Solve $\int_0^\infty\frac{\ln(2e^x-1)}{e^x-1}dx$

Question

In one of the final problems of the MIT integration bee for this year, $$I=\int_0^\infty\frac{\ln(2e^x-1)}{e^x-1}dx$$was one of the given problems. My try was to let $u=e^x-1$ to get $$I=\int_0^\infty\frac{\ln(2u+1)}{u(u+1)}du=\int_0^\infty\frac{\ln(x+1)}{x(x+2)}dx$$I don't know whether I would be right but I had a feeling this was a dead end. Turning the original integral into a geometric series doesn't seem promising either. How should I solve this?

Note: These problems are solved in 5 minutes so please come up with a solution that can be done in such a time limit.

Answer 1

Substitute $t=\frac{1}{2e^x-1}$ \begin{align} & \int_0^\infty\frac{\ln(2e^x-1)}{e^x-1}dx\\ =&\int_0^1 \frac{\ln t^2}{t^2-1}dt =\int_0^1 \int_0^\infty \frac{2y}{(1+y^2)(1+t^2y^2)}dy \ dt\\ =& \ 2\int_0^\infty \frac{\arctan y}{1+y^2}dy =\arctan^2 y\bigg|_0^\infty = \frac{\pi^2}4 \end{align}

Answer 2

Put $$I\left ( a \right ) =\int_{0}^{\infty } \frac{\log\left ( ax+1 \right ) }{x\left ( x+1 \right ) }\mathrm{d}x, $$ then $$I'\left ( a \right ) =\int_{0}^{\infty } \frac{\mathrm{d}x }{\left ( ax+1 \right )\left ( x+1 \right ) }=\frac{\log a}{a-1}.$$ Therefore $$\begin{align*} I\left ( 2 \right )&= \int^{1}_{-1}\frac{\log\left(x+1\right)}{x}\mathrm{d}x\\ & = -\int_{-1}^{1} \frac{1}{x} \sum _{n = 1}^{\infty}\frac{\left ( -x \right )^n }{n}\mathrm{d}x\\ &= \sum _{n = 0}^{\infty}\int_{-1}^{1}\frac{x^{2n}}{2n+1} \mathrm{d}x\\ &=\sum _{n = 0}^{\infty}\frac{2}{\left ( 2n+1 \right )^2 } \\ &=\frac{\pi^2}{4} \end{align*}. $$

Answer 3

SOLUTION 1 (Integration Bee Style)

Let $x=-\ln(1-u)$. Then

$$ \begin{align} \mathcal{I}&:= \int_{0}^{\infty}\frac{\ln\left(2e^{x}-1\right)}{e^{x}-1}dx \\ &= \int_{0}^{1}\frac{\ln\left(2e^{-\ln\left(1-u\right)}-1\right)}{e^{-\ln\left(1-u\right)}-1}\cdot\frac{1}{1-u}du \\ &= 2\int_{0}^{1}\frac{\operatorname{artanh}u}{u}du \\ &= 2\int_{0}^{1}\sum_{n=0}^{\infty}\frac{u^{2n}}{2n+1}du \\ &= \sum_{n=0}^{\infty}\frac{2}{2n+1}\int_{0}^{1}u^{2n}du \\ &= \sum_{n=0}^{\infty}\frac{2}{\left(2n+1\right)^{2}} \\ &= 2\left(\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\sum_{k=1}^{\infty}\frac{1}{4k^{2}}\right) \\ &= 2\left(\frac{\pi^{2}}{6}-\frac{\pi^{2}}{24}\right) \\ &= \frac{\pi^{2}}{4}\,.\\ \end{align} $$

We finish with

$$ \bbox[#FCFFE7,border:5px dotted#639E00,10px]{\int_{0}^{\infty}\frac{\ln\left(2e^{x}-1\right)}{e^{x}-1}dx=\frac{\pi^{2}}{4}} $$

and we're done!

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SOLUTION 2 (Contour Integration)

Here is another solution that doesn't use a series evaluation, but uses Euclidean geometry and contour integrals.

We do the same substitution as the previous solution. So

$$ \begin{align} \mathcal{I} &:= \int_{0}^{\infty}\frac{\ln\left(2e^{x}-1\right)}{e^{x}-1}dx \\ \overset{x\,\mapsto\,-\ln(1-x)}{=}&\int_{0}^{1}\frac{\ln\left(2e^{-\ln\left(1-x\right)}-1\right)}{e^{-\ln\left(1-x\right)}-1}\cdot\frac{1}{1-x}dx \\ &= 2\int_{0}^{1}\frac{\operatorname{artanh}x}{x}dx \\ &\overset{\clubsuit}{=} \int_{-1}^{1}\frac{\operatorname{artanh}x}{x}dx \\ \end{align} $$

where at $\clubsuit$ we use the fact that $\displaystyle \frac{\operatorname{artanh} x}{x}$ is even.

Next, we define a holomorphic function $f: \mathbb{C}\,\setminus ((-\infty,-1] \cup [1,\infty)) \to \mathbb{C}$ where $\displaystyle z \mapsto \frac{\operatorname{artanh}z}{z}$. This function employs the usual principal branch, specifically $\displaystyle\operatorname{Arg}\left(\frac{1+z}{1-z}\right) \in (-\pi, \pi]$. Additionally, this function's domain includes $z = 0$ because that element is a removable singularity. One can prove this by redefining $\operatorname{artanh}(z)$ as its Maclaurin series and making it analytic in a neighborhood of $z=0$. This removes the need for some circular indent around that point.

Next, we define a counterclockwise contour $\mathcal{C} = [r-1,1-r] \cup \gamma_1 \cup \Gamma \cup \gamma_{-1}$ where

$$ \begin{align} \gamma_1 &:= \left\{1-re^{-it}: t\in \left[0,\arcsin\left(\frac{\sqrt{4-r^{2}}}{2}\right)\right]\right\} \\ \Gamma &:= \left\{e^{it}: t \in \left[\arccos\left(1-\frac{r^{2}}{2}\right),\pi-\arccos\left(1-\frac{r^{2}}{2}\right)\right]\right\} \\ \gamma_{-1} &:= \left\{re^{-it}-1: t \in \left[-\arcsin\left(\frac{\sqrt{4-r^{2}}}{2}\right),0\right]\right\}\\ \end{align} $$

with the restriction $0 < r \ll 1$.

Here is a visual using the standard domain coloring with shading.

(IDK why my PC misaligns that "color break", or however you call it, on the left side. Must be a compiling glitch🤦‍♂️)

I'll briefly explain how we can come up with these subsets of the contour and skip the messy algebra and geometry.

(1) We find the intersections of the circles $x^2+y^2=1$, $(x-1)^2+y^2=r^2$, and $(x+1)^2+y^2=r^2$. They are $\left(\frac{r^{2}}{2}-1,\frac{r}{2}\sqrt{4-r^{2}}\right)$, $\left(1-\frac{r^{2}}{2},\frac{r}{2}\sqrt{4-r^{2}}\right)$, $\left(\frac{r^{2}}{2}-1,-\frac{r}{2}\sqrt{4-r^{2}}\right)$, and $\left(1-\frac{r^{2}}{2},-\frac{r}{2}\sqrt{4-r^{2}}\right)$. Only the first two are relevant.

(2) To find how many radians we need to connect $\gamma_1$ and $\gamma_{-1}$ to $\Gamma$, we construct four right triangles such that their top vertices touch where the contours should intersect.

(3) We assign each triangle a variable, say $\theta$ or something, and solve for each variable via trigonometry. E.G. one of the triangles on the far left can have a $\theta$ next to the point $z=-1$, and we find that the amount of radians needed to connect $\gamma_{-1}$ and $\Gamma$ is $\displaystyle \theta = \arcsin\left(\frac{\sqrt{4-r^{2}}}{2}\right)$ obtained from $\displaystyle \sin(\theta) = \frac{\frac{r}{2}\sqrt{4-r^{2}}}{r}$.

Here is a visual of what's happening.

To convince myself I didn't make some silly algebra mistake somewhere, I programmed a manual animation of how the contour behaves as each radius $r \to 0^+$. You can view it here. In that link, when you drag the $r$ slider, the contour maintains its shape as a closed contour along with the right triangles.

Getting that contour construction business out of the way, we express the contour integral over $\mathcal{C}$ as

$$ \oint_{\mathcal{C}} f = \int_{r-1}^{1-r}f + \int_{\gamma_1}f + \int_{\Gamma}f + \int_{\gamma_{-1}}f \,. $$

We recover $\mathcal{I}$ by equating the real part on both sides and then applying $r \to 0^+$ on both sides as follows:

$$ \lim_{r\,\to\,0^+}\Re\oint_{\mathcal{C}} f = \mathcal{I} + \lim_{r\,\to\,0^+}\Re\int_{\gamma_1}f + \lim_{r\,\to\,0^+}\Re\int_{\Gamma}f + \lim_{r\,\to\,0^+}\Re\int_{\gamma_{-1}}f \,. $$

To evaluate $\displaystyle \lim_{r\,\to\,0^+}\Re\int_{\gamma_1}f$, we'll prove that the modulus of $\displaystyle \Re\int_{\gamma_{-1}}f$ goes to $0$ as $r \to 0^+$.

Proof. Let $\displaystyle z \in \left\{re^{-it}-1: t \in \left[-\arcsin\left(\frac{\sqrt{4-r^{2}}}{2}\right),0\right]\right\}$ and bound $|f|$ like this:

$$ \begin{align} |f| &= \left|\frac{\operatorname{artanh}z}{z}\right| \\ &\leq \frac{1}{2\left|z\right|}\left(\left|\ln\left|\frac{1+z}{1-z}\right|\right|+\left|i\operatorname{Arg}\left(\frac{1+z}{1-z}\right)\right|\right) \\ &\overset{z\,\in\,\gamma_{-1}}{=} \frac{1}{2\left|re^{-it}-1\right|}\left(\left|\ln\left|\frac{1+re^{-it}-1}{1-re^{-it}+1}\right|\right|+\left|\operatorname{Arg}\left(\frac{1+re^{-it}-1}{1-re^{-it}+1}\right)\right|\right) \\ &= \frac{1}{2\left|re^{-it}-1\right|}\left(\left|\ln\left|\frac{re^{-it}}{2-re^{-it}}\right|\right|+\left|\operatorname{Arg}\left(\frac{re^{-it}}{2-re^{-it}}\right)\right|\right) \\ &\leq \frac{1}{2\left(1-r\right)}\left(\ln\sqrt{r^{2}\sin^{2}t+\left(2-r\cos t\right)^{2}}-\ln r+\pi\right)\,. \\ \end{align} $$

Note that the arc length of $\gamma_{-1}$ is $\displaystyle r \arcsin\left(\frac{\sqrt{4-r^{2}}}{2}\right)$, obtained from the set-builder notation of $\gamma_{-1}$ and the usual arc length formula for circles.

Using these two results, we use the Estimation Lemma and get

$$ 0 \leq \left|\Re \int_{\gamma_{-1}}f\right| \leq \left|\int_{\gamma_{-1}}f\right| \leq \frac{1}{2\left(1-r\right)}\left(\ln\sqrt{r^{2}\sin^{2}t+\left(2-r\cos t\right)^{2}}-\ln r+\pi\right)\cdot r\arcsin\left(\frac{\sqrt{4-r^{2}}}{2}\right)\,. \\ $$

Taking $r \to 0^+$ on the upper bound, we get

$$ \lim_{r\,\to\,0^+}\frac{1}{2\left(1-r\right)}\left(\ln\sqrt{r^{2}\sin^{2}t+\left(2-r\cos t\right)^{2}}-\ln r+\pi\right)\cdot r\arcsin\left(\frac{\sqrt{4-r^{2}}}{2}\right) = 0\,. $$

By the Squeeze Theorem, we get

$$ \lim_{r\,\to\,0^+}\left|\Re \int_{\gamma_{-1}}f\right| = 0\,. $$

This leads to

$$ \lim_{r\,\to\,0^+}\Re \int_{\gamma_{-1}}f = 0 $$

and we're done with the integral evaluation! Q.E.D.

Similarly, we can prove that

$$\lim_{r\,\to\,0^+}\Re\int_{\gamma_1}f = 0$$

which I'll leave as an exercise (More like I'm too lazy to type the proof :P).

Next, we evaluate the contour integral over $\Gamma$ as follows. Let $\displaystyle z \in \left\{e^{it}: t \in \left[\arccos\left(1-\frac{r^{2}}{2}\right),\pi-\arccos\left(1-\frac{r^{2}}{2}\right)\right]\right\}$. Then

$$ \begin{align} \lim_{r\,\to\,0^+} \Re \int_{\Gamma}f &\overset{z\,\in\,\Gamma}{=} \lim_{r\,\to\,0^+}\Re \int_{\arccos\left(1-\frac{r^{2}}{2}\right)}^{\pi-\arccos\left(1-\frac{r^{2}}{2}\right)}\frac{\operatorname{artanh}\left(e^{it}\right)}{e^{it}}\cdot ie^{it}dt \\ &= -\frac{1}{2}\lim_{r\,\to\,0^+} \Im \int_{\arccos\left(1-\frac{r^{2}}{2}\right)}^{\pi-\arccos\left(1-\frac{r^{2}}{2}\right)}\left(\ln\left|\frac{1+e^{it}}{1-e^{it}}\right|+i\operatorname{Arg}\left(\frac{1+e^{it}}{1-e^{it}}\right)\right)dt \\ &= -\frac{1}{2}\lim_{r\,\to\,0^+}\int_{\arccos\left(1-\frac{r^{2}}{2}\right)}^{\pi-\arccos\left(1-\frac{r^{2}}{2}\right)}\operatorname{Arg}\left(i\cot\left(\frac{t}{2}\right)\right)dt \\ &= -\frac{1}{2}\lim_{r\,\to\,0^+}\int_{\arccos\left(1-\frac{r^{2}}{2}\right)}^{\pi-\arccos\left(1-\frac{r^{2}}{2}\right)}\frac{\pi}{2}dt \\ &= -\frac{1}{2}\int_{0}^{\pi}\frac{\pi}{2}dt \\ &= -\frac{\pi^{2}}{4} \,. \end{align} $$

For $\displaystyle \oint_{\mathcal{C}}f$, it equals $0$ by Cauchy's Integral Theorem.

Gathering all the results together, we have

$$ \require{cancel}\cancelto{0}{\oint_{\mathcal{C}}f} = \mathcal{I} + \cancelto{0}{\lim_{r\,\to\,0^+}\Re\int_{\gamma_1}f} -\frac{\pi^{2}}{4} + \cancelto{0}{\lim_{r\,\to\,0^+}\Re\int_{\gamma_{-1}}f}\,. $$

We finally conclude with

$$ \bbox[#f2ebe6,border: 5px inset#765341,10px]{\int_{0}^{\infty}\frac{\ln\left(2e^{x}-1\right)}{e^{x}-1}dx=\frac{\pi^{2}}{4}} $$

and we're done!

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Solution 3 (Incomplete)

Let $\displaystyle x \mapsto \ln\left(\frac{e^{x}+1}{2}\right)$. Then

$$ \begin{align} \mathcal{I} &:= \int_{0}^{\infty}\frac{\ln\left(2e^{x}-1\right)}{e^{x}-1}dx \\ &\overset{x\,\mapsto\,\ln\left(\frac{e^{x}+1}{2}\right)}{=} \int_{0}^{\infty}\frac{\ln\left(2e^{\ln\left(\frac{e^{x}+1}{2}\right)}-1\right)}{e^{\ln\left(\frac{e^{x}+1}{2}\right)}-1}\cdot\frac{e^{x}}{e^{x}+1}dx \\ &= \int_{0}^{\infty}\frac{2xe^{x}}{e^{2x}-1}dx \\ &= \int_{-\infty}^{\infty}\frac{xe^{x}}{e^{2x}-1}dx \\ &= \frac{1}{2}\int_{-\infty}^{\infty}x\operatorname{csch}xdx\,. \\ \end{align} $$

Define a holomorphic function $f: \mathbb{C} \,\setminus (\mathbb{Z}\setminus \left\{0\right\}) \to \mathbb{C}$ where $z \mapsto z \operatorname{csch} z$. Here is a visual down below.

If you choose to try this out, I'll finish this answer. You would eventually finish with

$$ \bbox[#FFF6F7,border: 5px dashed#CC0B1B,11.5px]{\int_{0}^{\infty}\frac{\ln\left(2e^{x}-1\right)}{e^{x}-1}dx=\frac{\pi^{2}}{4}} $$

and be done!

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Answer 4

We are now evaluating the following integral: $$ \int_{0}^{\infty} \frac {\ln \left( 2 {e}^{x} - 1 \right)}{{e}^{x} - 1} \text {d} x = \, ? $$ My first observation is $$ 2 {e}^{x} - 1 = {e}^{2 x} - {\left( {e}^{x} - 1 \right)}^{2}. $$ I am not sure if this is relevant. Anyway, it looks like ${e}^{x} \mapsto x$ is a good substitution. So $x \mapsto \ln \left( x \right)$, and $\text {d} x \mapsto \text {d} x / x$. As a consequence, $$ \int_{0}^{\infty} \frac {\ln \left( 2 {e}^{x} - 1 \right)}{{e}^{x} - 1} \text {d} x = \int_{1}^{\infty} \frac {\ln \left( 2 x - 1 \right)}{x \left( x - 1 \right)} \text {d} x. $$ Next, we perform the substitution: $x \mapsto 1 / x$, so $\text {d} x \mapsto - \text {d} x / {x}^{2}$. Accordingly, $$ \int_{1}^{\infty} \frac {\ln \left( 2 x - 1 \right)}{x \left( x - 1 \right)} \text {d} x = \int_{0}^{1} \frac {\ln \left( 2 - x \right)}{1 - x} \text {d} x - \int_{0}^{1} \frac {\ln \left( x \right)}{1 - x} \text {d} x. $$ Let $x \mapsto 1 - x$, so $$ \int_{0}^{1} \frac {\ln \left( 2 - x \right)}{1 - x} \text {d} x = \int_{0}^{1} \frac {\ln \left( 1 + x \right)}{x} \text {d} x $$ and $$ \int_{0}^{1} \frac {\ln \left( x \right)}{1 - x} \text {d} x = \int_{0}^{1} \frac {\ln \left( 1 - x \right)}{x} \text {d} x. $$ Both of the remaining integrals are classic. Recall $$ \ln \left( 1 - t \right) = - \sum_{k \ge 1} \frac {{t}^{k}}{k}. $$ This is the Maclaurin series representation for the natural logarithm. Let $t = a x$, so $$ \frac {\ln \left( 1 - a x \right)}{x} = - \sum_{k \ge 1} \frac {{a}^{k}}{k} \cdot {x}^{k - 1}. $$ Integrate both sides: $$ \int_{0}^{1} \frac {\ln \left( 1 - a x \right)}{x} \text {d} x = - \sum_{k \ge 1} \frac {{a}^{k}}{{k}^{2}}. $$ If $a = 1$, then $$ \int_{0}^{1} \frac {\ln \left( 1 - x \right)}{x} \text {d} x = - \sum_{k \ge 1} \frac {1}{{k}^{2}} = - \frac {{\pi}^{2}}{6}. $$ Above, we have used the well-known answer to the Basel problem. Linked is a derivation; specifically, it is Tom Apostol's derivation, which is my personal favourite.

If $a = -1$, then $$ \begin{align} \int_{0}^{1} \frac {\ln \left( 1 + x \right)}{x} \text {d} x & = - \sum_{k \ge 1} \frac {{\left( - 1 \right)}^{k}}{{k}^{2}} \\ & = - \sum_{k \ge 1} \frac {1}{{\left( 2 k \right)}^{2}} + \sum_{k \ge 1} \frac {1}{{\left( 2 k + 1 \right)}^{2}} \\ & = - \sum_{k \ge 1} \frac {2}{{\left( 2 k \right)}^{2}} + \sum_{k \ge 1} \frac {1}{{k}^{2}} \\ & = \frac {1}{2} \sum_{k \ge 1} \frac {1}{{k}^{2}} \\ & = \frac {{\pi}^{2}}{12}. \end{align} $$ So, finally, $$ \int_{0}^{\infty} \frac {\ln \left( 2 {e}^{x} - 1 \right)}{{e}^{x} - 1} \text {d} x = \frac {{\pi}^{2}}{12} + \frac {{\pi}^{2}}{6} = \boxed {\frac {{\pi}^{2}}{4}}. $$ I am not sure if all of this can possibly be done within 5 minutes. (I personally can't: I am typically slow. I am not built for mathematical competitions.)

Answer 5

$$\begin{align} I&=\int_0^\infty\frac{\ln(2e^x-1)}{e^x-1}dx\\ &\stackrel{x=\ln(\tfrac{u+1}2)}{=}\int_1^\infty\frac{2\ln u}{u^2-1}du\\ &\stackrel{u=\frac1v}{=}\int_0^1\frac{2\ln v}{v^2-1}dv\\ &\stackrel{v=e^{-y}}{=}\int_0^\infty\frac{2ye^y}{e^{2y}-1}dy\\ &=\int_0^\infty\frac{2y}{e^y+1}dy+\int_0^\infty\frac{2y}{e^{2y}-1}dy\\ &=2\eta(2)+\frac12\zeta(2)\\ &=2(\tfrac{\pi^2}{12})+\frac12(\tfrac{\pi^2}6)\\ &=\frac{\pi^2}4. \end{align}$$ where $\eta(s)$ and $\zeta(s)$ are the famous Dirichlet eta and Rieamann zeta functions.

Answer 6

You can proceed with your last attempted integral as follows, though note that you had omitted a factor of $2$.

Substitutions:

$$\int_0^\infty \frac{\log(x+1)}{x(x+2)} \, dx \stackrel{x\to x-1}= \int_1^\infty \frac{\log x}{(x-1)(x+1)} \stackrel{x\to\tfrac1x}= \int_0^1 \frac{\log x}{x^2-1} \, dx$$

Series:

$$- \sum_{n\ge0} \int_0^1 x^{2n} \log x \, dx = \sum_{n\ge0} \frac1{(2n+1)^2}$$

The last sum is easy if you're familiar with the Basel problem.

Answer 7

Let’s consider the parametrised integral $$ I(a)=\int_0^{\infty} \frac{\ln \left(a\left(e^x-1\right)+1\right)}{e^x-1} d x $$ Differentiating $I(a)$ w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \frac{e^{-x}}{a+(1-a) e^{-x}} d x \\ & =\frac{1}{a-1}\left[\ln \left(a+(1-a) e^{-x}\right)\right]_0^{\infty} \\ & =\frac{\ln a}{a-1} \end{aligned} $$ Integrating back, we have $$ \int_0^{\infty} \frac{\ln \left(2e^x-1\right)}{e^x-1} d x =I(2)-I(0) =\int_0^2 \frac{\ln a}{a-1} d a \stackrel{u=a-1}{=} \int_{-1}^1 \frac{\ln (u+1)}{u} d u=\frac{\pi^2}{4} $$