Help with evaluating the integral $\int_0^{\infty} \frac{x \sinh x}{(1+\cosh^2(x))^2} dx$

Question

I was trying to compute the integral

$$\ I = \int_0^{\infty} \frac{x \sinh(x)}{(1+\cosh^2(x))^2} dx$$

So, I tried defining:

$$\ I(a) = \int_0^{\infty} \frac{ x \sinh(ax)}{(1+\cosh^2(x))^2}dx$$

Then

$$\ I'(a) = \int_0^{\infty} \frac{x\cosh(ax)}{(1+\cosh^2x)^2} dx$$

Then I substituted $\cosh(x) = t$, but I was stuck with this method and could not proceed any further.

So, I tried another approach.

I used the fact that we can express $\sinh(x)$ as:

$$ \sinh(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$

Then swapping the integral and summation signs, we get:

$$\ I = \sum_{n=0}^{\infty}\frac{1}{(2n+1)!} \cdot \int_0^{\infty} \frac{x^{2n+2}}{(1+\cosh^2(x))^2} dx$$

But I am unable to proceed with the evaluation of the inner integral.

Can someone help me evaluate the value of this integral?

Any help is appreciated. Thank you for reading!

Answer 1

Here is a way using the Feynman technique:

$$I=\int_0^{\infty} \frac{x \sinh(x)}{(1+\cosh^2(x))^2} dx=\int_1^\infty\frac{\cosh^{-1}(x)}{(x^2+1)^2}dx$$

Define:

$$I(a)= \int_1^\infty\frac{\cosh^{-1}(ax)}{(x^2+1)^2}dx\implies I’(a)=\int_1^\infty\frac x{\sqrt{(ax)^2-1}(x^2+1)^2}dx$$

It can be shown that:

$$I’(a)=\frac{a^2\pi}{4(a^2+1)^\frac32}\implies I(a)=\frac\pi4\left(\sinh^{-1}(a)-\frac a{\sqrt{a^2+1}}\right)+C$$

Clearly $I(0)=0$ giving $C=0$. Therefore, our integral is $I(1)$:

$$\boxed{\int_0^{\infty} \frac{x \sinh(x)}{(1+\cosh^2(x))^2} dx =\frac\pi8\left(2\sinh^{-1}(1)-\sqrt2\right)}$$

Answer 2

Let's consider $$I_1(a)=\frac12\int_{-\infty}^\infty\frac{x\sinh x}{\cosh^2x+a^2}dx$$ then $$I_2(a)=\frac12\int_{-\infty}^\infty\frac{x\sinh x}{(\cosh^2x+a^2)^2}dx=-\frac1{2a}\frac{d}{da}I_1(a)$$ $$I_3(a)=\frac12\int_{-\infty}^\infty\frac{x\sinh x}{(\cosh^2x+a^2)^3}dx=-\frac1{2\cdot 2\,a}\frac{d}{da}I_2(a),\,\,\text{etc.}$$ To evaluate $I_1(a)$ it is convenient to consider $$J(a, \beta)=\frac12\int_{-\infty}^\infty\frac{e^{\beta x}\sinh x}{\cosh^2x+a^2}dx$$ then $$I_1(a)=\frac{\partial}{\partial\beta}\,\bigg|_{\beta=0}J(a, \beta)$$ Denoting also $a=\sinh b$ $$J(a, \beta)=\frac12\int_{-\infty}^\infty\frac{e^{\beta x}\sinh x}{(\cosh x+i\sinh b)(\cosh x-i\sinh b)}dx$$ Using a rectangular contour in the complex plane $-R\to R\to R+\pi i\to-R+\pi i\to-R$ $$J(a, \beta)\big(1+e^{\pi i\beta}\big)=\frac{2\pi i}2\sum\operatorname{Res}\frac{e^{\beta z}\sinh z}{(\cosh z+i\sinh b)(\cosh z-i\sinh b)}$$ In fact we have only two simple poles inside the contour - at $z_{1,2}=\frac{\pi i}2\pm b$

The evaluation of residues is straightforward and gives $$J(a, \beta)\big(1+e^{\pi i\beta}\big)=\pi i\, e^{\frac{\pi i\beta}2}\frac{ie^{\beta b}\cosh b-ie^{-\beta b}\cosh(-b)}{2i^2\sinh b\cosh b}=\pi e^{\frac{\pi i\beta}2}\frac{\sinh\beta b}{\sinh b}$$ $$\boxed{\,\,J(a, \beta)=\frac\pi{2\cos\frac{\pi \beta}2}\,\frac{\sinh\beta b}{\sinh b}\,\,}$$ $$I_1(a)=\frac{\partial}{\partial\beta}\,\bigg|_{\beta=0}J(a, \beta)=\frac\pi 2\frac b{\sinh b}$$ or, coming back to the parameter $a$ $$\boxed{\,\,I_1(a)=\frac12\int_{-\infty}^\infty\frac{x\sinh x}{\cosh^2x+a^2}dx=\frac\pi 2\frac{\sinh^{-1}a}a=\frac\pi2\,\frac{\ln(a+\sqrt{1+a^2})}a\,\,}$$ Taking the first derivative with respect to $a$ we get the answer obtained by @Тyma Gaidash and @Random Variable

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PS taking derivatives of $J(a,\beta)$ with respect to $\beta$ we can easily evaluate integrals with different powers of $x$ in the nominator.

Answer 3

As an alternative contour-based solution to the one suggested in comments, substitute $y=e^x$, then $y\mapsto\dfrac1y$ and average the integrals over $[1,\infty)$ and $(0,1]$:

$$\begin{align*} I &= \int_0^\infty \frac{x \sinh x}{\left(1+\cosh^2x\right)^2} \, dx \\ &= \int_1^\infty \frac{8y^2 \left(y^2-1\right)}{\left(y^4+6y^2+1\right)^2} \log y \, dy \\ &= \int_0^1 \frac{8 y^2 \left(y^2-1\right)}{\left(y^4+6y^2+1\right)^2} \log y \, dy \\ &= 4 \underbrace{\int_0^\infty \frac{y^2 \left(y^2-1\right)}{\left(y^4+6y^2+1\right)^2} \log y \, dy}_{=:J} \end{align*}$$

While we're here, observe that the $J$-integral would be zero if the $\log y$ factor wasn't there. Now use the residue theorem to evaluate

$$\oint_C \frac{z^2\left(z^2-1\right)}{\left(z^4+6z^2+1\right)^2} \log^2z \, dz$$

along a keyhole contour, using the branch of the logarithm such that $\arg z\in(0,2\pi)$. Note the second-order poles at $\sqrt{3\pm2\sqrt2}\,e^{i \tfrac\pi2}$ and $\sqrt{3\pm2\sqrt2}\,e^{i \tfrac{3\pi}2}$. Omitting the gritty details, this leads to another somewhat tedious collection of residues.

$$\begin{align*} & i2\pi \sum_{\rm poles} \operatorname{Res} f(z) \\ &= \int_0^\infty \frac{t^2\left(t^2-1\right)}{\left(t^4+6t^2+1\right)^2} \log^2t \, dt + \int_\infty^0 \frac{t^2\left(t^2-1\right)}{\left(t^4+6t^2+1\right)^2} \left(\log t+i2\pi\right)^2 \, dt \\ &= 4\pi^2 \underbrace{\int_0^\infty \frac{t^2\left(t^2-1\right)}{\left(t^4+6t^2+1\right)^2} \, dt}_{=0} - i4\pi J \\[2ex] \implies I &= - 2 \sum_{\rm poles} \operatorname{Res} f(z) \end{align*}$$

Answer 4

Real Analysis Approach:

First make the substitution $t= \cosh(x)$. Then integrate by parts.

For the antiderivative of $\frac{1}{(1+t^{2})^{2}}$ choose $$\frac{1}{2} \left( \frac{t}{1+t^{2}} + \arctan(t) \right) - \frac{\pi}{4} = \frac{1}{2} \left( \frac{t}{1+t^{2}}- \arctan \left(\frac{1}{t} \right) \right). $$

This will make the boundary term vanish as $t \to \infty$.

$$ \begin{align} I &= \int_0^{\infty} \frac{x \sinh(x)}{\left(1+\cosh^2(x)\right)^2} \, \mathrm dx \\ &= \int_{1}^{\infty} \frac{\operatorname{arcosh}(t)}{(1+t^{2})^{2}} \, \mathrm dt \\ &= \small \frac{\operatorname{arcosh}(t)}{2} \left(\frac{t}{1+t^{2}} - \arctan \left(\frac{1}{t} \right) \right) \Bigg|_{1}^{\infty} - \frac{1}{2}\int_{1}^{\infty} \frac{1}{\sqrt{t^{2}-1}} \left(\frac{t}{1+t^{2}}- \arctan \left(\frac{1}{t} \right) \right)\, \mathrm dt \\ &= 0+ \frac{1}{2} \left( \int_{1}^{\infty}\frac{\arctan \left(\frac{1}{t} \right) }{\sqrt{t^{2}-1}} \, \mathrm dt - \int_{1}^{\infty} \frac{t}{\sqrt{t^{2}-1}\left(1+t^{2} \right)} \, \mathrm dt \right) \\ &= \frac{1}{2} \left(-\int_{1}^{0} \frac{\arctan(v)}{\sqrt{\left(\frac{1}{v} \right)^{2}-1}} \frac{\mathrm d v}{v^{2}} - \int_{0}^{\infty} \frac{w}{\sqrt{w^{2}} (1+1+w^{2})} \, \mathrm d w\right) \\ &= \frac{1}{2} \left(\int_{0}^{1} \frac{\arctan (v)}{v \sqrt{1-v^{2}}} \, \mathrm dv - \int_{0}^{\infty} \frac{1}{2+w^{2}} \, \mathrm dw \right) \\ &\overset{\spadesuit}{=} \frac{1}{2} \left(\frac{\pi}{2} \, \operatorname{arsinh}(1)- \frac{\pi}{2 \sqrt{2}} \right) \\&= \frac{\pi}{8} \left(2\ln(1+\sqrt{2})-\sqrt{2} \right). \end{align}$$

$\spadesuit$ Calculate $\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$

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Complex Analysis Approach:

The following approach is somewhat similar to Svyatoslav's answer.

In a previous answer I used a rectangular contour to show that $$\int_{0}^{\infty} \frac{x \sinh(x)}{\cos(2a)+ \cosh(2x)} \, \mathrm dx = \frac{\pi}{4} \frac{a}{\sin (a)} , \quad -\frac{\pi}{2} < \Re(a) < \frac{\pi}{2}.$$

(The value of the integral at $a=0$ is $\lim_{a \to 0} \frac{\pi}{4} \frac{a}{\sin (a)} = \frac{\pi}{4}.$)

Differentiating both sides of the equation with respect to $a$, we have $$ 2 \sin(2a) \int_{0}^{\infty} \frac{x \sinh (x)}{\left( \cos(2a)+\cosh(2x)\right)^{2}} \, \mathrm dx = \frac{\pi}{4} \left(\csc(a) - a \cos(a) \csc^{2}(a) \right). $$

Therefore,

$ \begin{align} \int_{0}^{\infty} \frac{x \sinh(x)}{\left(1+\cosh^{2}(x) \right)^{2}} \, \mathrm dx &= 4\int_{0}^{\infty} \frac{x \sinh (x)}{ \left(3+ \cosh(2x) \right)^{2}} \, \mathrm dx \\ &= \small4 \, \frac{1}{2 \sin \left(\arccos(3) \right)} \frac{\pi}{4} \left(\csc \left(\frac{\arccos(3)}{2} \right) - \frac{\arccos(3)}{2} \cos \left(\frac{\arccos(3)}{2} \right) \csc^{2} \left(\frac{\arccos(3)}{2} \right)\right) \\ &\overset{\diamondsuit}{=} 4 \, \frac{1}{2 \sqrt{1-3^{2}}} \frac{\pi}{4} \left(\frac{\sqrt{2}}{\sqrt{1-3}} - \frac{\arccos(3)}{2} \sqrt{\frac{1+3}{2}} \frac{2}{1-3}\right) \\ &= \frac{\pi}{4 i \sqrt{2}} \left(\frac{1}{i} + \frac{\arccos(3) \sqrt{2}}{2} \right) \\ &\overset{\clubsuit}{=} \frac{\pi}{4 i \sqrt{2}} \left(\frac{1}{i} -\frac{\sqrt{2}}{2} i \ln \left(3-2 \sqrt{2} \right) \right) \\ &= \frac{\pi}{8} \left(- \ln \left(3-2 \sqrt{2} \right)-\sqrt{2} \right) \\ &= \frac{\pi}{8} \left( -\ln \left((1+\sqrt{2})^{-2} \right)-\sqrt{2} \right) \\ &= \frac{\pi}{8} \left(2\ln \left(1+\sqrt{2} \right) -\sqrt{2} \right). \end{align}$

$\diamondsuit$ Use the half-angle identities $\cos^{2} (\frac{z}{2}) = \frac{1+\cos (z)}{2}$ and $\sin^{2} (\frac{z}{2}) = \frac{1-\cos (z)}{2}. $

$\clubsuit$ https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms

Answer 5

If, as you did, let $x=\cosh ^{-1}(t)$, $$I=\int_0^{\infty} \frac{x \sinh (x)}{(1+\cosh^2(x))^2} \,dx=\int_1^{\infty}\frac{\cosh ^{-1}(t)}{\left(t^2+1\right)^2}\,dt$$ which is a bit more pleasant.

If we want to use series, for large values of $t$ $$\cosh ^{-1}(t)=\log(2t)-\sum_{n=1}^\infty \frac {a_n}{b_n} \,t^{-2n}$$ where the $a_n$ and $b_n$ form sequences $A052468$ and $A052469$ in $OEIS$.

$$J_0=\int_1^\infty \frac{\log (2 t)}{\left(t^2+1\right)^2}\,dt=\frac{C}{2}-\frac{\log (2)}{4}-\frac \pi 8 (1-\log (2))$$ $$J_n=\int_1^\infty \frac{t^{-2n}}{\left(t^2+1\right)^2}\,dt=\frac 14+\frac{2n+1}{8} \left(H_{\frac{2n-1}{4}}-H_{\frac{2n+1}{4}}\right)$$

The summation would converge quite fast since $$\frac {a_n}{b_n}\,J_n \sim \frac{1}{50 \,n^{9/4}} $$

Computing the partial sums and converting to decimals $$\left( \begin{array}{cc} 10 & 0.137534 \\ 20 & 0.137117 \\ 30 & 0.137007 \\ 40 & 0.136959 \\ 50 & 0.136934 \\ 60 & 0.136918 \\ 70 & 0.136908 \\ 80 & 0.136901 \\ 90 & 0.136896 \\ 100 & 0.136892 \\ \end{array} \right)$$ while the "exact" value is $0.136869$ which is the result given by @RandomVariable in comments.