# Closed form of $\int_{0}^{\infty} \frac{\tanh(x)\,\tanh(2x)}{x^2}\;dx$

I have homework to evaluate this integral $$I=\int_{0}^{\infty} \frac{\tanh(x)\,\tanh(2x)}{x^2}\;dx$$

Here is what I have done so far. I tried integration by parts using $u=\tanh(x)\,\tanh(2x)$ and $dv=\frac{dx}{x^2}$, I got \begin{align}\int_{0}^{\infty} \frac{\tanh(x)\,\tanh(2x)}{x^2}\;dx&=-\left.\frac{\tanh(x)\,\tanh(2x)}{x}\right|_{0}^{\infty}+2\int_{0}^{\infty}\frac{\tanh(2x)\,\text{sech}(2x)}{x}\;dx\\&=2\int_{0}^{\infty}\frac{\tanh(2x)\,\text{sech}(2x)}{x}\;dx\end{align} At this part I'm stuck. I'm thinking of using Frullani's integral but I'm having trouble to find a relation as such $\tanh(2x)\,\text{sech}(2x)=f(ax)-f(bx)$.

I also tried using differentiation under integral sign by considering $$I(a,b)=\int_{0}^{\infty} \frac{\tanh(ax)\,\tanh(bx)}{x^2}\;dx$$ then $$\frac{dI}{da}=\int_{0}^{\infty} \frac{\text{sech}^2(ax)\,\tanh(bx)}{x}\;dx=\int_{0}^{\infty} \frac{\tanh(bx)-\tanh^2(ax)\tanh(bx)}{x}\;dx$$ Again I tried to use Frullani's integral but I'm having trouble to find the sufficient $f(x)$. Integrating again with respect to $b$, I got $$\frac{d^2I}{da\;db}=\int_{0}^{\infty} \text{sech}^2(ax)\text{sech}^2(bx)\;dx$$ It's obviously a dead end to me. At this rate my friends and I contacted my professor to confirm whether the integral can be evaluated in terms of elementary functions or not because W|A cannot find it (I know that W|A cannot do everything). He only said, "Sure! The answer is only 3 characters" and then he left us. Assuming he is right, so $I$ must have a nice closed form, but I'm unable to find it.

Would you help me? Any help would be appreciated. Thanks in advance.

• just as comment: The substitution $x=arctanh(y)$ together with $\tanh(2x)=\frac{2 \tanh(x)}{1+\tanh(x)^2}$ is maybe worth to give it a try. – tired Nov 11 '14 at 17:47
• @tired I did, but it got complicated – Venus Nov 11 '14 at 17:51
• at which point? – tired Nov 11 '14 at 17:54
• @tired At the denominator: $\text{arctanh}\,(y)^2$ – Venus Nov 11 '14 at 17:55
• i tried a litte bit, and indeed things didn't get really easier...:-( – tired Nov 11 '14 at 19:04

Your professor is right. Note that $$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}, \tanh(2x)=\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}=\frac{(e^x-e^{-x})(e^x+e^{-x})}{e^{2x}+e^{-2x}}$$ and hence \begin{eqnarray*} \int_0^\infty\frac{\tanh(x)\tanh(2x)}{x^2}dx&=&\int_0^\infty\frac{(e^{x}-e^{-x})^2}{x^2(e^{2x}+e^{-2x})}dx\\ &=&\int_0^\infty\frac{e^{2x}-2+e^{-2x}}{x^2(e^{2x}+e^{-2x})}dx. \end{eqnarray*} Now define $$I(a)=\int_0^\infty\frac{e^{ax}-2+e^{-ax}}{x^2(e^{2x}+e^{-2x})}dx$$ to get \begin{eqnarray} I''(a)&=&\int_0^\infty\frac{e^{(-a-2)x}+e^{(-a+2)x}}{1+e^{-4x}}dx,\\ &=&\int_0^\infty\sum_{n=0}^\infty(-1)^n(e^{(-a-2)x}+e^{(-a+2)x})e^{-4nx}dx\\ &=&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)\\ &=&\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right). \end{eqnarray} So \begin{eqnarray} I'(a)&=&\int_0^a\frac{\pi}{4}\sec\left(\frac{\pi t}{4}\right)dt\\ &=&\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)-\ln\cos\left(\frac{\pi a}{4}\right) \end{eqnarray} and hence \begin{eqnarray} I(2)&=&\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da-\int_0^2\ln\cos\left(\frac{\pi a}{4}\right)da. \end{eqnarray} Note $$\int_0^2\ln\cos\left(\frac{\pi a}{4}\right)da=-2\ln2$$ from Evaluating $\int_0^{\large\frac{\pi}{4}} \log\left( \cos x\right) \, \mathrm{d}x$ and it should not be hard to get $$\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{8G}{\pi}-2\ln 2$$ and thus $$I(2)=\frac{8G}{\pi}.$$

$\bf{Update}$ 1: Let us first work on $\sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)=\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right)$. In fact \begin{eqnarray*} &&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)\\ &=&\sum_{n=0}^\infty\left(\frac{1}{8n-a+2}-\frac{1}{8n-a+6}\right)+\sum_{n=0}^\infty\left(\frac{1}{8n+a+2}-\frac{1}{8n+a+6}\right)\\ &=&\sum_{n=0}^\infty\frac{4}{(8n-a+2)(8n-a+6)}+\sum_{n=0}^\infty\frac{4}{(8n+a+2)(8n+a+6)}\\ &=&\sum_{n=-\infty}^\infty\frac{4}{(8n-a+2)(8n-a+6)}=\sum_{n=-\infty}^\infty\frac{4}{(8n-a+4)^2-2^2}\\ &=&\frac{1}{16}\sum_{n=-\infty}^\infty\frac{1}{(n+\frac{4-a}{8})^2-(\frac{1}{4})^2}. \end{eqnarray*} Now using a result from Closed form for $\sum_{n=-\infty}^\infty \frac{1}{(z+n)^2+a^2}$ for $a=\frac{i}{4}$ and $z=\frac{4-a}{8}$ and after some basic calculation we can get this result. Also see An alternative proof for sum of alternating series evaluates to $\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right)$ for a short proof.

$\bf{Update}$ 2: We work on $\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{8G}{\pi}-2\ln 2$. In fact \begin{eqnarray} \int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{4}{\pi}\int_0^{\pi/2}\ln\left(1+\sin(a)\right)da=\frac{4}{\pi}\int_0^{\pi/2}\ln(1+\cos(a))da. \end{eqnarray} Using $2\cos^2\frac{a}{2}=1+\cos a$ and a result $G=\int_0^{\pi/4}\ln(\cos(t))dt$ from http://en.wikipedia.org/wiki/Catalan%27s_constant, it is easy to obtain $$\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da= \frac{8G}{\pi}-2\ln 2.$$

• Amazing! This very simple! You've posted since 19 hrs ago but why I didn't notice it earlier? +1 – Venus Nov 12 '14 at 18:19
• Anyway, could you elaborate this part $$\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{8G}{\pi}-2\ln 2$$Thanks... – Venus Nov 12 '14 at 18:40
• Never mind, I can handle it. Thanks for your simple & nice solution. :-) – Venus Nov 12 '14 at 18:43
• Sorry, last question. How I can proof this series $$\sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)=\frac{ \pi }{4}\sec\left(\frac{a \pi }{4}\right)$$ Do you have any references? Or did you use residue method? Can you elaborate using real analysis method? Thanks... – Venus Nov 12 '14 at 18:49
• @Venus: Let $u=\dfrac\pi4 a$ and $\sin u=\cos\bigg(\dfrac\pi2-u\bigg)$ and $\dfrac{1+\cos t}2=\cos^2\dfrac t2$ – Lucian Nov 12 '14 at 22:54

Let $f(x) = \tanh(x)\tanh(2x)$. Rewrite the integral at hand as $$I = \int_0^\infty \frac{f(x)}{x^2}dx = \frac12 \int_{-\infty}^{\infty} \frac{f(x)}{x^2} dx = \frac12 \lim_{k\to\infty} \int_{-k\pi}^{k\pi} \frac{f(x)}{x^2} dx\\$$ For each positive integer $k$, consider the rectangular region $$D_k = \big\{ u + iv \in \mathbb{C} : |u| \le k\pi, 0 \le v \le k\pi \big\}$$ and the contour integral over its boundary:

$$C_k \stackrel{def}{=} \int_{\partial D_k} \frac{f(z)}{z^2} dz = \left( \int_{-k\pi}^{k\pi} + \int_{k\pi}^{k\pi(1+i)} + \int_{k\pi(1+i)}^{k\pi(-1+i)} + \int_{k\pi(-1+i)}^{-k\pi} \right) \frac{f(z)}{z^2} dz$$

$C_k$ split into 4 pieces, one for each edges of the $D_k$. It is not hard to show as $k \to \infty$, the contribution from the three edges (the top, left and bottom) goes to zero. This implies

$$I = \frac12 \lim_{k\to\infty} \int_{\partial D_k} \frac{f(z)}{z^2} dz\tag{*1}$$

Let $\phi = e^{2z}$, we have

\begin{align}f(z) &= \tanh(z) \tanh(2z) = \left(\frac{\phi - 1}{\phi+1}\right)\left(\frac{\phi^2-1}{\phi^2+1}\right) = \frac{(\phi-1)^2}{\phi^2+1} = 1 - \frac{2\phi}{\phi^2+1} \\ &= 1 - \frac{1}{\cosh 2z}\end{align}

Recall the well known? expansion of $\cot z$

$$\cot z = \sum_{n=-\infty}^\infty \frac{1}{z - n\pi}$$

We have

$$\begin{array}{rrl} & \frac{1}{\sin z} &= \frac{1+\cos z}{\sin z} - \frac{\cos z}{\sin z} = \cot\frac{z}{2} - \cot z = \sum_{n=-\infty}^\infty \frac{(-1)^n}{z - n\pi}\\ \implies & \frac{1}{\cos z} &= -\frac{1}{\sin(z - \frac{\pi}{2})} = \sum_{n=-\infty}^\infty \frac{(-1)^{n-1}}{z - (n+\frac12)\pi}\\ \implies & \frac{1}{\cosh z} &= \frac{1}{\cos(-i z)} = -i \sum_{n=-\infty}^\infty \frac{(-1)^n}{z - (n+\frac12)\pi i}\\ \implies & f(z) & = 1 + \frac{i}{2} \sum_{n=-\infty}^\infty \frac{(-1)^n}{z - (n+\frac12)\frac{\pi}{2} i}\tag{*2} \end{array}$$

Inside $D_k$, $f(z)$ has poles at $(n + \frac12)\frac{\pi}{2} i$ for each $n \in \mathbb{N}$. We can evaluate the contour integral in $(*1)$ by summing the residues at these poles. As a result,

$$I = \frac12 (2\pi i)(\frac{i}{2})\sum_{n=0}^\infty \frac{(-1)^n}{((n+\frac12)\frac{\pi}{2}i)^2} = \frac{8K}{\pi}$$

where $\;\displaystyle K = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}\;$ is the Catalan's constant.

Update

If one want to minimize the use of complex analysis, an alternate starting point is the expansion of $f(z)$ in $(*2)$. Let $\alpha_n = (n+\frac12)\frac{\pi}{2}$ and notice $\alpha_{-1-n} = -\alpha_n$, we have

$$f(x) = 1 + \frac{i}{2}\sum_{n=0}^\infty (-1)^n\left(\frac{1}{x -\alpha_n i} - \frac{1}{x + \alpha_n i}\right) = 1 - \sum_{n=0}^\infty \frac{(-1)^n \alpha_n}{x^2 + \alpha_n^2}$$ Since $f(0) = 0$, we have

$$\frac{f(x)}{x^2} = \frac{f(x)-f(0)}{x^2} = \frac{1}{x^2} \sum_{n=0}^\infty (-1)^n \alpha_n\left( \frac{1}{\alpha_n^2} - \frac{1}{x^2 + \alpha_n^2}\right) = \sum_{n=0}^\infty\frac{(-1)^n}{\alpha_n(x^2+\alpha_n^2)}$$ If you look at the terms of this series in units of pair, we find the value of any pair is always non-negative. This means we can integrate this series pair by pair and get:

$$\int_0^\infty \frac{f(x)}{x^2}dx = \sum_{n=0}^\infty \frac{(-1)^n}{\alpha_n}\int_0^\infty \frac{dx}{x^2+\alpha_n^2} = \frac{\pi}{2}\sum_{n=0}^\infty \frac{(-1)^n}{\alpha_n^2} = \frac{8}{\pi}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = \frac{8K}{\pi}$$ The same result we obtained using contour integrals.

• Thank you for your answer. I hope someone here will give me an answer with a real analysis method. +1. :-) – Venus Nov 11 '14 at 19:46
• Nice answer. +1 – Start wearing purple Nov 11 '14 at 20:59
• There is a typo in the first line: The second integral 'start' at $-\infty$. Otherwise, it is quite fine => +1. – Felix Marin Nov 11 '14 at 23:26
• Could this problem be tackled by similar means ? – Lucian Nov 12 '14 at 1:51
• @FelixMarin thanks, I fixed that. – achille hui Nov 12 '14 at 3:43

Here is a method that avoids complex analysis. That being said however, I think achille hui's approach is far superior as this answer uses too much (unnecessarily) heavy machinery.

Let $x\mapsto-\ln{x}$. We get \begin{align} \mathcal{I} =\int^1_0\frac{1-x^2}{1+x^2}\frac{1-x^4}{1+x^4}\frac{{\rm d}x}{x\ln^2{x}} =\int^1_0\left(\frac{1}{x}-\frac{2x}{1+x^4}\right)\frac{{\rm d}x}{\ln^2{x}} \end{align} Define $$\mathcal{I}(\alpha)=\int^1_0\left(\frac{1}{x}-\frac{2x}{1+x^4}\right)\frac{x^{\alpha}{\rm d}x}{\ln^2{x}}$$ Differentiating twice, one obtains \begin{align} \mathcal{I}''(\alpha) =&\frac{1}{\alpha}-2\sum^\infty_{n=0}(-1)^n\int^1_0x^{4n+\alpha+1}{\rm d}x\\ =&\frac{1}{\alpha}-2\sum^\infty_{n=0}\frac{(-1)^n}{4n+\alpha+2}\\ =&\frac{1}{\alpha}-\frac{1}{4}\sum^\infty_{n=0}\frac{1}{n+\frac{\alpha+2}{8}}+\frac{1}{4}\sum^\infty_{n=0}\frac{1}{n+\frac{\alpha+6}{8}}\\ =&\frac{1}{\alpha}+\frac{1}{4}\psi_0\left(\frac{\alpha+2}{8}\right)-\frac{1}{4}\psi_0\left(\frac{\alpha+6}{8}\right) \end{align} Integrate back once. \begin{align} \mathcal{I}'(\alpha) =&\ln{\alpha}+2\ln{\Gamma\left(\frac{\alpha+2}{8}\right)}-2\ln{\Gamma\left(\frac{\alpha+6}{8}\right)}+C_1 \end{align} The constant of integration is, by Striling's asymptotic series, $-3\ln{2}$. Now, recall the fact that $$\int\ln{\Gamma(x)}\ {\rm d}x=\frac{x(1-x)}{2}+\frac{x}{2}\ln(2\pi)+x\ln{\Gamma(x)}-\ln{G(x+1)}+C_2$$ You may find a proof in this answer. Hence, integrating once again, we obtain \begin{align} \mathcal{I}(\alpha) =&\alpha\ln{\alpha}-\alpha-\frac{(\alpha+2)(\alpha-6)}{8}+(\alpha+2)\ln(2\pi)+2(\alpha+2)\ln{\Gamma\left(\frac{\alpha+2}{8}\right)}\\ &-16\ln{G\left(\frac{\alpha+10}{8}\right)}+\frac{(\alpha-2)(\alpha+6)}{8}-(\alpha+6)\ln(2\pi)\\ &-2(\alpha+6)\ln{\Gamma\left(\frac{\alpha+6}{8}\right)}+16\ln{G\left(\frac{\alpha+14}{8}\right)}-3\alpha\ln{2}+C_2\\ =&C_2-4\ln(2\pi)-3\alpha\ln{2}+\alpha\ln{\alpha}+2(\alpha+2)\ln{\Gamma\left(\frac{\alpha+2}{8}\right)}\\ &-2(\alpha+6)\ln{\Gamma\left(\frac{\alpha+6}{8}\right)}-16\ln{G\left(\frac{\alpha+10}{8}\right)}+16\ln{G\left(\frac{\alpha+14}{8}\right)} \end{align} One may use the asymptotic series of the Barnes G (derived from the log-gamma integral and Stirling's) to check that $C_2=4\ln(2\pi)$. Letting $\alpha\to 0$, \begin{align} \mathcal{I} =&\mathcal{I}(0)\\ =&4\ln{\Gamma\left(\frac{1}{4}\right)}-12\ln{\Gamma\left(\frac{3}{4}\right)}-16\ln{G\left(\frac{5}{4}\right)}+16\ln{G\left(\frac{7}{4}\right)} \end{align} It remains to simplify this result. Using the Gamma reflection formula $\Gamma(z)\Gamma(1-z)=\pi\csc(\pi z)$, as well as the Barnes G functional equation $G(z+1)=G(z)\Gamma(z)$, we get $$\mathcal{I}=-6\ln{2}-12\ln{\pi}-16\ln\left(\frac{G\left(\frac{1}{4}\right)}{G\left(\frac{7}{4}\right)}\right)$$ Now recall that $G(z+1)$ has the infinite product representation $$G(z+1)=\left(\sqrt{2\pi}\right)^{z}e^{-\frac{(1+\gamma)z^2+z}{2}}\prod^\infty_{k=1}e^{\frac{z^2}{2k}-z}\left(1+\frac{z}{k}\right)^k$$ Taking the logarithm, \begin{align} \ln{G(z+1)} =&-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}+\sum^\infty_{k=1}\left\{k\ln\left(1+\frac{z}{k}\right)+\frac{z^2}{2k}-z\right\}\\ =&-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}+\sum^\infty_{k=1}\left\{\frac{z^2}{2k}-z+\sum^\infty_{m=1}\frac{(-1)^{m-1}}{m}\frac{z^m}{k^{m-1}}\right\}\\ =&-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}+\sum^\infty_{k=1}\sum^\infty_{m=3}\frac{(-1)^{m-1}}{m}\frac{z^m}{k^{m-1}}\\ =&-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}+\sum^\infty_{m=3}\frac{(-1)^{m-1}\zeta(m-1)}{m}z^m\\ =&-\frac{z}{2}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}+\sum^\infty_{m=2}\frac{(-1)^{m}\zeta(m)}{m+1}z^{m+1}\\ \end{align} Letting $z\mapsto -z$, $$\ln{G(1-z)}=\frac{z}{2}-\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}-\sum^\infty_{m=2}\frac{\zeta(m)}{m+1}z^{m+1}$$ It follows that \begin{align} \ln{\frac{G(1-z)}{G(1+z)}} =&z-z\ln(2\pi)-2\sum^\infty_{m=1}\frac{\zeta(2m)}{2m+1}z^{2m+1}\\ =&-z\ln(2\pi)+\int^z_0\pi x\cot(\pi x)\ {\rm d}x \end{align} Let $z=\frac{3}{4}$. \begin{align} \ln\left(\frac{G\left(\frac{1}{4}\right)}{G\left(\frac{7}{4}\right)}\right) =&-\frac{3}{4}\ln(2\pi)+2\sum^\infty_{n=1}\int^\frac{3}{4}_0\pi x\sin(2n\pi x)\ {\rm d}x\\ =&-\frac{3}{4}\ln(2\pi)+\frac{1}{2\pi}\sum^\infty_{n=1}\frac{\sin(3n\pi/2)}{n^2}-\frac{3}{4}\sum^\infty_{n=1}\frac{\cos(3n\pi/2)}{n}\\ =&-\frac{3}{4}\ln(2\pi)-\frac{1}{2\pi}\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}+\frac{3}{8}\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\\ =&-\frac{3}{8}\ln{2}-\frac{3}{4}\ln{\pi}-\frac{\mathbf{G}}{2\pi} \end{align} Finally, $$\mathcal{I}=-6\ln{2}-12\ln{\pi}-16\left(-\frac{3}{8}\ln{2}-\frac{3}{4}\ln{\pi}-\frac{\mathbf{G}}{2\pi}\right)=\Large{\frac{8\mathbf{G}}{\pi}}$$

• Thank you so much for the real method approach. It's ok if it's too complicated, I haven't learned contour integration yet. +1 – Venus Nov 12 '14 at 10:27
• @Venus I am glad you found this answer useful in some way. Thanks. – M.N.C.E. Nov 12 '14 at 12:22
• +1 because you takes care of writing the result at the very end. Usually we have to guess what was the result. Good. – Felix Marin Dec 21 '14 at 4:25

Write $$I=\frac{1}{2}\int^{\infty}_{-\infty}\frac{\tanh(x)\tanh(2x)}{x^2}dx=\int^{\infty}_{-\infty}\frac{f(x)}{\cosh(\pi x)}dx,$$ where $f(x)=\frac{2(-1+\cosh(\pi x))}{\pi^2 x^2}$.

Lemma.

If the function $f(z)$ is analytic in $Im(z)>0$ and continuous at $Im(z)\geq 0$. It also satisfies $$|f(z)|\leq C (1+|z|)^{N} e^{b |Re(z)|}\textrm{, }0\leq b<\pi$$ for all $z$ with $Im(z)\geq 0$ (note: here is $b=\pi$, but $N=-2$), then $$\int^{\infty}_{-\infty}\frac{f(t)}{\cosh(\pi t)}e^{ita}dt=2\sum^{\infty}_{k=0}(-1)^k f(i(k+1/2))$$ Proof. The function $g(z)=\frac{f(z)}{\cosh(\pi z)}$ is meromorphic in $Im(z)>0$ and has poles at $z_k=i(k+1/2)$ with $$Res(g;i(k+1/2))=f(i(k+1/2))\frac{(-1)^k}{\pi i}$$ a in the real line and meromorphic at $Im(z)>0$. Hence if $m$ is natural number, $R=m$ and $\gamma_R$ is the upper semicircle of diameter $[-R,R]$, then $$\frac{1}{2\pi i}\int_{\gamma_R}g(z)dz=\sum^{\infty}_{k=0}\frac{(-1)^k}{\pi i}f(i(k+1/2))$$ For we have $|\cos(\pi R e^{i \theta})|\geq c e^{\pi |\cos(\theta)|}$ and $$\left|\int^{\pi}_{0}\frac{f(Re^{i\theta})}{\cosh(\pi R e^{i\theta})}iRe^{i\theta}\right|\leq C \left|\int^{\pi}_{0}(1+R)^{-2}Re^{(b-\pi)R|\cos(\theta)|}d\theta\right|\leq C$$ $$\leq C \int^{\pi}_{0}(1+R)^{-1}d\theta\rightarrow 0\textrm{, when }R\rightarrow \infty.$$ $$QED$$

For the case $$f(x)=\frac{2(-1+\cosh(\pi x))}{\pi^2 x^2}$$
we have $$I=\frac{\pi}{2}2\sum^{\infty}_{k=0}(-1)^k\frac{8(1+\sin(k\pi))}{\pi^2(2k+1)^2}=\frac{8C}{\pi}$$