\int x/(1+x^2)*1/(tanh(\pi*x/2))dx

Today´s post we will prove the challenging result posted by @integralsbot here:

$\int_{0}^{\infty} \frac{x}{\left(1+x^{2}\right)^{2}} \frac{d x}{\tanh \frac{\pi x}{2}}=\frac{\pi^{2}}{8}-\frac{1}{2}$

To this end, we will first establish the following three results:

$\int_0^\infty\frac{e^{-st}}{\cos(bx)+\cosh(bt)}\,dt&=\frac{2}{\sin(bx)}\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin\left(bkx \right)}{s+bk}\\ & \\ & \\ \int_0^\infty\frac{s}{(s^2+x^2)}\frac{\sinh(bx)}{\sinh(cx)}\,dx&=\pi\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin\left(\frac{\pi kb}{c} \right)}{c s+\pi k}\\ & \\ & \\ \int_0^\infty\frac{x}{(s^2+x^2)}\frac{\cosh(bx)}{\sinh(cx)}\,dx&=\frac{\pi}{2cs}+\pi\sum_{k=1}^\infty \frac{(-1)^{k}\cos\left(\frac{\pi kb}{c} \right) }{c s+\pi k}$

We begin proving two Lemmas:

Lemma 1:

$\frac{1}{\cos(bx)+\cosh(bt)}=\frac{2}{\sin(bx)}\sum_{k=1}^\infty (-1)^{k-1} e^{-bkt}\sin\left(bkx \right)$

proof:

$\begin{aligned} \frac{1}{\cos(bx)+\cosh(bt)}&=\frac{2}{e^{ibx}+e^{-ibx}+e^{bt}+e^{-bt}}\\ &=\frac{2}{e^{bt}\left(1+e^{-2bt}+e^{-bt+ibx}+e^{-bt-ibx}\right)}\\ &=\frac{2e^{-bt}}{\left(1+e^{-bt+ibx}\right)\left(1+e^{-bt-ibx}\right)}\\ &=\frac{2e^{-bt}}{2ie^{-bt}\sin(bx)}\left(\frac{1}{1+e^{-bt-ibx}}-\frac{1}{1+e^{-bt+ibx}} \right)\\ &=\frac{2e^{-bt}}{2ie^{-bt}\sin(bx)}\left(\sum_{k=0}^\infty (-1)^k e^{-bkt-ibkx}-\sum_{k=0}^\infty (-1)^k e^{-bkt+ibkx} \right)\\ &=\frac{2}{2i\sin(bx)}\sum_{k=0}^\infty (-1)^{k+1} e^{-bkt}\left(e^{ibkx}-e^{-ibkx} \right)\\ &=\frac{2}{\sin(bx)}\sum_{k=0}^\infty (-1)^{k+1} e^{-bkt}\sin\left(bkx \right) \\ &=\frac{2}{\sin(bx)}\sum_{k=1}^\infty (-1)^{k-1} e^{-bkt}\sin\left(bkx \right) \qquad \blacksquare\\ \end{aligned}$

Lemma 2:

$\sum_{k=1}^\infty (-1)^{k-1} \cos\left(xk \right)=\frac12$

Proof:

$\begin{aligned} \sum_{k=1}^\infty (-1)^{k-1} \cos\left(xk \right)&=\frac12\sum_{k=1}^\infty (-1)^{k-1} \left(e^{xk}+e^{-xk} \right)\\ &=\frac12\left(\sum_{k=1}^\infty (-1)^{k-1} e^{xk}+\sum_{k=1}^\infty (-1)^{k-1}e^{-xk} \right)\\ &=\frac12\left(\frac{1}{1+e^{xk}}+\frac{1}{1+e^{-xk}}\right)\\ &=\frac12\left(\frac{2+e^{xk}+e^{-xk}}{\left(1+e^{xk}\right)\left(1+e^{-xk}\right)}\right)\\ &=\frac12\left(\frac{2+e^{xk}+e^{-xk}}{2+e^{xk}+e^{-xk}}\right)\\ &=\frac12 \qquad \blacksquare\\ \end{aligned}$

Now, let´s evaluate the first integral

$\begin{aligned} I&=\int_0^\infty\frac{e^{-st}}{\cos(bx)+\cosh(bt)}\,dt\\ &=\frac{2}{\sin(bx)}\sum_{k=1}^\infty (-1)^{k-1} \sin\left(bkx \right)\int_0^\infty e^{-(s+bk)t}\,dt \qquad (*)\\ &=\frac{2}{\sin(bx)}\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin\left(bkx \right)}{s+bk} \qquad \blacksquare \end{aligned}$

where in (*) we applied Lemma 1

Now recall the Laplace transform of the cosine function, a proof can be found in this post:

$\int_0^\infty e^{-st}\cos(xt)\,dt=\frac{s}{(s^2+x^2)}$

Then

$\begin{aligned} I(s)&=\int_0^\infty\frac{s}{(s^2+x^2)}\frac{\sinh(bx)}{\sinh(cx)}\,dx\\ &=\int_0^\infty\left(\int_0^\infty e^{-st}\cos(xt)\,dt \right)\frac{\sinh(bx)}{\sinh(cx)}\,dx\\ &=\int_0^\infty e^{-st} \int_0^\infty \frac{\cos(xt)\sinh(bx)}{\sinh(cx)}\,dx\,dt\\ &=\int_0^\infty e^{-st} \left(\frac{\pi}{2 c} \cdot \frac{\sin\left(\frac{b \pi}{c}\right)}{\cosh\left(\frac{t \pi}{c}\right)+\cos\left(\frac{b \pi}{c}\right)} \right)\,dt \qquad (**)\\ &=\frac{\pi}{2 c} \sin\left(\frac{b \pi}{c}\right)\int_0^\infty \frac{e^{-st}}{\cosh\left(\frac{t \pi}{c}\right)+\cos\left(\frac{b \pi}{c}\right)} \\ &=\frac{\pi \sin\left(\frac{b \pi}{c}\right)}{c\sin\left(\frac{b \pi}{c}\right)}\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin\left(\frac{\pi kb}{c} \right)}{s+\frac{\pi k}{c}}\\ &=\pi\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin\left(\frac{\pi kb}{c} \right)}{c s+\pi k} \qquad \blacksquare \end{aligned}$

In (**) we used the following result proved in this post

$\int_0^\infty \frac{\cos(xt)\sinh(bx)}{\sinh(cx)}\,dx =\frac{\pi}{2 c} \cdot \frac{\sin\left(\frac{b \pi}{c}\right)}{\cosh\left(\frac{t \pi}{c}\right)+\cos\left(\frac{b \pi}{c}\right)}$

Therefore, we have that

$\int_0^\infty\frac{s}{(s^2+x^2)}\frac{\sinh(bx)}{\sinh(cx)}\,dx=\pi\sum_{k=1}^\infty \frac{(-1)^{k-1} \sin\left(\frac{\pi kb}{c} \right)}{c s+\pi k}$ (1)

Differentiating (1) w.r. to gives us

$\begin{aligned} \int_0^\infty\frac{x}{(s^2+x^2)}\frac{\cosh(bx)}{\sinh(cx)}\,dx&=\frac{\pi}{c s }\sum_{k=1}^\infty \frac{(-1)^{k-1} \pi k }{c s+\pi k}\cos\left(\frac{\pi kb}{c} \right)\\ &=\frac{\pi}{cs}\sum_{k=1}^\infty \frac{(-1)^{k-1} \left(cs+\pi k-cs\right) }{c s+\pi k}\cos\left(\frac{\pi kb}{c} \right)\\ &=\frac{\pi}{cs}\sum_{k=1}^\infty (-1)^{k-1} \cos\left(\frac{\pi kb}{c} \right)-\pi\sum_{k=1}^\infty \frac{(-1)^{k-1}\cos\left(\frac{\pi kb}{c} \right) }{c s+\pi k}\\ &=\frac{\pi}{2cs}+\pi\sum_{k=1}^\infty \frac{(-1)^{k}\cos\left(\frac{\pi kb}{c} \right) }{c s+\pi k} \qquad \blacksquare\\ \end{aligned}$

Where in the last line we used Lemma 2, then

$\int_0^\infty\frac{x}{(s^2+x^2)}\frac{\cosh(bx)}{\sinh(cx)}\,dx=\frac{\pi}{2cs}+\pi\sum_{k=1}^\infty \frac{(-1)^{k}\cos\left(\frac{\pi kb}{c} \right) }{c s+\pi k}$ (2)

If we let b=c in (2) we obtain

$\int_0^\infty\frac{x}{(s^2+x^2)}\frac{1}{\tanh(cx)}\,dx=\frac{\pi}{2cs}+\pi\sum_{k=1}^\infty \frac{1 }{c s+\pi k}$ (3)

Letting $c=\frac{\pi}{2}$ in (3)

$\begin{aligned} \int_0^\infty\frac{x}{(s^2+x^2)}\frac{1}{\tanh\left(\frac{\pi x }{2}\right)}\,dx&=\frac{1}{s}+\sum_{k=1}^\infty \frac{1 }{ \frac{s}{2}+ k}\\ &=\frac{1}{s}+2\sum_{k=1}^\infty \frac{1 }{ s+ 2k}\\ \end{aligned}$ (4)

Now, if we differentiate (4) w.r. to s we obtain

$\int_0^\infty\frac{x}{(s^2+x^2)^2}\frac{1}{\tanh\left(\frac{\pi x }{2}\right)}\,dx &=\frac{1}{2s^3}+\frac{1}{s}\sum_{k=1}^\infty \frac{1 }{ (s+ 2k)^2}\\$ (5)

And now setting s=1 in (5)

$\begin{aligned} \int_0^\infty\frac{x}{(1+x^2)^2}\frac{1}{\tanh\left(\frac{\pi x }{2}\right)}\,dx &=\frac{1}{2}+\sum_{k=1}^\infty \frac{1 }{ ( 2k+1)^2}\\ &=\frac{1}{2}+1-1+\sum_{k=1}^\infty \frac{1 }{ ( 2k+1)^2}\\ &=-\frac{1}{2}+\sum_{k=0}^\infty \frac{1 }{ ( 2k+1)^2}\\ &=\frac{\pi^2}{8}-\frac{1}{2} \qquad \blacksquare \end{aligned}$

Proof of $\sum_{k=0}^\infty \frac{1 }{ ( 2k+1)^2}=\frac{\pi^2}{8}$

First note that

$\sum_{k=0}^\infty \frac{1 }{ ( 2k+1)^2}=\sum_{k=1}^\infty \frac{1 }{ ( 2k-1)^2}$ (A.1)

Than recall the well known result (a simple proof can be found in this post)

$\sum_{k=1}^\infty \frac{1 }{ k^2}=\frac{\pi^2}{6}$ (A.2)

We can split the L.H.S. of (A.2) in it´s even and odd terms, namely

$\begin{aligned} \sum_{k=1}^\infty \frac{1 }{ k^2}&=\sum_{k=1}^\infty \frac{1 }{ (2k)^2}+\sum_{k=1}^\infty \frac{1 }{ ( 2k-1)^2}\\ &=\frac14\sum_{k=1}^\infty \frac{1 }{ k^2}+\sum_{k=1}^\infty \frac{1 }{ ( 2k-1)^2}\\ \end{aligned}$

$\begin{aligned} \sum_{k=1}^\infty \frac{1 }{ ( 2k-1)^2}&=\zeta(2)-\frac14\zeta(2)\\ &=\frac34\zeta(2)\\ &=\frac34 \frac{\pi^2}{6}\\ &=\frac{\pi^2}{8} \qquad \blacksquare \end{aligned}$

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