Evaluating Accelerator's integral $\int_0^{\infty} \frac{e^{4x}-e^{2x}}{x(e^{2x}+1)^3}dx$

Question

I met this integral in a post of Accelerator (Contour Integral Involving $e^z$, a Semicircle, and Triangle Inequality). WA instantly evaluated it as $I=\int_0^{\infty} \frac{e^{4x}-e^{2x}}{x(e^{2x}+1)^3}dx =\frac{7\zeta(3)}{4\pi^2}$. Then, it is stuck in my mind.

I posted a simpler version of this integral here: Definite integral by Malmsten's formula

I evaluated this integral trickly but unfortunately not so rigorously in the following way:

Taking the second derivative of both sides of $\frac{1}{e^{2x}+1}=\sum_{k=1}^{\infty} (-1)^{k-1}e^{-2kx}$ and simplying we get $$\frac{e^{4x}-e^{2x}}{(e^{2x}+1)^3}=\sum_{k=1}^{\infty} (-1)^{k-1}k^{2}e^{-2kx}.$$ Since, $$\sum_{k=1}^{\infty} (-1)^{k-1}k^{2}=\eta(-2)=(1-2^{3})\zeta(-2)=0,\tag{1}$$ we have $$I=\sum_{k=1}^{\infty} (-1)^{k-1}k^{2}\left(\int_0^{\infty}\frac{e^{-2kx}-e^{-2x}}{x}\right).\tag{2}$$ Then, by Frullani's theorem (Proof of Frullani's theorem), we get $$I=\sum_{k=1}^{\infty} (-1)^{k}k^{2}\ln k=\eta'(-2)=(1-2^3)\zeta'(-2)=(-7)\left(-\frac{\zeta(3)}{4\pi^2}\right)=\frac{7\zeta(3)}{4\pi^2}.$$ How can I make this way more rigorous? In $(2)$, interchange of the summation and integration is not obvious. $(1)$ is divergent so is $(2)$!

Can you suggest another solution? Thanks in advance.

Answer 1

Substitute $t=e^{2x}$

$$I=\int_0^{\infty} \frac{e^{4x}-e^{2x}}{x(e^{2x}+1)^3}dx =\int_1^\infty \frac{t-1}{\ln t (t+1)^3}\overset{t\to\frac1t}{dt} = \frac12\int_0^\infty \frac{t-1}{\ln t \ (t+1)^3}dt $$ Let $J(a)= \int_0^\infty \frac{t^a}{\ln t \ (t+1)^3}dt$ $$J’(a)=\int_0^\infty \frac{t^a}{(1+t)^3}dt=\frac{\pi a(1-a)}{2\sin \pi a} $$ Then \begin{align} I=& \ \frac12(J(1)-J(0))\\ =& \ \frac\pi4\int_0^1 \frac{a(1-a)}{\sin \pi a}da \overset{ibp}=\frac 12 \int_0^1 x\ln \tan \frac{\pi x}2 \overset{t=\tan\frac{\pi x}2}{dx}\\ =& \ \frac2{\pi^2}\int_0^\infty \frac{\ln t\tan^{-1}t}{1+t^2}dt = \frac2{\pi^2}\int_0^\infty \int_0^1 \frac{t \ln t}{(1+t^2)(1+y^2 t^2)} \overset{t\to \frac1{yt}}{ dy} \ dt\\ = &\ \frac1{\pi^2}\int_0^\infty \int_0^1 \frac{-t\ln y}{(1+t^2)(1+y^2 t^2)} dt\ dy = \frac1{\pi^2}\int_0^1 \frac{\ln^2 y}{1-y^2}dy \\ = &\ \frac1{\pi^2}\cdot \frac{7\zeta(3)}4 = \frac{7 \zeta(3)}{4\pi^2} \end{align}

Answer 2

Let the original integral equal $I$. Notice that

$$I = \int_{0}^{\infty}\frac{e^{4x}-e^{2x}}{x\left(e^{2x}+1\right)^{3}}dx = \int_{0}^{\infty}\frac{\tanh\left(x\right)}{4x\cosh^{2}\left(x\right)}dx$$

Integrating by parts yields

$$I = \frac{1}{8}\int_{0}^{\infty}\frac{\tanh^{2}\left(x\right)}{x^{2}}dx.$$

This popular link here provides a way to evaluate the integral above, but I'll add my own method.

Because the integrand is an even function, we can say

$$I = \frac{1}{16}\int_{-\infty}^{\infty}\frac{\tanh^{2}\left(x\right)}{x^{2}}dx.$$

Next, we will prove that $$ \frac{1}{16}\int_{-\infty}^{\infty}\frac{\tanh^{2}\left(z\right)}{z^{2}}dz = \frac{1}{2\pi^{2}}\int_{0}^{1}\frac{1}{z}\ln^{2}\left(\frac{1-z}{1+z}\right)dz. $$

Recall the Weierstrauss Product representation of $\cosh{(z)}$ as

$$ \cosh{(z)} = \prod_{n=0}^{\infty}\left(1+\left(\frac{2z}{\pi\left(2n+1\right)}\right)^{2}\right). $$ Taking the logarithm and the derivative on both sides, we get $$ \eqalign{ \ln\left(\cosh\left(z\right)\right)\ &=\ \ln\prod_{n=0}^{\infty}\left(1+\left(\frac{2z}{\pi\left(2n+1\right)}\right)^{2}\right) \\ &= \sum_{n=0}^{\infty}\ln\left(1+\left(\frac{2z}{\pi\left(2n+1\right)}\right)^{2}\right) \\ \frac{d}{dz}\ln\left(\cosh\left(z\right)\right)\ &=\ \frac{d}{dz}\sum_{n=0}^{\infty}\ln\left(1+\left(\frac{2z}{\pi\left(2n+1\right)}\right)^{2}\right) \\ \tanh\left(z\right) & = z\sum_{n=0}^{\infty}\frac{8}{4z^{2}+\left(\pi\left(2n+1\right)\right)^{2}} \\ \frac{\tanh\left(z\right)}{z}\ &=\ \sum_{n=0}^{\infty}\frac{8}{4z^{2}+\left(\pi\left(2n+1\right)\right)^{2}} \\ \frac{\tanh^{2}\left(z\right)}{z^{2}} &= \left(\sum_{n=0}^{\infty}\frac{8}{4z^{2}+\left(\pi\left(2n+1\right)\right)^{2}}\right)^{2} \\ &= \left(\sum_{n=0}^{\infty}\frac{8}{4z^{2}+\left(\pi\left(2n+1\right)\right)^{2}}\right)\left(\sum_{m=0}^{\infty}\frac{8}{4z^{2}+\left(\pi\left(2m+1\right)\right)^{2}}\right) \\ &= \sum_{n,m=0}^{\infty}\frac{64}{\left(4z^{2}+\left(\pi\left(2n+1\right)\right)^{2}\right)\left(4z^{2}+\left(\pi\left(2m+1\right)\right)^{2}\right)} \\ &= 64\sum_{n,m=0}^{\infty}\left(\frac{1}{4\left(z^{2}+\frac{\left(\pi\left(2n+1\right)\right)^{2}}{4}\right)}\frac{1}{4\left(z^{2}+\frac{\left(\pi\left(2m+1\right)\right)^{2}}{4}\right)}\right). \\ } $$

On both sides, we integrate over $\mathbb{R}$ and multiply by $\frac{1}{16}$ so that $I$ becomes $$ \eqalign{ &\frac{1}{16}\int_{-\infty}^{\infty}64\sum_{n,m=0}^{\infty}\left(\frac{1}{4\left(z^{2}+\frac{\left(\pi\left(2n+1\right)\right)^{2}}{4}\right)}\frac{1}{4\left(z^{2}+\frac{\left(\pi\left(2m+1\right)\right)^{2}}{4}\right)}\right)dz \\ &= \frac{2}{\pi^{2}}\sum_{n,m=0}^{\infty}\left(\frac{1}{\left(2n+1\right)\left(2m+1\right)}\frac{1}{2n+1+2m+1}\right) \\ &= \frac{2}{\pi^{2}}\sum_{n,m=0}^{\infty}\left(\frac{1}{\left(2n+1\right)\left(2m+1\right)}\int_{0}^{1}\frac{z^{2n+1+2m+1}}{z}dz\right) \\ &= \frac{2}{\pi^{2}}\int_{0}^{1}\frac{1}{z}\sum_{n,m=0}^{\infty}\frac{z^{2n+1}}{2n+1}\frac{z^{2m+1}}{2m+1}dz \\ &= \frac{1}{2\pi^{2}}\int_{0}^{1}\frac{1}{z}\ln^{2}\left(\frac{1-z}{1+z}\right)dz,\\ } $$ where we used the power series representation $$ \ln\left(\frac{1-z}{1+z}\right)=-2\sum_{k=0}^{\infty}\frac{z^{2k+1}}{2k+1}, $$ derived here in this YouTube video.

Next, we will prove that $$ \int_{0}^{1}\frac{1}{z}\ln^{2}\left(\frac{1-z}{1+z}\right)dz = \frac{7}{2}\zeta{(3)}, $$ where Apery's Constant is defined as $$ \zeta{(3)} = \sum_{n=1}^{\infty}\frac{1}{n^{3}}. $$ Let that integral equal $J$ and let $f(z) = \frac{1}{z}\log^{2}\left(\frac{1-z}{1+z}\right)$. Notice we have the singularities $z_a = 0$ and $z_b = 1$. For small $\epsilon > 0$, let's define a closed contour $C$ as $$ C = \left[1-\epsilon, \epsilon\right] \cup \gamma_1 \cup \left[i\epsilon, i\right] \cup \Gamma \cup \gamma_2. $$ This contour travels clockwise. It closely resembles a quarter circle in the first quadrant of the complex plane such that $\gamma_1$ and $\gamma_2$ are small, quarter-circle, and indented arcs (with radius $\epsilon$) around the singularities $z_a = 0$ and $z_b = 1$, respectively. Also, $\Gamma$ closely resembles a quarter-circle arc (with radius $1$) such that a small radian, call it $t_{\epsilon}$, is cut off.

Below is a visual of $C$. One of my best friends Dairah coded this, so credit to her.

By Cauchy's Residue Theorem, we get $\oint_C f(z)dz$ to be $$ \eqalign{ 0 &= \int_{1-\epsilon}^{\epsilon}f(x)dx + \int_{\gamma_1}f(z)dz + \int_{i}^{i\epsilon}f(z)dz + \int_{\Gamma}f(z)dz + \int_{\gamma_2}f(z)dz \\ \int_{\epsilon}^{1-\epsilon}f(x)dx &= \int_{\gamma_1}f(z)dz + \int_{i}^{i\epsilon}f(z)dz + \int_{\Gamma}f(z)dz + \int_{\gamma_2}f(z)dz. \\ } $$ Taking the limit as $\epsilon,t_{\epsilon} \to 0$ on both sides, we get $$ J = \lim_{\epsilon \to 0}\int_{\gamma_1}f(z)dz + \lim_{\epsilon \to 0}\int_{i}^{i\epsilon}f(z)dz + \lim_{t_{\epsilon} \to 0}\int_{\Gamma}f(z)dz + \lim_{\epsilon \to 0}\int_{\gamma_2}f(z)dz. $$

We will call each of those four limits $J_1$, $J_2$, $J_3$, and $J_4$, respectively.

First, we will prove $J_1 = 0$.

Let $z = \epsilon e^{i\phi}$ for $\phi \in \left[0, \frac{\pi}{2}\right]$. Then $$ \eqalign{ J_1 &= \lim_{\epsilon \to 0}\int_{0}^{\frac{\pi}{2}} f\left(\epsilon e^{i\phi}\right)i \epsilon e^{i\phi}d\phi \\ &= i\lim_{\epsilon \to 0}\int_{0}^{\frac{\pi}{2}}\frac{1}{\epsilon e^{i\phi}}\log^2{\left(\frac{1-\epsilon e^{i\phi}}{1+\epsilon e^{i\phi}}\right)}\epsilon e^{i\phi}d\phi \\ &= i\int_{0}^{\frac{\pi}{2}}\log^2{\left(\frac{1}{1}\right)}d\phi \\ &= 0. } $$

Second, we will prove $J_4 = 0$ also.

Let $z = 1 + \epsilon e^{i\theta}$ where $\theta \in \left[\frac{\pi}{2}, \pi\right]$. Then using a chain of inequalities, observe that

$$ \eqalign{ \left|\int_{\frac{\pi}{2}}^{\pi}f\left(1+ \epsilon e^{i\theta}\right)i\epsilon e^{i\theta}\right| &= \left|\int_{\frac{\pi}{2}}^{\pi} \frac{1}{1+\epsilon e^{i\theta}}\log^2{\left(\frac{1-1-\epsilon e^{i\theta}}{1+1+\epsilon e^{i\theta}}\right)}i\epsilon e^{i\theta}d\theta\right| \\ &\leq \int_{\frac{\pi}{2}}^{\pi}\left|\frac{1}{1+\epsilon e^{i\theta}}\right|\cdot\epsilon\left|\log{\left(\frac{-\epsilon e^{i\theta}}{2+\epsilon e^{i\theta}}\right)}\right|^2 d\theta \\ &=\int_{\frac{\pi}{2}}^{\pi}\frac{\epsilon\left|\ln{\left|\frac{-\epsilon e^{i\theta}}{2+\epsilon e^{i\theta}}\right|}+i\arg{\left(\frac{-\epsilon e^{i\theta}}{2+\epsilon e^{i\theta}}\right)}\right|^2}{\sqrt{\epsilon^2+2\epsilon\cos{(\theta)}+1}} d\theta \\ &= \int_{\frac{\pi}{2}}^{\pi}\frac{\epsilon\left|\ln{\left(\frac{\epsilon}{\sqrt{\epsilon^2+4\epsilon\cos{(\theta)}+4}}\right)}+i\arg{\left(\frac{-\epsilon e^{i\theta}}{2+\epsilon e^{i\theta}}\right)}\right|^2}{\sqrt{\epsilon^2+2\epsilon\cos{(\theta)}+1}} d\theta \\ &= \int_{\frac{\pi}{2}}^{\pi}\frac{\left|\sqrt{\epsilon}\ln{(\epsilon)}-\sqrt{\epsilon}\ln{\left(\sqrt{\epsilon^2+4\epsilon\cos{(\theta)}+4}\right)}+\sqrt{\epsilon}i\arg{\left(\frac{-\epsilon e^{i\theta}}{2+\epsilon e^{i\theta}}\right)}\right|^2}{\sqrt{\epsilon^2+2\epsilon\cos{(\theta)}+1}} d\theta \\ } $$ Taking $\epsilon \to 0$ of that last integral, we get $$ \lim_{\epsilon \to 0}\int_{\frac{\pi}{2}}^{\pi}\frac{\left|\sqrt{\epsilon}\ln{(\epsilon)}-\sqrt{\epsilon}\ln{\left(\sqrt{\epsilon^2+4\epsilon\cos{(\theta)}+4}\right)}+\sqrt{\epsilon}i\arg{\left(\frac{-\epsilon e^{i\theta}}{2+\epsilon e^{i\theta}}\right)}\right|^2}{\sqrt{\epsilon^2+2\epsilon\cos{(\theta)}+1}} d\theta = \int_{\frac{\pi}{2}}^{\pi}\frac{\left|0-0+0\right|^2}{\sqrt{0+0+1}}d\theta = 0. $$ (Notice that as $\epsilon \to 0$, we get $\sqrt{\epsilon}\ln{(\epsilon)} \to 0$ by L'Hôpital's Rule.)

So,

$$ 0 \leq \lim_{\epsilon \to 0} \left|\int_{\frac{\pi}{2}}^{\pi}f\left(1+ \epsilon e^{i\theta}\right)i\epsilon e^{i\theta}\right| \leq 0. $$ By the Squeeze Theorem, we get that $$ \lim_{\epsilon \to 0} \left|\int_{\frac{\pi}{2}}^{\pi}f\left(1+ \epsilon e^{i\theta}\right)i\epsilon e^{i\theta}\right| = 0, $$ implying that $$ \lim_{\epsilon \to 0} \int_{\frac{\pi}{2}}^{\pi}f\left(1+ \epsilon e^{i\theta}\right)i\epsilon e^{i\theta} = 0. $$

Third, we will prove $J_2 = \frac{7}{2}\zeta{(3)} - 2\pi C$, where $\zeta{(3)} := \sum_{n=0}^{\infty}\frac{1}{n^3}$ is Apery's Constant and $C := \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$ is Catalan's Constant.

We get $J_2$ to be $$ \eqalign{ \int_{0}^{i}\frac{1}{z}\log^{2}\left(\frac{1-z}{1+z}\right)dz &= \int_{0}^{-1}\frac{1}{iz}\log^{2}\left(\frac{1-iz}{1+iz}\right)idz \text{ (using $z \to iz$)}\\ &= \int_{0}^{-1}\frac{\left(2i\arctan\left(z\right)\right)^{2}}{z}dz \\ &= 8\int_{0}^{1}\frac{\arctan\left(z\right)\ln\left(z\right)}{1+z^{2}}dz \text{ (using IBP)} \\ &= -4\pi\int_{1}^{\infty}\frac{\ln\left(z\right)}{z^{2}+1}dz+8\int_{1}^{\infty}\frac{\arctan\left(z\right)\ln\left(z\right)}{z^{2}+1}dz \text{ (using $z \to \frac{1}{z}$)} \\ &= -2\pi\int_{1}^{\infty}\frac{\ln\left(z\right)}{z^{2}+1}dz+4\int_{0}^{\infty}\frac{\arctan\left(z\right)\ln\left(z\right)}{z^{2}+1}dz, \\ } $$ where we used the fact that $\int_0^1f(z)dz + \int_1^{\infty}f(z)dz = \int_0^{\infty}f(z)dz$ and did some algebra. Thus, $$ -2\pi\int_{1}^{\infty}\frac{\ln\left(z\right)}{z^{2}+1}dz+4\int_{0}^{\infty}\frac{\arctan\left(z\right)\ln\left(z\right)}{z^{2}+1}dz = -2\pi C + 4\left(\frac{7}{8}\zeta(3)\right). \tag{1} $$

Fourth, we will prove that $J_3 = 2\pi C$.

We parameterize $z = e^{it}$ for $t \in \left[t_{\epsilon}, \frac{\pi}{2}\right]$, for $t_{\epsilon} \to 0$. Then $J_3$ becomes $$ \eqalign{ -\int_{0}^{\frac{\pi}{2}}\frac{1}{e^{it}}\log^{2}\left(\frac{1-e^{it}}{1+e^{it}}\right)ie^{it}dt &= -i\int_{0}^{\frac{\pi}{2}}\left(\ln\left|-i\tan\left(\frac{t}{2}\right)\right|+i\arg{\left(-i\tan\left(\frac{t}{2}\right)\right)}\right)^{2}dt \\ &= -2i\int_{0}^{\frac{\pi}{4}}\left(\ln\left(\tan\left(t\right)\right)-\frac{i\pi}{2}\right)^{2}dt \text{ (using $\frac{t}{2} \to t$)} \\ &= -2i\int_{0}^{\frac{\pi}{4}}\ln^{2}\left(\tan\left(t\right)\right)dt-2\pi\int_{0}^{\frac{\pi}{4}}\ln\left(\tan\left(t\right)\right)dt+\frac{i\pi^{3}}{8} \\ } $$

Thus,

$$J_3 = -2i\left(\frac{\pi^{3}}{16}\right)-2\pi\left(-C\right)+\frac{i\pi^{3}}{8}. \tag{2} \\$$

Hence, $J$ simplifies to $$ J = 0 -2\pi C + 4\left(\frac{7}{8}\zeta(3)\right) -2i\left(\frac{\pi^{3}}{16}\right)-2\pi\left(-C\right)+\frac{i\pi^{3}}{8} + 0 = \frac{7}{2}\zeta(3). $$

Therefore, we multiply by $\frac{1}{2\pi^2}$ on both sides to solve for $I$.

In conclusion, the integral $I$ is

$$\int_{0}^{\infty}\frac{e^{4x}-e^{2x}}{x\left(e^{2x}+1\right)^{3}}dx = \frac{7\zeta(3)}{4\pi^2}.$$

If anyone wants to, I can solve $(1)$ and $(2)$ in the comments.

I greatly appreciate you taking the time to solve an integral I was trying to figure out for a long time. I attempted this integral a while back and used this attempt as part of another problem I was solving, so I've copied and pasted part of my solution from Overleaf and made a few adjustments. It is a long solution and there are plenty of easier ways to solve it like what Quanto provided in his answer, but I think using a crazy contour path is salvageable.

I hope I don't have complex argument/branch issues with my logarithms. I understand it is a lot to read, but please let me know if you catch any errors so I can learn more.

Answer 3

Here, we use contour integration. Proceeding we have

$$\begin{align} I&=\int_0^\infty \frac1x\frac{e^{4x}-e^{2x}}{(1+e^{2x})^3}\,dx\\\\ &=\frac18\int_{-\infty}^\infty \frac1x\frac{\sinh(x)}{\cosh^3(x)}\,dx \\\\ &=\frac{\pi i}{4} \sum_{n=0}^\infty \text{Res}\left(\frac1z\frac{\sinh(z)}{\cosh^3(z)}, z=i(2n+1)\pi/2\right)\\\\ &=\frac{\pi i}{4} \sum_{n=0}^\infty\left(-\frac{8i}{\pi^3(2n+1)^3}\right)\\\\ &=\frac{7\zeta(3)}{4\pi^3} \end{align}$$

And we are done!

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There is a question regarding whether contour integration works here. In particular, the question is whether the integration over a semi-circular arc approaches $0$ as the radius of the semicircle approaches $0$. We proceed now to show that contour integration is applicable.

In THIS ANSWER, I showed that $|\cot(\pi z)|$ is uniformly bounded on the circle $|z|=(N+1/2)$ for $N\in \mathbb{N^+}$. Analogously, it is straightforward to show that $|\tanh(z)|$ is uniformly bounded on the circle $N\pi$, $N\in \mathbb{N^+}$.

Let $B$ represent an upper bound of $|\tanh(z)|$ on the circle $N\pi$, $C_N$ be the semi-circular arc with radius $N\pi$ in the upper-half plane, and $I_N=\int_{C_N}\frac{\sinh(z)}{z\cosh^3(z)}\,dz$. Then we have the estimates on the semi-circular arc of radius $N\pi$

$$\begin{align} |I_n|&=\left|\int_{C_N}\frac{\sinh(z)}{z\cosh^3(z)}\,dz\right|\\\\ &\le \int_{C_N}\frac{1}{N\pi}\left|\tanh(z)\right|\,\left|\text{sech}^2(z)\right|\,|dz|\\\\ &\le B \int_0^\pi \left|\text{sech}(N\pi e^{i\phi})\right|^2\,d\phi\\\\ &=4B \int_0^{\pi} \frac{|e^{-N\pi e^{i\phi}}|^2}{|1+e^{-2N\pi e^{i\phi}}|^2}\,d\phi\\\\ &\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\int_0^{\pi/2} e^{-2N\pi \cos(\phi)}\,d\phi\\\\ &\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\int_0^{\pi/2} e^{-2N\pi (1-2\phi/\pi)}\,d\phi\\\\ &\le \frac{8B}{\inf_{\phi\in[0,\pi], N\in\mathbb{N^+}} {|1+e^{-2N\pi e^{i\phi}}|^2}}\frac{1-e^{-2N\pi}}{4N} \end{align}$$

which clearly approaches $0$ as $N\to \infty$.

So, by judiciously selecting the radius of the semi-circle to be $N\pi$, we construct a sequence of integrals $I_N$ such that $\lim_{N\to\infty}I_N=0$.