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Nielsen Polylogarithms and the Generalized Logsine Integral

DreamWeaver

Well-known member
Sep 16, 2013
337
In this tutorial we will be exploring the Nielsen Polylogarithm, and it's close relative, the Generalized Logsine Integral. Both of these functions have numerous applications in maths and physics, not least of all in Quantum Theory. The main purpose here, however, will be employing them in the evaluation of certain types of integrals - particularly, but not exclusively, those involving logarithms of trigonometric functions.


For non-negative integers \(\displaystyle m\) and \(\displaystyle n\), where \(\displaystyle n-m \ge 1\), the Generalized Logsine Integral is defined by:


\(\displaystyle \mathscr{Ls}_n^{(m)} (\theta) = -\int_0^{\theta} x^m \log^{n-m-1} \Bigg| 2 \sin \frac{x}{2} \Bigg| \, dx\)


By the First Fundamental Theorem of Calculus, we immediately have


\(\displaystyle \frac{d}{d\theta} \mathscr{Ls}_n^{(m)} (\theta)= - \theta^m \log^{n-m-1} \Bigg| 2 \sin \frac{\theta}{2} \Bigg| \)


Next, we consider the integrated form of the Generalized Logsine Integral, and perform an integration by parts:


\(\displaystyle \int_0^{\theta} \mathscr{Ls}_n^{(m)} (x) \, dx = \theta \mathscr{Ls}_n^{(m)} (\theta) - \int_0^{\theta} x \, \left[ \frac{d}{dx} \mathscr{Ls}_n^{(m)} (x) \right] \,dx=\)


\(\displaystyle \theta \mathscr{Ls}_n^{(m)} (\theta) + \int_0^{\theta} x^{m+1} \log^{n-m-1} \Bigg| 2 \sin \frac{\theta}{2} \Bigg| \, dx =\)


\(\displaystyle \theta \mathscr{Ls}_n^{(m)} (\theta) + \int_0^{\theta} x^{m+1} \log^{(n+1)-(m+1)-1} \Bigg| 2 \sin \frac{\theta}{2} \Bigg| \, dx \)


Hence we have


\(\displaystyle (01) \quad \int_0^{\theta} \mathscr{Ls}_n^{(m)} (x) \, dx = \theta \mathscr{Ls}_n^{(m)} (\theta) - \mathscr{Ls}_{n+1}^{(m+1)} (\theta)\)


Next, in light of the standard integral definition of the Clausen function, namely


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


we have the special case:


\(\displaystyle (02) \quad \text{Cl}_2(\theta) = \mathscr{Ls}_2^{(0)} (\theta)\)


As I've shown on various other threads, the following integral evaluation holds:


\(\displaystyle \int_0^{\theta} \text{Cl}_2(x) \,dx = \zeta (3)- \text{Cl}_3(\theta)\)


Conversely, by appealing to \(\displaystyle (01)\) and \(\displaystyle (02)\) above, we have


\(\displaystyle \int_0^{\theta} \text{Cl}_2(x) \,dx = \int_0^{\theta} \mathscr{Ls}_2^{(0)} (x) \,dx = \theta \mathscr{Ls}_2^{(0)} (\theta) - \mathscr{Ls}_{3}^{(1)} (\theta)=\)


\(\displaystyle \theta \text{Cl}_2(\theta) - \mathscr{Ls}_{3}^{(1)} (\theta) = \zeta (3)- \text{Cl}_3(\theta)\)


Hence


\(\displaystyle (03) \quad \mathscr{Ls}_{3}^{(1)} (\theta) = \text{Cl}_3(\theta) + \theta \text{Cl}_2(\theta) -\zeta(3)
\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
We will develop many more relations and special values for the Generalized Logsine Integral, but first, and application.


The Inverse Sine integral of second order, \(\displaystyle \text{Si}_2(z)\), has a simple closed form in terms of the Clausen function. Higher order Inverse Sine Integrals, however, require Nielsen Polylogarithms or Generalized Sine Integrals for their closed form.

Define


\(\displaystyle \text{Si}_1(z) = \sin^{-1}z\)

\(\displaystyle \text{Si}_2(z) = \int_0^z \frac{\sin^{-1} x}{x}\,dx\)

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


Then


\(\displaystyle \text{Si}_3(z) = \int_0^z \frac{ \text{Si}_2(x) }{x}\,dx = \text{Si}_2(x) \log x \, \Bigg|_0^z-\int_0^z \frac{\log x \sin^{-1} x}{x}\,dx=\)


\(\displaystyle \text{Si}_2(z) \log z - \left[ \frac{1}{2} (\log x)^2 \sin^{-1} x \, \Bigg|_0^z - \frac{1}{2} \int_0^z \frac{(\log x)^2}{ \sqrt{1-x^2} } \, dx \right]\)


Now, if we write \(\displaystyle z= \sin \theta\), and apply the substitution \(\displaystyle x = \sin y\) to that last integral, we get


\(\displaystyle \text{Si}_3(\sin \theta) = \text{Si}_2(\sin \theta) \log (\sin \theta) - \frac{\theta}{2} \log^2(\sin \theta) +\frac{1}{2} \int_0^{\theta} \log^2(\sin y) \, dy\)


Next, we apply the substitution \(\displaystyle y=x/2\) to that last logsine integral to obtain


\(\displaystyle \text{Si}_3(\sin \theta) = \text{Si}_2(\sin \theta) \log (\sin \theta) - \frac{\theta}{2} \log^2(\sin \theta) +\frac{1}{4} \int_0^{2\theta} \log^2 \left( \sin \frac{x}{2} \right) \, dx\)


Finally, we consider the integral


\(\displaystyle \int_0^{2\theta} \log^2 \Bigg| 2 \sin \frac{x}{2} \Bigg| \,dx= \int_0^{2\theta} \left[ \log 2 + \log \Bigg| \sin \frac{x}{2} \Bigg| \, \right]^2 \,dx= \)


\(\displaystyle \int_0^{2\theta} \left[ (\log 2)^2 + 2\log 2 \, \log \Bigg| \sin \frac{x}{2} \Bigg| + \log^2 \Bigg| \sin \frac{x}{2} \Bigg| \, \right]^2 \,dx=\)


\(\displaystyle 2\theta (\log 2)^2 +2 \log 2 \int_0^{2\theta} \log \Bigg| \sin \frac{x}{2} \Bigg| \, dx + \int_0^{2\theta} \log^2 \Bigg| \sin \frac{x}{2} \Bigg| \, dx\)


Clearly, by the definition of the Generalized Logsine Integral, we also have


\(\displaystyle \int_0^{2\theta} \log^2 \Bigg| 2 \sin \frac{x}{2} \Bigg| \,dx = - \mathscr{Ls}_3^{(0)} (2\theta)\)


Furthermore,


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


\(\displaystyle -\text{Cl}_2(2\theta)-2\theta \log 2\)


Putting all of this together, we have


\(\displaystyle - \mathscr{Ls}_3^{(0)} (2\theta) = 2\theta (\log 2)^2 +2 \log 2 \Bigg[ -\text{Cl}_2(2\theta)-2\theta \log 2 \Bigg] \)

\(\displaystyle + \int_0^{2\theta} \log^2 \Bigg| \sin \frac{x}{2} \Bigg| \, dx\)


Hence


\(\displaystyle \int_0^{2\theta} \log^2 \Bigg| \sin \frac{x}{2} \Bigg| \, dx= - \mathscr{Ls}_3^{(0)} (2\theta) + 2\theta (\log 2)^2 + 2 \text{Cl}_2(2\theta) \, \log 2\)


Within the interval \(\displaystyle 0 < \theta < \pi\), we can drop the absolute value sign in that last integral, and write


\(\displaystyle \int_0^{2\theta} \log^2 \left( \sin \frac{x}{2} \right) \, dx= - \mathscr{Ls}_3^{(0)} (2\theta) + 2\theta (\log 2)^2 + 2 \text{Cl}_2(2\theta) \, \log 2\)




Applying this to our partial evaluation of \(\displaystyle \text{Si}_3(\sin \theta)\) above gives the closed form:


\(\displaystyle (04) \quad \text{Si}_3(\sin \theta) = \)


\(\displaystyle \text{Si}_2(\sin \theta) \log (\sin \theta) - \frac{\theta}{2} \log^2(\sin \theta) -\frac{1}{4} \mathscr{Ls}_3^{(0)} (2\theta) +\frac{\theta}{2}(\log 2)^2\)


\(\displaystyle + \frac{\log 2}{2} \text{Cl}_2(2\theta)\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
A quick calculation gives the following closed form for the second order Inverse Sine Integral, in terms of the Clausen function:


\(\displaystyle \text{Si}_2(\sin \theta) = \theta \log 2 +\theta \log(\sin \theta) + \frac{1}{2} \text{Cl}_2(2\theta)\)


Substituting this back into \(\displaystyle (04)\), and then simplifying gives



\(\displaystyle (05) \quad \text{Si}_3(\sin \theta) = \)


\(\displaystyle \frac{\theta}{2} \log^2 (2 \sin \theta ) + \frac{1}{2} \log (2 \sin \theta) \, \text{Cl}_2(2\theta) -\frac{1}{4} \mathscr{Ls}_3^{(0)} (2\theta)\)




Examples:




Setting \(\displaystyle \theta = \pi/4\) gives


\(\displaystyle \text{Si}_3\left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{2}(\log 2)^2 +\frac{1}{4} \log 2\, \text{Cl}_2 \left( \frac{\pi}{2} \right) -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{\pi}{2} \right)=\)


\(\displaystyle \frac{\pi}{32}(\log 2)^2 +\frac{G}{4} \log 2 -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{\pi}{2} \right)\)




Setting \(\displaystyle \theta = \pi/6\) gives


\(\displaystyle \text{Si}_3\left( \frac{1}{2} \right) = -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{\pi}{3} \right)\)




Setting \(\displaystyle \theta = \pi/3\) gives


\(\displaystyle \text{Si}_3\left( \frac{\sqrt{3}}{2} \right)= \frac{\pi}{24} (\log 3)^2 + \frac{1}{4}\log 3 \, \text{Cl}_2 \left( \frac{2\pi}{2} \right) -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{2\pi}{3} \right)\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Further values of the third order Inverse Sine Integral, evaluated by means of (05), are given below:




\(\displaystyle \theta = \pi/10 \quad \Rightarrow\)


\(\displaystyle \quad \text{Si}_3 \left( \frac{\sqrt{5}-1}{4} \right) =\)

\(\displaystyle \frac{\pi}{20} \log^2 \left( \frac{ \sqrt{5}-1}{2} \right) + \frac{1}{2} \log \left( \frac{\sqrt{5}-1}{2} \right) \, \text{Cl}_2 \left( \frac{\pi}{5} \right) -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{\pi}{5} \right)
\)


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


\(\displaystyle \theta = \pi/8 \quad \Rightarrow\)


\(\displaystyle \quad \text{Si}_3 \left( \frac{\sqrt{2 - \sqrt{2} }}{2} \right) =\)

\(\displaystyle \frac{\pi}{64} \log^2 \left( 2-\sqrt{2} \right) + \frac{1}{4} \log \left( 2-\sqrt{2} \right) \, \text{Cl}_2 \left( \frac{\pi}{4} \right) -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{\pi}{4} \right)
\)


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


\(\displaystyle \theta = 3\pi/8 \quad \Rightarrow\)


\(\displaystyle \quad \text{Si}_3 \left( \frac{\sqrt{2 + \sqrt{2} }}{2} \right) =\)

\(\displaystyle \frac{3\pi}{64} \log^2 \left( 2+\sqrt{2} \right) + \frac{1}{4} \log \left( 2+\sqrt{2} \right) \, \text{Cl}_2 \left( \frac{3\pi}{4} \right) -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{3\pi}{4} \right)
\)


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


\(\displaystyle \theta = \pi/5 \quad \Rightarrow\)


\(\displaystyle \quad \text{Si}_3 \left( \frac{1}{2} \sqrt{ \frac {5 - \sqrt{5} }{2} } \right) =\)

\(\displaystyle \frac{\pi}{40} \log^2 \left( \frac{5-\sqrt{5}}{2} \right) + \frac{1}{4} \log \left( \frac{5-\sqrt{5}}{2} \right) \, \text{Cl}_2 \left( \frac{2\pi}{5} \right) -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{2\pi}{5} \right)
\)


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


\(\displaystyle \theta = 2\pi/5 \quad \Rightarrow\)


\(\displaystyle \quad \text{Si}_3 \left( \frac{1}{2} \sqrt{ \frac {5 + \sqrt{5} }{2} } \right) =\)

\(\displaystyle \frac{\pi}{20} \log^2 \left( \frac{5+\sqrt{5}}{2} \right) + \frac{1}{4} \log \left( \frac{5+\sqrt{5}}{2} \right) \, \text{Cl}_2 \left( \frac{4\pi}{5} \right) -\frac{1}{4} \mathscr{Ls}_3^{(0)} \left( \frac{4\pi}{5} \right)
\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
A little later on, we will explore the higher order Inverse Sine Integrals \(\displaystyle \text{Si}_4(x)\) and \(\displaystyle \text{Si}_5(x)\), and find closed form expressions for them in terms of the Generalized Logsine Integrals. Hopefully, however, the evaluation of \(\displaystyle \text{Si}_3(x)\) given above will be sufficient to convince the reader that all higher order Inverse Sine Integrals can be expressed as finite sums of Clausen functions and Logsine Integrals alone.

The higher order Inverse Tangent Integrals, \(\displaystyle \text{Ti}_m(z)\), as well as other related functions we will explore later, require a little more than just Logsine |Integrals and Clausen functions, so we introduce a number of other Generalized Log-Trig Integrals/ functions. In each case, I'll use what seems to be the standard notation, but it's worth noting that some authors use alternative notations.

Analogous to \(\displaystyle \mathscr{Ls}_n^{(m)} (\theta)\), we define:


\(\displaystyle \mathscr{Lc}_n^{(m)} (\theta) = -\int_0^{\theta} x^m \log^{n-m-1} \Bigg| 2 \cos \frac{x}{2} \Bigg| \, dx\)


\(\displaystyle \mathscr{Lt}_n^{(m)} (\theta) = -\int_0^{\theta} x^m \log^{n-m-1} \Bigg| 2 \tan \frac{x}{2} \Bigg| \, dx\)


\(\displaystyle \mathscr{Lsc}_{(m,n)} (\theta) = -\int_0^{\theta} \log^{m-1} \Bigg| 2 \sin \frac{x}{2} \Bigg| \, \log^{n-1} \Bigg| 2 \cos \frac{x}{2} \Bigg| \, dx\)



As with the Logsine Integrals, by the First Fundamental Theorem of Calculus, we immediately have


\(\displaystyle (06) \quad \frac{d}{d\theta} \mathscr{Lc}_n^{(m)} (\theta)= - \theta^m \log^{n-m-1} \Bigg| 2 \cos \frac{\theta}{2} \Bigg| \)


\(\displaystyle (07) \quad \frac{d}{d\theta} \mathscr{Lt}_n^{(m)} (\theta)= - \theta^m \log^{n-m-1} \Bigg| 2 \tan \frac{\theta}{2} \Bigg| \)


\(\displaystyle (08) \quad \frac{d}{d\theta} \mathscr{Lsc}_{(m,n)} (\theta) = - \log^{m-1} \Bigg| 2 \sin \frac{\theta}{2} \Bigg| \, \log^{m-1} \Bigg| 2 \cos \frac{\theta}{2} \Bigg| \)



We now turn our attention to developing various relations between the four main types of Generalized Log-Trig integrals, starting with a duplication formula for the Logsine Integral.

Consider the Integral


\(\displaystyle \mathscr{Ls}_n^{(0)} (2\theta) = -\int_0^{2\theta} \log^{n-1} \Bigg| 2 \sin \frac{x}{2} \Bigg| \, dx\)


Use of the duplication formula for the Sine function \(\displaystyle \sin 2 x= 2\sin x \cos x\) gives


\(\displaystyle \mathscr{Ls}_n^{(0)} (2\theta) = -\int_0^{2\theta} \log^{n-1} \Bigg| 2 \sin \frac{x}{4} + 2 \sin \frac{x}{4} \Bigg|^{n-1} \, dx\)


Now, provided that \(\displaystyle 0 < \theta < \pi \), this can be rewritten in the form


\(\displaystyle - \int_0^{2 \theta} \left[ \log \Bigg| 2 \sin \frac{x}{4} \Bigg| + \log \Bigg| 2 \cos \frac{x}{4} \Bigg| \right]^{n-1} \, dx \)


Next, apply the substitution \(\displaystyle x=2y\), and expand the logarithmic power, by use of the finite Binomial Theorem:


\(\displaystyle -2 \, \sum_{j=0}^{n-1} \binom{n-1}{j} \int_0^{\theta} \log^{n-j-1} \Bigg| 2 \sin \frac{y}{2} \Bigg| \, \log^{j} \Bigg| 2 \cos \frac{y}{2} \Bigg| \, dy= \)


\(\displaystyle 2\, \sum_{j=0}^{n-1} \binom{n-1}{j} \mathscr{Lsc}_{(n-j, j+1)} (\theta) \)



Putting all of this together gives us the desired Duplication formula for the Logsine Integral:


\(\displaystyle (09) \quad \mathscr{Ls}_n^{(0)} (2\theta) = 2\, \sum_{j=0}^{n-1} \binom{n-1}{j} \mathscr{Lsc}_{(n-j, j+1)} (\theta) \)


In the sum above, when \(\displaystyle j=n-1\) we end up with a term that is a pure Logcosine Integral. Similarly, when \(\displaystyle j=0\) we end up with a term that is a pure Logcosine Integral, so the duplication formula can be written in the form:



\(\displaystyle (10) \quad \mathscr{Ls}_n^{(0)} (2\theta) = \)


\(\displaystyle 2\, \mathscr{Ls}_n^{(0)} (\theta) + 2\, \mathscr{Lc}_n^{(0)} (\theta)+ 2\, \sum_{j=1}^{n-2} \binom{n-1}{j} \mathscr{Lsc}_{(n-j, j+1)} (\theta) \)
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
To obtain the reflection formulae for the Logsine and Logcosine integrals, where each is expressed as a finite sum of the other, we simply apply the transformation \(\displaystyle x=\pi-y\) to both functions:


\(\displaystyle \mathscr{Lc}_n^{(m)} ( \theta ) = -\int_0^{\theta} x^m \log^{n-m-1} \Bigg| 2 \cos \frac{x}{2} \Bigg| \, dx=\)


\(\displaystyle \int_{\pi}^{\pi-\theta} (\pi - y)^m \log^{n-m-1} \Bigg| 2 \cos \frac{(\pi - y)}{2} \Bigg| \, dy=\)


\(\displaystyle \sum_{j=0}^m (-1)^j \binom{m}{j} \pi^{m-j} \int_{\pi}^{\pi-\theta} y^j \log^{n-m-1} \Bigg| 2 \sin \frac{y}{2} \Bigg| \, dy=\)


\(\displaystyle \sum_{j=0}^m (-1)^{j} \binom{m}{j} \pi^{m-j} \int_{\pi}^{\pi-\theta} y^j \log^{(n-m+j)-j-1} \Bigg| 2 \sin \frac{y}{2} \Bigg| \, dy=\)





\(\displaystyle \sum_{j=0}^m (-1)^{j} \binom{m}{j} \pi^{m-j} \int_{0}^{\pi-\theta} y^j \log^{(n-m+j)-j-1} \Bigg| 2 \sin \frac{y}{2} \Bigg| \, dy - \)


\(\displaystyle \sum_{j=0}^m (-1)^{j} \binom{m}{j} \pi^{m-j} \int_{0}^{\pi} y^j \log^{(n-m+j)-j-1} \Bigg| 2 \sin \frac{y}{2} \Bigg| \, dy= \)



\(\displaystyle \sum_{j=0}^m (-1)^{j+1} \binom{m}{j} \pi^{m-j} \Bigg[ \mathscr{Ls}_{n-m+j}^{(j)} ( \pi - \theta ) - \mathscr{Ls}_{n-m+j}^{(j)} ( \pi ) \Bigg]\)


The equivalent form for the Logsine Integral is proven in precisely the same way, so I'll skip the proof.



Reflection formulae:



\(\displaystyle (11) \quad \mathscr{Ls}_n^{(m)} ( \theta ) =\)


\(\displaystyle \sum_{j=0}^m (-1)^{j+1} \binom{m}{j} \pi^{m-j} \Bigg[ \mathscr{Lc}_{n-m+j}^{(j)} ( \pi - \theta ) - \mathscr{Lc}_{n-m+j}^{(j)} ( \pi ) \Bigg]\)



\(\displaystyle (12) \quad \mathscr{Lc}_n^{(m)} ( \theta ) = \)


\(\displaystyle \sum_{j=0}^m (-1)^{j+1} \binom{m}{j} \pi^{m-j} \Bigg[ \mathscr{Ls}_{n-m+j}^{(j)} ( \pi - \theta ) - \mathscr{Ls}_{n-m+j}^{(j)} ( \pi ) \Bigg]\)
 
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DreamWeaver

Well-known member
Sep 16, 2013
337
The reflection formula for the Logtangent Integral is found by applying the same substitution as above, \(\displaystyle x= \pi-y\):


\(\displaystyle \mathscr{Lt}_n^{(m)} (\theta) = -\int_0^{\theta} x^m \log^{n-m-1} \Bigg| 2 \tan \frac{x}{2} \Bigg| \, dx=\)


\(\displaystyle \int_{\pi}^{\pi-\theta} (\pi -y)^m \log^{n-m-1} \Bigg| 2 \tan \frac{ (\pi - y) }{2} \Bigg| \, dy=\)


\(\displaystyle \int_{\pi}^{\pi-\theta} (\pi -y)^m \log^{n-m-1} \Bigg| 2 \cot \frac{ y }{2} \Bigg| \, dy=\)


\(\displaystyle \int_{\pi}^{\pi-\theta} (\pi -y)^m \log^{n-m-1} \Bigg| -2 \tan \frac{ y }{2} \Bigg| \, dy=\)


\(\displaystyle \int_{\pi}^{\pi-\theta} (\pi -y)^m \log^{n-m-1} \Bigg| 2 \tan \frac{ y }{2} \Bigg| \, dy=\)


\(\displaystyle \sum_{j=0}^m (-1)^j \binom{m}{j} \pi^{m-j} \, \int_{\pi}^{\pi-\theta} \log^{n-m-1} \Bigg| 2 \tan \frac{ y }{2} \Bigg| \, dy=\)


\(\displaystyle \sum_{j=0}^m (-1)^j \binom{m}{j} \pi^{m-j} \, \int_{\pi}^{\pi-\theta} \log^{(n-m+j)-j-1} \Bigg| 2 \tan \frac{ y }{2} \Bigg| \, dy=\)


\(\displaystyle \sum_{j=0}^m (-1)^{j+1} \binom{m}{j} \pi^{m-j} \Bigg[ \mathscr{Lt}_{n-m+j}^{(j)} (\pi-\theta) - \mathscr{Lt}_{n-m+j}^{(j)} (\pi) \Bigg] \)




Reflection formula:



\(\displaystyle (13) \quad \mathscr{Lt}_n^{(m)} (\theta) = \)


\(\displaystyle \sum_{j=0}^m (-1)^{j+1} \binom{m}{j} \pi^{m-j} \Bigg[ \mathscr{Lt}_{n-m+j}^{(j)} (\pi-\theta) - \mathscr{Lt}_{n-m+j}^{(j)} (\pi) \Bigg] \)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
In the special case \(\displaystyle m=0\), the Logtangent can be expressed as a finite sum of Lsc-integrals:




\(\displaystyle \mathscr{Lt}_n^{(0)} ( \theta ) = -\int_0^{\theta} \log^{n-1} \Bigg| 2 \tan \frac{x}{2} \Bigg| \, dx=\)



\(\displaystyle -\int_0^{\theta} \left[ \Big| \log 2 \Big| + \log \Bigg| \tan \frac{x}{2} \Bigg| \right]^{n-1} \, dx= \)



\(\displaystyle -\int_0^{\theta} \left[ \log 2 + \log \Bigg| \frac{ \sin (x/2) }{ \cos (x/2) } \Bigg| \right]^{n-1} \, dx=\)



\(\displaystyle -\int_0^{\theta} \left[ \log 2 + \log \Bigg| \frac{ 2 \sin (x/2) }{ 2 \cos (x/2) } \Bigg| \right]^{n-1} \, dx=\)



\(\displaystyle - \sum_{j=0}^{n-1} \binom{n-1}{j} (\log 2)^{n-j-1} \, \int_0^{\theta} \left[ \log \Bigg| \frac{ 2\sin (x/2) }{ 2\cos (x/2) } \Bigg| \right]^{j} \, dx=
\)



\(\displaystyle - \sum_{j=0}^{n-1} \binom{n-1}{j} (\log 2)^{n-j-1} \, \int_0^{\theta} \left[ \log \Bigg| 2 \sin \frac{x}{2} \Bigg| - \log \Bigg| 2 \sin \frac{x}{2} \Bigg| \right]^{j} \, dx=\)



\(\displaystyle - \sum_{j=0}^{n-1} \binom{n-1}{j} (\log 2)^{n-j-1} \, \sum_{k=0}^j (-1)^k \binom{j}{k} \, \int_0^{\theta} \log^{j-k} \Bigg| 2 \sin \frac{x}{2} \Bigg| \, \log^k \Bigg| 2 \cos \frac{x}{2} \Bigg| \, dx=\)



\(\displaystyle \sum_{j=0}^{n-1} \binom{n-1}{j} (\log 2)^{n-j-1} \, \sum_{k=0}^j (-1)^k \binom{j}{k} \, \mathscr{Lsc}_{(j-k+1, k+1)} (\theta)\)



Hence



\(\displaystyle (14) \quad \mathscr{Lt}_n^{(0)} ( \theta ) =\)


\(\displaystyle \sum_{j=0}^{n-1} \binom{n-1}{j} (\log 2)^{n-j-1} \, \sum_{k=0}^j (-1)^k \binom{j}{k} \, \mathscr{Lsc}_{(j-k+1, k+1)} (\theta)\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
In the interval \(\displaystyle 0 < z \le 1\), the Inverse Tangent Integrals are defined by


\(\displaystyle \text{Ti}_1(z) = \sum_{k=0}^{\infty} (-1)^k \frac{z^{2k+1}}{(2k+1)} = \tan^{-1} z
\)


\(\displaystyle \text{Ti}_{m+1}(z) = \sum_{k=0}^{\infty} (-1)^k \frac{z^{2k+1}}{(2k+1)^{m+1}} = \int_0^z \frac{\text{Ti}_m(t)}{t}\,dt\)


As I've shown on various other threads, the following result holds:


\(\displaystyle (15) \quad \int_0^{\theta} \log^m(\tan x)\, dx = \int_0^{\tan \theta} \frac{(\log x)^m}{(1+x^2)}\,dx = \)


\(\displaystyle m! \, \sum_{j=0}^m (-1)^j \frac{ \log^{m-j} (\tan \theta) }{(m-j)!} \, \text{Ti}_{j+1} (\tan \theta)\)



Similarly, on the Inverse Sine / Tangent Integral tutorial I proved the following evaluation of the second order Inverse Tangent Integral:


\(\displaystyle (16) \quad \text{Ti}_2(\tan \theta) = \theta \log(\tan \theta) + \frac{1}{2} \text{Cl}_2(2\theta) + \frac{1}{2} \text{Cl}_2(\pi- 2\theta)\)


Setting \(\displaystyle m=3\) in the LHS of (15) gives


\(\displaystyle \int_0^{\theta} \log^3(\tan x)\,dx = \int_0^{\theta} \Bigg[ \log (2 \tan x) - \log 2 \Bigg]^3 \, dx=\)


\(\displaystyle \int_0^{\theta} \Bigg[ \log^3 (2 \tan x) - \log 2 \log^2 (2 \tan x) + (\log 2)^2 \log (2 \tan x) - (\log 2)^3 \Bigg]\, dx\)


The substitution \(\displaystyle y=2x\) gives


\(\displaystyle \frac{1}{2} \int_0^{2\theta} \log^3 \left(2 \tan \frac{y}{2} \right) \, dy - \frac{\log 2}{2} \, \int_0^{2\theta} \log^2 \left(2 \tan \frac{y}{2} \right) \, dy + \)


\(\displaystyle \frac{(\log 2)^2}{2} \, \int_0^{2\theta} \log \left(2 \tan \frac{y}{2} \right) \, dy - \theta (\log 2)^3 \)


Since we've halved the variable, and yet double the range of integration, the condition \(\displaystyle 0 < \theta \le \pi/4\) still holds. Furthermore, within this range we can add (or remove) the absolute value notation on the terms inside each logarithm without changing the value. Multiplying both sides by a factor of \(\displaystyle 2\), we have:


\(\displaystyle 2 \, \int_0^{\theta} \log^3(\tan x)\,dx = \)


\(\displaystyle \int_0^{2\theta} \log^3 \Bigg| 2 \tan \frac{y}{2} \Bigg| \, dy -\log 2 \, \int_0^{2\theta} \log^2 \Bigg| 2 \tan \frac{y}{2} \Bigg| \, dy + \)


\(\displaystyle (\log 2)^2 \, \int_0^{2\theta} \log \Bigg| 2 \tan \frac{y}{2} \Bigg| \, dy - \theta (\log 2)^3 =\)


\(\displaystyle - \mathscr{Lt}_4^{(0)} (2\theta) + \mathscr{Lt}_3^{(0)} (2\theta) \, \log 2 - \mathscr{Lt}_2^{(0)} (2\theta) \, (\log 2)^2 - \theta (\log 2)^3 \)


Conversely, setting \(\displaystyle m=3\) in the RHS of (15), and then multiplying by a factor of \(\displaystyle 2\) gives:


\(\displaystyle 2 \theta \, \log^3(\tan \theta) -6\, \text{Ti}_2(\tan \theta) \, \log^2(\tan \theta) + 12 \, \text{Ti}_3(\tan \theta) \, \log(\tan \theta) -12\, \text{Ti}_4(\tan \theta) \)




Thus


\(\displaystyle (17) \quad - \mathscr{Lt}_4^{(0)} (2\theta) + \mathscr{Lt}_3^{(0)} (2\theta) \, \log 2 - \mathscr{Lt}_2^{(0)} (2\theta) \, (\log 2)^2 - \theta (\log 2)^3 = \)


\(\displaystyle 2 \theta \, \log^3(\tan \theta) -6\, \text{Ti}_2(\tan \theta) \, \log^2(\tan \theta) + 12 \, \text{Ti}_3(\tan \theta) \, \log(\tan \theta) -12\, \text{Ti}_4(\tan \theta) \)




More to follow shortly, including various applications of that last result... Think it's been enough for one day. Phew! :D