Solve $\int_0^\infty\frac x{e^x-e^{\frac x2}}dx$

Question

I was able to solve the integral $$\int_0^\infty\frac x{e^x-e^\frac x2}dx=4\left(\frac{\pi^2}6-1\right)$$ I want to see other approaches to solving it. Here is my solution: $$\int_0^\infty\frac x{e^x-e^\frac x2}dx=\int_0^\infty\frac{xe^{-x}}{1-e^{-\frac x2}}dx=\int_0^\infty xe^{-x}\sum_{k=0}^\infty e^{-\frac{kx}2}dx=\sum_{k=0}^\infty\int_0^\infty xe^{-(1+\frac{kx}2)x}dx=\sum_{k=0}^\infty\frac1{(1+\frac k2)^2}\int_0^\infty xe^{-x}dx=\sum_{k=0}^\infty\frac1{(1+\frac k2)^2}=\sum_{k=0}^\infty\frac{4}{(k+2)^2}=4\left(\frac{\pi^2}6-1\right)$$I came up with this integral when I was trying to prove that $\int_0^\infty\frac x{e^x-1}dx$ converges using the comparison test. I replaced the $1$ with $e^x/2$ while checking on Desmos whether this integral exists or not, but I put $e^{x/2}$ instead. I decided to have a go at this integral.

Answer 1

Less simple that @Svyatoslav's solution.

Working the antiderivative $$I=\int\frac x{e^x-e^{\frac x2}}\,dx$$ $$x=2\log(t)\quad \implies \quad I=4\int \frac{ \log (t)}{(t-1) t^2} \,dt$$ that is to say $$I=4\int \left(\frac{1}{t-1}-\frac{1}{t^2}-\frac{1}{t} \right)\log(t)\,dt$$ Using integration by parts $$I=4 \left(\frac{1}{t}-\frac{1}{2} \log ^2(t)+\frac{\log (t)}{t}-\text{Li}_2(1-t)\right)$$ Back to $x$ $$I=-4 \text{Li}_2\left(1-e^{x/2}\right)-\frac{x^2}{2}+2 e^{-x/2} (x+2)$$

$$J=\int_0^\infty\frac x{e^x-e^{\frac x2}}\,dx=\frac{2 \pi ^2}{3}-4=4 \left(\frac{\pi ^2}{6}-1\right)$$

Working asymptotics $$K=\int_0^t\frac x{e^x-e^{\frac x2}}\,dx$$ $$K=4 \left(\frac{\pi ^2}{6}-1\right)+2 t\,e^{-t/2}+e^{-t}-\frac{4}{9} e^{-3 t/2}+\frac 14 e^{-2t}+\cdots $$ whose absolute error is less than $0.01$ if $t \geq 16$ and less than $0.001$ if $t \geq 22$.

Answer 2

$$\int_0^\infty\frac x{e^x-{e^{x/2}}}dx=4\int_0^\infty \frac{xe^{-x}}{e^x-1}dx\\ =4\int_0^\infty\left(\frac x{e^x-1}-xe^{-x}\right)dx=4\left(\frac{\pi^2}{6}-1\right)$$

Answer 3

Map $x \mapsto 2x$ so that

$$I := \int_{0}^{\infty}\frac{x}{e^{x}-e^{\frac{x}{2}}}dx = 4\int_{0}^{\infty}\frac{x}{e^{2x}-e^{x}}dx\,.$$

Let $\displaystyle f(z) = \frac{z^2}{e^{2z}-e^z}$. Its set of poles is $\left\{z \in \mathbb{C} \text{ } \wedge \text{ } n \in \mathbb{Z} \backslash \left\{0\right\}: z = 2\pi ni\right\}$. There is a removable pole at the origin because the principal part of the Laurent expansion of $f(z)$ equals $0$. Let $r \ll \pi \ll R$. We integrate along the following contour in the positive direction.

Cauchy's Integral Theorem allows us to write $\displaystyle \oint_C f(z)dz$ as

$$ 0 = \left(\int_{0}^{R}+\int_{R}^{R+2\pi i}+\int_{R+2\pi i}^{r+2\pi i}+\int_{r+2\pi i}^{2\pi i-ir}+\int_{2\pi i-ir}^{0}\right)f(z)dz \,. $$

We equate the imaginary part to avoid divergence issues and then apply $r \to 0^+$ and $R \to \infty$ on both sides.

The first integral vanishes:

$$ I_1 := \lim_{R\to\infty} \Im \int_{0}^{R}\frac{z^{2}}{e^{2z}-e^{z}}dz=0\,. $$

The second integral also vanishes. Note that

$$ \left|\Im \int_{R}^{R+2\pi i}\frac{z^{2}}{e^{2z}-e^{z}}dz\right| \leq \left|\int_{R}^{R+2\pi i}\frac{z^{2}}{e^{2z}-e^{z}}dz\right|\le MR $$

where $R$ is the length of the contour and $M$ is a sufficiently tight upper bound of $|f(z)|$. Parameterize $z=R+iy$ where $y \in [0,2\pi]$. We get that $M$ by

$$ \begin{align} |f(z)| &= \frac{\left|z\right|^{2}}{\left|e^{2z}-e^{z}\right|} \\ &\leq \frac{\left|z\right|^{2}}{\left|\left|e^{2z}\right|-\left|e^{z}\right|\right|} \\ &= \frac{\left|R+iy\right|^{2}}{\left|\left|e^{2\left(R+iy\right)}\right|-\left|e^{R+iy}\right|\right|} \\ &\leq \frac{R^{2}+2Ry+y^{2}}{e^{2R}-e^{R}}\,. \\ \end{align} $$

We can use the Estimation Lemma and the Squeeze Theorem to prove that

$$ I_2 := \lim_{R\to\infty} \Im \int_{R}^{R+2\pi i}\frac{z^{2}}{e^{2z}-e^{z}}dz=0\,. $$

We can retrieve the integral we want from evaluating the third integral. Let $z = x + 2\pi i$ so that

$$ \begin{align} I_3 &= \lim_{R\to\infty}\lim_{r\to 0^+}\Im\int_{R+2\pi i}^{r+2\pi i}\frac{z^{2}}{e^{2z}-e^{z}}dz \\ &= -\lim_{R\to\infty}\lim_{r\to 0^+}\Im\int_{r}^{R}\frac{\left(x+2\pi i\right)^{2}}{e^{2\left(x+2\pi i\right)}-e^{x+2\pi i}}dx \\ &= 4\pi^2 \lim_{R\to\infty}\lim_{r\to 0^+}\Im\int_{r}^{R}\frac{dx}{e^{2x}-e^{x}} - 4\pi\lim_{R\to\infty}\lim_{r\to 0^+}\Im i \int_{r}^{R}\frac{xdx}{e^{2x}-e^{x}} - \lim_{R\to\infty}\lim_{r\to 0^+}\Im\int_{r}^{R}\frac{x^{2}dx}{e^{2x}-e^{x}} \\ &= 0 - \pi I - 0 \\ &= -\pi I \,. \\ \end{align} $$

We evaluate the fourth integral through a residue: $$ \begin{align} I_4 &:= \lim_{r\to0^+}\Im \int_{r+2\pi i}^{2\pi i-ir}\frac{z^{2}}{e^{2z}-e^{z}}dz \\ &= -\Im \frac{i\pi}{2} \mathop{\mathrm{Res}}_{z = 2\pi i} \frac{z^2}{e^{2z}-e^z} \\ &= -\frac{\pi}{2} \Re \lim_{z \to 2\pi i} \frac{z^2}{\frac{d}{dz} \left(e^{2z}-e^z\right)} \\ &= 2\pi^3 \,. \end{align} $$ For the last integral, parameterize $z=iy$ so that $$ \begin{align} I_5 &:= \lim_{r \to 0^+}\Im \int_{2\pi i-ir}^{0}\frac{z^{2}}{e^{2z}-e^{z}}dz \\ &= -\lim_{r \to 0^+}\Im i \int_{0}^{2\pi-r}\frac{\left(iy\right)^{2}}{e^{2iy}-e^{iy}}dy \\ &= \lim_{r \to 0^+} \Re \int_{0}^{2\pi-r}\frac{y^{2}}{e^{2iy}-e^{iy}}dy \\ &= \lim_{r \to 0^+} \Re\int_{0}^{2\pi-r}\left(-\frac{1}{2}y^{2}\left(2\cos y+1\right)-\frac{i}{2}y^{2}\cos\left(\frac{3y}{2}\right)\csc\left(\frac{y}{2}\right)\right)dy \\ &= -\frac{1}{2}\int_{0}^{2\pi}y^{2}\left(2\cos y+1\right)dy-0 \\ &= -\frac{4}{3}\pi^{3}-4\pi \,. \\ \end{align} $$

We gather the contributions together and conclude with

$$ 0 = 0+0-\pi I + 2\pi^3 - \frac{4}{3}\pi^3-4\pi $$ $$ \implies I = \frac{2}{3} \pi^2 - 4 $$

and we're done!