\log\Gamma(x)\,dx\)


Where \(\displaystyle 0 < q \le 1\). The special case where \(\displaystyle q=1\) appears in a number of contemporary papers, but - thus far - I've not encountered the general case for \(\displaystyle q\).

In the process of evaluating the above integral, we are led to consider a two generalized Clausen Functions, that in turn, can be expressed as finite sums of derivatives of the Hurwitz Zeta function \(\displaystyle \zeta(z,a)\), where:

\(\displaystyle \zeta(z,a)=\sum_{k=0}^{\infty}\frac{1}{(k+a)^z}\)


Definition:

For the purposes of this tutorial, we define the (possibly new? - although I doubt it) generalized Poly-Clausen Functions:


\(\displaystyle C_m(n;\theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\cos k\theta}{k^m}=\lim_{\,z\to m}\frac{d^n}{dz^n}\sum_{k=1}^{\infty} \frac{\cos k\theta}{k^z}\)


\(\displaystyle S_m(n;\theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\sin k\theta}{k^m}=\lim_{\,z\to m}\frac{d^n}{dz^n}\sum_{k=1}^{\infty} \frac{\sin k\theta}{k^z}\)


Following the usual convention for CL and SL-type Clausen Functions, where applicable, we will use the following notation, which is entirely consistent with regular Clausen Functions:


\(\displaystyle \text{Cl}_{2m}(n;\theta)=\sum_{k=1}^{\infty} \frac{(\log k)^n \sin k \theta}{k^{2m}}\)

\(\displaystyle \text{Cl}_{2m+1}(n;\theta)=\sum_{k=1}^{\infty} \frac {(\log k)^n \cos k \theta}{k^{2m+1}}\)

\(\displaystyle \text{Sl}_{2m}(n;\theta)=\sum_{k=1}^{\infty} \frac{(\log k)^n \cos k \theta}{k^{2m}}\)

\(\displaystyle \text{Sl}_{2m+1}(n;\theta)=\sum_{k=1}^{\infty} \frac {(\log k)^n \sin k \theta}{k^{2m+1}}\)


Evaluations of the Generalized Clausen Functions:


With regards to evaluating our parametric loggamma integral, \(\displaystyle \Gamma i_m(q)\), we will have a particular need for the Generalized Clausen functions at \(\displaystyle q=1/p\), and \(\displaystyle p \in \mathbb{Z}^{+}\). In other words, we will need to evaluate the functions:


\(\displaystyle C_m \left(n;\frac{2\pi}{p}\right)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\cos \left( \frac{2\pi k}{p}\right)}{k^m}\)

\(\displaystyle S_m \left(n;\frac{2\pi}{p}\right)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\sin \left( \frac{2\pi k}{p}\right)}{k^m}\)


Actually, we will only really need the case where \(\displaystyle n=1\), but there's no harm in evaluating the general case for \(\displaystyle n\).

As a starting point, we want to split those sums on the RHS into a finite composition of infinite sums. Indeed, we want each sum to split into exactly p-parts, with denominators corresponding to \(\displaystyle (kp+1)\), \(\displaystyle (kp+2)\), etc, up to \(\displaystyle (kp +p)\).

For example, splitting the cosine-type generalized Poly-Clausen Function in such a way, we obtain:

\(\displaystyle C_m \left(n;\frac{2\pi}{p}\right)=(-1)^n\sum_{j=1}^p \left( \sum_{k=0}^{\infty}\frac{\log^n(kp+j) \cos \left( \frac{2\pi (kp+j)}{p}\right)}{(kp+j)^m} \right)=\)

\(\displaystyle \frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \left( \sum_{k=0}^{\infty} \frac{\log^n \left[p(k+j/p) \right]}{(k+j/p)^m} \right)\)


Next, we expand the logarithmic term via the binomial theorem, like so:

\(\displaystyle \log^n[p(k+j/p)]=\sum_{l=0}^n \binom{n}{l}(\log p)^{n-l}\log^l(k+j/p)\)


Which gives us the partial evaluation:

\(\displaystyle \frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \sum_{l=0}^n \binom{n}{l}(\log p)^{n-l} \left( \sum_{k=0}^{\infty} \frac{\log^l (k+j/p) }{(k+j/p)^m} \right)\)


That final, rightmost (doubly 'nested') sum - within the large brackets - can be expressed in terms of derivatives of the Hurwitz Zeta function.

\(\displaystyle \frac{d^l}{dz^l}\zeta(z,a)=\frac{d^l}{dz^l}\sum_{k=0}^{\infty}\frac{1}{(k+a)^z}=(-1)^l\sum_{k=0}^{\infty} \frac{\log^l (k+a) }{(k+a)^z}\)


So, with the understanding that - henceforth - ALL derivatives of the Hurwitz Zeta function included here are derivatives with respect to the first parameter, we may write:


\(\displaystyle C_m \left(n;\frac{2\pi}{p}\right)=\)


\(\displaystyle \frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \sum_{l=0}^n \binom{n}{l}(\log p)^{n-l} (-1)^l\zeta^{(l)}\left(m, \frac{j}{p} \right)\)

Which - with a re-labelling of the summation index - is the result we required:

\(\displaystyle C_m \left(n;\frac{2\pi}{p}\right)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\cos \left( \frac{2\pi k}{p}\right)}{k^m}=\)

\(\displaystyle \frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \sum_{k=0}^n \binom{n}{k}(\log p)^{n-k} (-1)^k\zeta^{(k)}\left(m, \frac{j}{p} \right)\)


An equivalent calculation gives us the result for the other Poly-Clausen Function:


Results:


\(\displaystyle C_m \left(n;\frac{2\pi}{p}\right)=\)


\(\displaystyle (-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\cos \left( \frac{2\pi k}{p}\right)}{k^m}=\)


\(\displaystyle \frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \sum_{k=0}^n \binom{n}{k}(\log p)^{n-k} (-1)^k\zeta^{(k)}\left(m, \frac{j}{p} \right)\)




\(\displaystyle S_m \left(n;\frac{2\pi}{p}\right)=\)


\(\displaystyle (-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\sin \left( \frac{2\pi k}{p}\right)}{k^m}=\)


\(\displaystyle \frac{(-1)^n}{p^m} \sum_{j=1}^{p-1} \sin \left( \frac{2\pi j}{p} \right) \sum_{k=0}^n \binom{n}{k}(\log p)^{n-k} (-1)^k\zeta^{(k)}\left(m, \frac{j}{p} \right)\)




**Notice that, when \(\displaystyle j=p\) in that last result, we get a vanishing term corresponding to \(\displaystyle \sin 2\pi\), hence the reduction in the upper limit of the summation index, with \(\displaystyle p\) being replaced by \(\displaystyle p-1\).

Comments and/or questions should be posted here:

http://mathhelpboards.com/commentar...lausen-function-associated-loggamma-7506.html
 

DreamWeaver

Well-known member
Sep 16, 2013
337
For certain small, rational arguments of the second parameter, derivatives of the Hurwitz Zeta function can variously be expressed in terms of (slightly simpler) related functions. These relations will be explored more thoroughly later on, but for now, I simply give their definitions, and a few basic functional relations.



(1) The Riemann Zeta function:

\(\displaystyle \zeta(z)=\sum_{k=0}^{\infty}\frac{1}{(k+1)^z} \equiv \sum_{k=1}^{\infty}\frac{1}{k^z}\)


(2) The Dirichlet Beta function:

\(\displaystyle \beta(z)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^z}\)


(3) The Eta Function:

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


(4) The Legendre Chi function:

\(\displaystyle \chi(z)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^z}\)


(5) Functional relations:

\(\displaystyle \eta(z)=\left(1-\frac{2}{2^z}\right)\zeta(z)\)

\(\displaystyle \chi(z)=\left(1- \frac{1}{2^z}\right)\zeta(z)\)


\(\displaystyle \chi(z)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^z}= \frac{1}{2^z} \sum_{k=0}^{\infty}\frac{1}{\left( k+\frac{1}{2} \right)^z}= \frac{1}{2^z} \zeta\left(z, \frac{1}{2} \right) \)


\(\displaystyle \eta(z) + \zeta(z) = 2 \chi (z)\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
We almost have all of the requisite results needed to evaluate the integral function \(\displaystyle \Gamma i_m(q)\), but a tiny bit more groundwork is required first...

For reasons that will soon become apparent, we will need closed form evaluations of the following trigonometric integrals (\(\displaystyle m \in \mathbb{Z}^{+}\)):


\(\displaystyle \int x^m\sin x\,dx\)

\(\displaystyle \int x^m\cos x\,dx\)


These can be found in many texts on elementary indefinite integrals. Indeed, the proof requires nothing more than repeated integration by parts, and a modicum of induction, so the proofs are omitted. We simply state the results:


\(\displaystyle (1) \quad \int x^{2n}\sin x\,dx=\)

\(\displaystyle (2n)! \, \Bigg\{ \sum_{j=0}^n(-1)^{j+1}\frac{x^{2n-2j}}{(2n-2j)!} \cos x + \sum_{j=0}^{n-1}(-1)^j\frac{x^{2n-2j-1}}{(2n-2j-1)!} \sin x \Bigg\}\)



\(\displaystyle (2) \quad \int x^{2n+1}\sin x\,dx=\)

\(\displaystyle (2n+1)! \, \Bigg\{ \sum_{j=0}^n(-1)^{j+1}\frac{x^{2n-2j+1}}{(2n-2j+1)!} \cos x + \sum_{j=0}^{n}(-1)^j\frac{x^{2n-2j}}{(2n-2j)!} \sin x \Bigg\}\)



\(\displaystyle (3) \quad \int x^{2n}\cos x\,dx=\)

\(\displaystyle (2n)! \, \Bigg\{ \sum_{j=0}^n(-1)^{j}\frac{x^{2n-2j}}{(2n-2j)!} \sin x + \sum_{j=0}^{n-1}(-1)^j\frac{x^{2n-2j-1}}{(2n-2j-1)!} \cos x \Bigg\}\)



\(\displaystyle (4) \quad \int x^{2n+1}\cos x\,dx=\)

\(\displaystyle (2n+1)! \, \Bigg\{ \sum_{j=0}^n(-1)^{j}\frac{x^{2n-2j+1}}{(2n-2j+1)!} \sin x + \sum_{j=0}^{n}(-1)^j\frac{x^{2n-2j}}{(2n-2j)!} \cos x \Bigg\}\)


Use of the floor function allows us to condense (1) and (2), respectively (3) and (4) into the following:



\(\displaystyle (5) \int x^m\sin x \, dx=\)


\(\displaystyle m! \, \Bigg\{ \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j+1} \frac{x^{m-2j}}{(m-2j)!} \cos x +
\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{x^{m-2j-1}}{(m-2j-1)!} \sin x \Bigg\}\)



\(\displaystyle (6) \int x^m\cos x \, dx=\)


\(\displaystyle m! \, \Bigg\{ \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j} \frac{x^{m-2j}}{(m-2j)!} \sin x +
\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{x^{m-2j-1}}{(m-2j-1)!} \cos x \Bigg\}\)



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


In terms of the above trigonometric integrals, the particular examples we will need later on are as follows:


\(\displaystyle \int_0^{q}x^m\sin 2\pi kx \, dx\)


\(\displaystyle \int_0^{q}x^m\cos 2\pi kx \, dx\)



The substitution \(\displaystyle y=2\pi k x\) then yields:


\(\displaystyle \int_0^{q}x^m\sin 2\pi kx \, dx=\frac{1}{(2\pi k)^{m+1}}\int_0^{2\pi kq}x^m\sin x\, dx\)


\(\displaystyle \int_0^{q}x^m\cos 2\pi kx \, dx=\frac{1}{(2\pi k)^{m+1}}\int_0^{2\pi kq}x^m\cos x\, dx\)


A little caution is required here, when evaluating the cosine series in (5) and (6) at \(\displaystyle x=0\), the reason being that these cosine series vanish except for the term where the exponent of \(\displaystyle x\) is equal to \(\displaystyle 0\). In both cases, we treat the term \(\displaystyle x^n\cos(0)\) as a limiting value, and define:


\(\displaystyle x^n\cos x \, \Bigg|_{x=0}=\cos (0) \, \lim_{\epsilon \to 0^{+}}\epsilon ^n =1\)


Referring back to integrals (1) and (2), we note that the exponent of \(\displaystyle x\) is only \(\displaystyle 0\) in the final term of the cosine series when \(\displaystyle m=2n\) is even. Conversely, looking back at integrals (3) and (4), we note that the cosine series vanishes except for the final term in (4), when \(\displaystyle m=2n+1\) is odd. For this reason, we have need of two generalized integer functions, where one is zero for odd numbers, 1 for even numbers, and the second is the reverse. Such integer functions are easily found. Define


\(\displaystyle \mathcal{F}(m) = \frac{[1+(-1)^m]}{2}=
\begin{cases}
0, & \text{if }m\text{ is odd} \\
1, & \text{if }m\text{ is even}
\end{cases}\)


\(\displaystyle \mathcal{G}(m) = \frac{[1+(-1)^{m+1}]}{2}=
\begin{cases}
1, & \text{if }m\text{ is odd} \\
0, & \text{if }m\text{ is even}
\end{cases}\)


In light of these generalized integer functions, we can now find a neat closed form for our two key trigonometric integrals.

\(\displaystyle \int_0^{q}x^m\sin 2\pi kx \, dx=\frac{1}{(2\pi k)^{m+1}}\int_0^{2\pi kq}x^m\sin x\, dx=\)


\(\displaystyle \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j+1} \frac{x^{m-2j}}{(m-2j)!} \cos x \, \Bigg|_0^{2\pi k q} +\)

\(\displaystyle \frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{x^{m-2j-1}}{(m-2j-1)!} \sin x \, \Bigg|_0^{2\pi k q} =\)


\(\displaystyle \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j+1} \frac{(2\pi kq)^{m-2j}}{(m-2j)!} \cos 2\pi kq \, +\)


\(\displaystyle \frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{(2\pi kq)^{m-2j-1}}{(m-2j-1)!} \sin 2\pi kq \, -\mathcal{F}(m) (-1)^{ \lfloor \frac{m}{2} \rfloor +1} \frac{m!}{(2\pi k)^{m+1}}\)


Equivalently, our cosine integral evaluates to:


\(\displaystyle \int_0^{q}x^m\cos 2\pi kx \, dx=\frac{1}{(2\pi k)^{m+1}}\int_0^{2\pi kq}x^m\cos x\, dx=\)


\(\displaystyle \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j} \frac{x^{m-2j}}{(m-2j)!} \sin x \, \Bigg|_0^{2\pi k q} +\)

\(\displaystyle \frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{x^{m-2j-1}}{(m-2j-1)!} \cos x \, \Bigg|_0^{2\pi k q} =\)



\(\displaystyle \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j} \frac{(2\pi kq)^{m-2j}}{(m-2j)!} \sin 2\pi kq +\)

\(\displaystyle \frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{(2\pi kq)^{m-2j-1}}{(m-2j-1)!} \cos 2\pi kq -\mathcal{G}(m) (-1)^{ \lfloor \frac{m-1}{2} \rfloor } \frac{m!}{(2\pi k)^{m+1}} \)





More to follow shortly... (Heidy)
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
Our goal is now in sight...! Or, to put it another way, the end is nigh (if you're a pessimist, that is lol (Hug) )

Just a tiny bit more housekeeping, and then we derive a first evaluation for \(\displaystyle \Gamma i_m(q)\), in the special case where \(\displaystyle q=1/p\) and \(\displaystyle p \in \mathbb{Z}^{+}\). The general case for \(\displaystyle \Gamma i_m (q/p)\) will follow a little later.

We just need a few more results first...

Firstly - and I really should have added these nearer the start - we define the (regular) Clausen functions as follows:


\(\displaystyle \text{Cl}_{2m}(\theta)=\sum_{k=1}^{\infty}\frac{ \sin k\theta}{k^{2m}}\)

\(\displaystyle \text{Cl}_{2m+1}(\theta)=\sum_{k=1}^{\infty}\frac{ \cos k\theta}{k^{2m+1}}\)

\(\displaystyle \text{Sl}_{2m}(\theta)=\sum_{k=1}^{\infty}\frac{ \cos k\theta}{k^{2m}}\)

\(\displaystyle \text{Sl}_{2m+1}(\theta)=\sum_{k=1}^{\infty}\frac{ \sin k\theta}{k^{2m+1}}\)


Although these definitions will be of considerable use later on, of particular interest are the CL-type Clausen Functions \(\displaystyle \text{Cl}_2(\theta)\) and \(\displaystyle \text{Cl}_1(\theta)\). The former has the integral representation:

\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\log \Bigg| 2\sin \frac{x}{2} \Bigg| \, dx\)


Differentiation with respect to \(\displaystyle \theta\) gives:

\(\displaystyle \frac{d}{d \theta} \text{Cl}_2(\theta)=-\log \Bigg| 2\sin \frac{x}{2} \Bigg| \equiv \sum_{k=1}^{\infty}\frac{\cos k\theta}{k}\)


The final piece in the puzzle is Kummer's Fourier expansion for the loggamma function:


\(\displaystyle \log \left( \frac{ \Gamma(x)}{ \sqrt{2\pi} } \right)=\)

\(\displaystyle \frac{1}{2}(\gamma +\log 2\pi)(1-2x)-\frac{1}{2}\log(2\sin \pi x) + \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \sin 2\pi kx\)


Kummer's expansion is valid for \(\displaystyle 0 < x < 1\), and at \(\displaystyle x=1\) we take the limiting value. It is no understatement to say that everything we develop below is dependent upon Kummer's result, so it would be remiss of me to omit a proof. However, I would quite like to move things along a little, so I'll add a proof a bit later on...

Re-write Kummer's expansion as:

\(\displaystyle \log \Gamma(x) = \frac{1}{2}(\gamma+ 2 \log 2\pi)-(\gamma + \log 2\pi)x \)

\(\displaystyle -\frac{1}{2}\log(2\sin \pi x) + \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \sin 2\pi kx
\)

In light of the previous definitions of the (regular) Clausen functions, we note that the logsine term in Kummer's expansion may be written as:

\(\displaystyle -\frac{1}{2}\log(2\sin \pi x) = -\frac{1}{2}\log \left( 2\sin \frac{2\pi x}{2}\right)=\frac{1}{2}\text{Cl}_1(2\pi x)=\frac{1}{2} \sum_{k=1}^{\infty}\frac{\cos 2\pi k x}{k}\)


So Kummer's expansion becomes:


\(\displaystyle \log \Gamma(x) = \frac{1}{2}(\gamma+ 2 \log 2\pi)-(\gamma + \log 2\pi)x +\frac{1}{2} \sum_{k=1}^{\infty}\frac{\cos 2\pi k x}{k}
+ \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \sin 2\pi kx\)


Next, we multiply both sides by \(\displaystyle x^m\), and then integrate over the interval \(\displaystyle [0,q]\):


\(\displaystyle \Gamma i_m(q)=\int_0^qx^m\log\Gamma(x) \, dx=\)


\(\displaystyle \frac{1}{2}(\gamma+ 2 \log 2\pi)\int_0^q x^m \, dx -(\gamma + \log 2\pi) \int_0^q x^{m+1}

\, dx\)

\(\displaystyle +\frac{1}{2} \sum_{k=1}^{\infty}\frac{1}{k} \int_0^q x^m \cos 2\pi k x \, dx
+ \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \int_0^q x^m \sin 2\pi kx
\)


The first two integrals on the RHS are trivial, and give the partial result:

\(\displaystyle \frac{q^{m+1}}{2(m+1)}(\gamma+ 2 \log 2\pi) - \frac{q^{m+2}}{(m+2)}(\gamma + \log 2\pi)\)

There is an obvious temptation to split these terms, and write them in the form

\(\displaystyle A \gamma + B \log 2\pi\)

Where \(\displaystyle A\) and \(\displaystyle B\) are constants in terms of \(\displaystyle m\) and \(\displaystyle q\). Take my word for it: it's not worth it... The results are far from elegant, or efficient, so I'll just stick with the form above.

The third integral on the RHS is:

\(\displaystyle \sum_{k=1}^{\infty} \frac{1}{k} \left( \int_0^q x^m \cos 2\pi kx \right)\)

We already have a neat closed form for this integral, and may write

\(\displaystyle \sum_{k=1}^{\infty} \frac{1}{k} \left( \int_0^q x^m \cos 2\pi kx \right)=\)

\(\displaystyle \sum_{k=1}^{\infty} \frac{1}{k} \Bigg\{ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{(2\pi kq)^{m-2j}}{(m-2j)!} \sin 2\pi kq \Bigg\}+\)

\(\displaystyle \sum_{k=1}^{\infty} \frac{1}{k} \Bigg\{ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{(2\pi kq)^{m-2j-1}}{(m-2j-1)!} \cos 2\pi kq \Bigg\}+\)

\(\displaystyle (-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \frac{m!}{(2\pi)^{m+1}} \sum_{k=1}^{\infty} \frac{1}{k^{m+2}}=\)



\(\displaystyle m! \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{q^{m-2j}}{(2\pi)^{2j+1} (m-2j)!} \sum_{k=1}^{\infty} \frac{\sin 2\pi kq}{k^{2j+2} } +\)

\(\displaystyle m! \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{q^{m-2j-1}}{(2\pi)^{2j+2} (m-2j-1)!} \sum_{k=1}^{\infty} \frac{\cos 2\pi kq}{k^{2j+3} } +\)

\(\displaystyle (-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \frac{m!}{(2\pi)^{m+1}} \zeta (m+2)\)


In light of the series definitions for the regular Clausen functions, this becomes:


\(\displaystyle \sum_{k=1}^{\infty} \frac{1}{k} \left( \int_0^q x^m \cos 2\pi kx \right)=\)

\(\displaystyle m! \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{q^{m-2j}}{(2\pi)^{2j+1} (m-2j)!} \text{Cl}_{2j+2} (2\pi q) +\)

\(\displaystyle m! \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{q^{m-2j-1}}{(2\pi)^{2j+2} (m-2j-1)!} \text{Cl}_{2j+3} (2\pi q) +\)

\(\displaystyle (-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \frac{m!}{(2\pi)^{m+1}} \zeta (m+2)\)





Three down, one to go... Be right back! (Heidy)
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
Combining all of the previous results, we have the following partial (almost complete) evaluation:


Theorem 1.0:


\(\displaystyle \Gamma i_m (q)=\)

\(\displaystyle \frac{q^{m+1}}{2(m+1)}(\gamma+ 2 \log 2\pi) - \frac{q^{m+2}}{(m+2)}(\gamma + \log 2\pi)+\)

\(\displaystyle \frac{1}{2} m! \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{q^{m-2j}}{(2\pi)^{2j+1} (m-2j)!} \text{Cl}_{2j+2} (2\pi q) +\)

\(\displaystyle \frac{1}{2} m! \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{q^{m-2j-1}}{(2\pi)^{2j+2} (m-2j-1)!} \text{Cl}_{2j+3} (2\pi q) +\)

\(\displaystyle \frac{1}{2} (-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \frac{m!}{(2\pi)^{m+1}} \zeta (m+2)+\)

\(\displaystyle \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \int_0^q x^m \sin 2\pi kx\)


As mentioned before, we will evaluate the general q-case shortly, for \(\displaystyle q \in \mathbb{Q}^{+}\), where \(\displaystyle 0 < q \le 1\). For now, however, let's set \(\displaystyle q=1/p\), \(\displaystyle p \in \mathbb{Z}^{+}\) in Theorem 1.0 to obtain:

\(\displaystyle \Gamma i_m(1/p)=\int_0^{1/p}x^m\log\Gamma(x)\,dx=\)


\(\displaystyle \frac{(\gamma+ 2 \log 2\pi)}{2p^{m+1}(m+1)} - \frac{(\gamma + \log 2\pi)}{p^{m+2}(m+2)}+
\frac{m!}{2} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{ \text{Cl}_{2j+2} \left( \frac{2\pi}{p} \right) }{(2\pi)^{2j+1} p^{m-2j} (m-2j)!} +\)

\(\displaystyle \frac{m!}{2} \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{ \text{Cl}_{2j+3} \left( \frac{2\pi}{p} \right) }{(2\pi)^{2j+2} p^{m-2j-1} (m-2j-1)!} +
\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \int_0^{1/p} x^m \sin 2\pi kx
\)


\(\displaystyle + \frac{ (-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \, m! }{2 (2\pi)^{m+1}} \zeta (m+2)+\)


In light of all the groundwork we've already done, that last integral is relatively - (Drunk) -straightforward...



\(\displaystyle \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \int_0^{1/p} x^m \sin 2\pi kx \, dx=\)

\(\displaystyle \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \Bigg\{ \frac{1}{(2\pi k)^{m+1}} \int_0^{2\pi k/p} x^m \sin x \, dx \Bigg\} =\)



\(\displaystyle \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \Bigg\{ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j+1} \frac{(2\pi k / p)^{m-2j}}{(m-2j)!} \cos (2\pi k / p) \, \Bigg\} +\)


\(\displaystyle \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \Bigg\{ \frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{(2\pi k / p)^{m-2j-1}}{(m-2j-1)!} \sin (2\pi k /p) \, \Bigg\} +\)


\(\displaystyle \frac{1}{\pi} \, \sum_{k=1}^{\infty}\frac{\log k}{k} \Bigg\{ -\mathcal{F}(m) (-1)^{ \lfloor \frac{m}{2} \rfloor +1} \frac{m!}{(2\pi k)^{m+1}} \Bigg\}=\)



\(\displaystyle \frac{m!}{\pi} \, \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } \frac{(-1)^{j+1}}{p^{m-2j} (2\pi)^{2j+1} (m-2j)! } \sum_{k=1}^{\infty} \frac{\log k}{k^{2j+2}} \cos \left( \frac{2\pi k}{p} \right) +\)


\(\displaystyle \frac{m!}{\pi} \, \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } \frac{(-1)^{j}}{p^{m-2j-1} (2\pi)^{2j+2} (m-2j-1)! } \sum_{k=1}^{\infty} \frac{\log k}{k^{2j+3}} \sin \left( \frac{2\pi k}{p} \right) +\)


\(\displaystyle \frac{m!}{\pi (2\pi)^{m+1}} \mathcal{F}(m) (-1)^{ \lfloor \frac{m}{2} \rfloor +1} \zeta'(m+2)=\)


In terms of the generalized Poly-Clausen Functions we introduced earlier, this becomes:



\(\displaystyle -\frac{m!}{\pi} \, \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } \frac{(-1)^{j+1}}{p^{m-2j} (2\pi)^{2j+1} (m-2j)! } C_{2j+2} \left(1; \frac{2\pi }{p} \right) +\)


\(\displaystyle -\frac{m!}{\pi} \, \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } \frac{(-1)^{j}}{p^{m-2j-1} (2\pi)^{2j+2} (m-2j-1)! } S_{2j+3} \left(1; \frac{2\pi }{p} \right) +\)


\(\displaystyle \frac{m!}{\pi (2\pi)^{m+1}} \mathcal{F}(m) (-1)^{ \lfloor \frac{m}{2} \rfloor +1} \zeta'(m+2)\)


Need a break... More to follow shortly... (Smoking)
 
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