Closed Form for  ∫10arctanh xtan(π2 x) dx~\int_0^1\frac{\text{arctanh }x}{\tan\left(\frac\pi2~x\right)}~dx

Does  10arctanh xtan(π2 x) dx  0.4883854771179872995286585433480  possess a closed form expression ?


This recent post, in conjunction with my age-old interest in Gudermannian functions, have inspired me to ask this question. The reason I suspect that such a closed form might possibly exist is because the integration interval is “meaningful” for both functions used in the integrand. However, none of the various approaches that I can think of seem to be of any help. Perhaps I’m missing something ?

Answer

Here is an approach.

We give a preliminary result.


A series of squares of logarithms


Let us consider the poly-Hurwitz zeta function initially defined by the series
ζ(s,ta,b):=+n=11(n+a)s(n+b)t,a>1,b>1,(s+t)>1.
The function ζ(,a,b) extends to a meromorphic function on C2 with only singularities on the set {(s,t)C2,(s+t)=1}. It clearly generalizes the classic Hurwitz zeta function initially defined by the series
ζ(s,a):=+n=01(n+a)s,a>0,s>1.

We have the following new result.

Theorem. Let a,b be complex numbers such that a>1 and b>1.

Then
+n=1log2(n+an+b)=ζ

where \log (z) denotes the principal value of the logarithm defined for all z \neq 0 by

\log (z) = \ln |z|+i \arg z, \quad -\pi<\arg z\leq \pi,

\displaystyle \zeta(\cdot,a) and \displaystyle \zeta(\cdot,\cdot \mid a,b) denoting the Hurwitz zeta function and the poly-Hurwitz zeta function respectively and where \zeta’’(0, a)=\partial_{s}^2\left.\zeta(s,a)\right|_{s=0},\qquad \zeta^{1,1}(0,0\mid a,b)=\partial_{st}^2\left.\zeta(s,t\mid a,b)\right|_{(s,t)=(0,0)}.

Proof.
On the one hand, one has

\begin{align}
&\partial_a \left(\zeta''(0,a+1)+\zeta''(0,b+1)-2\zeta^{1,1}(0,0\mid a,b)\right)\\\\
&= \left.\partial_s^2 \left(\partial_a \zeta(s,a+1)\right)\right|_{s=0}-2\left.\partial_{st}^2 \left(\partial_a \zeta(s,t\mid a,b)\right)\right|_{(s,t)=(0,0)}\\\\
&= \left.\partial_s^2 \left(-s\zeta(s+1,a+1)\right)\right|_{s=0}-2\left.\partial_{st}^2 \left(-s\zeta(s+1,t\mid a,b)\right)\right|_{(s,t)=(0,0)}\\\\
&= -\left.\left(2\zeta'(s+1,a+1)+s\zeta''(s+1,a+1)\right)\right|_{s=0}+2\left.\partial_s \!\left(s\zeta^{0,1}(s+1,t\mid a,b)\right)\right|_{(s,t)=(0,0)}\\\\
&=2\gamma_1(a+1)-2\gamma_1(b,a),
\end{align}
using Theorem 1 here.

On the other hand, one has


\begin{align}
\partial_a\! \left(\sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{n+a}{n+b}\!\right)\right)
\!= 2\sum_{n=1}^{+\infty} \frac{\log (n+a)-\log (n+b)}{n+a}
=2\gamma_1(a+1)-2\gamma_1(b,a),
\end{align}

using Theorem 2 here.

Observing that

\zeta(s,t\mid 0,0)=\zeta(s+t), \quad \zeta(s,1)=\zeta(s),
where \zeta(\cdot) is the Riemann zeta function, then

\zeta’’(0,1)-\zeta^{1,1}(0,0\mid 0,0)=0
and both sides of (3) vanish at a=b=0.

Thus (3) holds true. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box


Lucian's integral


We prove that Lucian's integral is related to the preceding family of logarithmic series.

Proposition 1. We have

\begin{align}
\int_0^1 \frac{\text{arctanh}\: x}{\tan \left( \frac{\pi}2x\right)}\:{\rm d}x=\frac\pi4-\frac1{2\pi}\sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{2n-1 }{ 2n+1}\!\right).\tag{4}
\end{align}

Proof. Let us proceed on Jack D'Aurizio's route which starts by using the standard expansion

\frac1{\tan \left( \frac{\pi}2x\right)}=\frac{2}{\pi x}-\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{\zeta(2n+2)}{2^{2n}}x^{2n+1},\quad 0<x<1,\tag{5}
then integrating termwise using

\begin{align}
&\int_{0}^{1}x^{2n+1}\:\text{arctanh} \:x \:{\rm d}x\\
&=\frac1{2(n+1)(2n+1)}+\frac{\ln2}{2(n+1)}+\frac1{4(n+1)}\left(\gamma+\psi \left(n+\frac12 \right) \right)\tag{6}
\end{align}

to get

\begin{align}
\int_0^1 \frac{\text{arctanh}\: x}{\tan \left( \frac{\pi}2x\right)}\:{\rm d}x &=\frac{\pi }{4}+\frac{2}{\pi }(1-\ln 2)\ln\left(\frac{\pi }{2}\right)\\\\&-\frac{1}{\pi }\sum_{n=0}^{\infty}\frac{\zeta(2n+2)}{(2n+1)2^{2n}}-\frac{1}{\pi }\sum_{n=0}^{\infty}\frac{\zeta(2n+2)\left(\psi\left(n+\frac12\right)+\gamma\right)}{(n+1)2^{2n+2}}. \tag7
\end{align}

We are left with two non trivial series to evaluate.

We prove that each series may be evaluated using the poly-Stieltjes constants.

One may write

\require{cancel}
\begin{align}
\sum_{n=0}^{\infty}\frac{\zeta(2n+2)}{(2n+1)2^{2n}}&=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac1{k^{2n+2}}\frac1{(2n+1)2^{2n}}\\
&=4\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac1{(2n+1)}\frac1{(2k)^{2n+2}}\\
&=\sum_{k=1}^{\infty}\frac1k\left(\log \left(1 + \frac1{2k}\right)-\log \left(1 - \frac1{2k}\right)\right)\\
&=\sum_{k=1}^{\infty}\frac1k\left(\log \left(k + \frac12\right)-\log \left(k - \frac12\right)\right)\\
&=\gamma_1\Big({\small\frac12,0}\Big)-\gamma_1\Big({\small-\frac12,0}\Big) \tag{8}
\end{align}
using Theorem 2 here.

To evaluate the last series on the right hand side of (7), one may check with some algebra that, for any complex number z satisfying |z|<1, the following identity holds true:

\begin{align}
&\sum_{n=0}^{\infty}\frac{\psi\left(n+\frac12\right)+\gamma}{n+1}z^{2n+2}\\
&=2z\log\left(\frac{1-z}{1+z} \right)-2\left(1- \ln 2 \right)\log (1-z^2)+\frac12\log^2\left(\frac{1-z}{1+z} \right). \tag9
\end{align}

Then

\require{cancel}
\begin{align}
&\sum_{n=0}^{\infty}\frac{\zeta(2n+2)\left(\psi\left(n+\frac12\right)+\gamma\right)}{(n+1)2^{2n+2}} \\
&=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac1{k^{2n+2}}\frac{\psi\left(n+\frac12\right)+\gamma}{(n+1)2^{2n+2}}\\
&=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac{\psi\left(n+\frac12\right)+\gamma}{n+1}\frac1{(2k)^{2n+2}}\\
&=\sum_{k=1}^{\infty}\frac1k \log \left(\frac{1-\frac1{2k}}{1+\frac1{2k}}\right)-2\left(1- \ln 2 \right)\sum_{k=1}^{\infty}\log \left(1 - \frac1{4k^2}\right)+\frac12\sum_{k=1}^{\infty}\log^2 \left(\frac{1-\frac1{2k}}{1+\frac1{2k}}\right)\\
&=\sum_{k=1}^{\infty}\frac1k \log \left(\frac{k-\frac12}{k+\frac12}\right)+2\left(1- \ln 2 \right)\ln\left(\frac{\pi }{2}\right)+\frac12\sum_{k=1}^{\infty}\log^2\!\left(\! \frac{2k-1 }{ 2k+1}\!\right)\\
&=\gamma_1\Big({\small-\frac12,0}\Big)-\gamma_1\Big({\small\frac12,0}\Big)+2\left(1- \ln 2 \right)\ln\left(\frac{\pi }{2}\right)+\frac12\sum_{k=1}^{\infty}\log^2\!\left(\! \frac{2k-1 }{ 2k+1}\!\right).\tag{10}
\end{align}

Inserting (10) and (8) into (7) gives the announced result (4). \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box

We deduce the following closed form.

Proposition 2. We have


\begin{align}
\int_0^1 \frac{\text{arctanh} x}{\tan \left( \frac{\pi}2x\right)}\:{\rm d}x=\frac\pi4+\frac2\pi\ln^2 2+\frac1\pi\ln 2\ln \pi+\frac1\pi\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big). \tag{11}
\end{align}

Proof. One may observe that

\begin{align}
\zeta\left(s,\frac12 \right) & = \left(2^s-1 \right)\zeta(s) \tag{12}\\
\zeta\left(s,\frac32 \right) & = \left(2^s-1 \right)\zeta(s)-2^s, \tag{13}
\end{align}
and recalling that \zeta'(0)=-\frac12 \ln (2 \pi), one may obtain

\begin{align}
\zeta''\left(0,\frac12 \right) & = -\frac32 \ln^2 2 - \ln 2 \ln \pi \tag{14}\\
\zeta''\left(0,\frac32 \right) & = -\frac52 \ln^2 2 - \ln 2 \ln \pi. \tag{15}
\end{align}

From (4) and (3), we have

\require{cancel}
\begin{align}
\int_0^1 \frac{\text{arctanh}\: x}{\tan \left( \frac{\pi}2x\right)}\:{\rm d}x
&=\frac\pi4-\frac1{2\pi}\sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{2n-1 }{ 2n+1}\!\right)\\\\
&=\frac\pi4-\frac1{2\pi}\sum_{n=1}^{+\infty}\log^2\!\left(\! \frac{n-\frac12 }{ n+\frac12}\!\right)\\\\
&=\frac\pi4-\frac1{2\pi}\left( \zeta''\left(0,-\frac12+1 \right)+ \zeta''\left(0,\frac12+1 \right)-2\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big)\right)\\\\
&=\frac\pi4-\frac1{2\pi}\left( \zeta''\left(0,\frac12 \right)+ \zeta''\left(0,\frac32 \right)-2\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big)\right),
\end{align}
by appealing to (14) and (15), we get (11). \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box

By combining Proposition 2 and pisco125's derivation we obtain the following new closed forms.

Proposition 3. We have


\begin{align}
\int_{0}^{\infty} {\ln (1+x^2)\over {e^{2\pi x}+1}}\:{\rm d}x =&\:\frac\pi4+\frac1{2\pi}\ln^2 2+\frac1\pi\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big) \tag{16}
\\\\
\sum_{k=1}^{\infty} {(-1)^k\over k} \text{Ci} (2k\pi)=&\:\frac{\pi^2}4+\frac12\ln^2 2+\:\zeta^{1,1}\Big(0,0 \:\Bigr\rvert {\small-\frac12,\frac12}\Big) \tag{17}
\end{align}

where \text{Ci} (\cdot) is the cosine integral and where \displaystyle \zeta(\cdot,\cdot\mid a,b) is the poly-Hurwitz zeta function.

Attribution
Source : Link , Question Author : Lucian , Answer Author : Community

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