Evaluate $\int_{0}^{\infty} \frac{x^2 + \sin(\pi x)}{1 + \exp(\pi x)} \,dx$

Question

\begin{align*} \int_{0}^{\infty} \frac{x^2 + \sin(\pi x)}{1 + \exp(\pi x)} \,dx \end{align*}

\begin{align*} &\int_{0}^{\infty} \frac{x^2 + \sin(\pi x)}{1 + e^{\pi x}} \,dx \\ &= \int_{0}^{\infty} \frac{x^2}{1 + e^{\pi x}} \,dx + \int_{0}^{\infty} \frac{\sin(\pi x)}{1 + e^{\pi x}} \,dx \\ &= \Omega_{1} + \Omega_{2} \end{align*}

\begin{align*} \Omega_{1} &= \int_{0}^{\infty} \frac{x^2}{1 + e^{\pi x}} \,dx \cdot \left\{e^{\pi x} = \frac{1}{t}, \, \pi x = -\ln(t), \, dx = -\frac{1}{\pi t}\right\} \\ &= -\frac{1}{\pi^3} \int_{0}^{1} \frac{\ln^2(t)}{1 + \frac{1}{t}} \cdot \frac{1}{t} \,dt \\ &= -\frac{1}{\pi^3} \int_{0}^{1} \frac{\ln^2(t)}{1 + t} \,dt \\ &= -\frac{1}{\pi^3} \sum_{n=0}^{\infty} (-1)^n \int_{0}^{1} t^n \ln^2(t) \,dt \end{align*}

\begin{align*} \text{cong} &= -\frac{2}{\pi^3} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n + 1)^3} \\ &= \frac{3\zeta(3)}{2\pi^3} \end{align*}

Answer 1

$$ \Omega_2=\frac{1}{\pi}\int_0^\infty \frac{\sin x}{1+e^x}dx =\frac{1}{\pi}\int_0^\infty \frac{e^{-x}\sin x}{1+e^{-x}}dx $$ the geometric series is $$ =\frac{1}{\pi}\sum_{k\ge 0}(-)^k\int_0^\infty e^{-(k+1)x}\sin x dx =\frac{1}{\pi}\sum_{k\ge 0}(-)^k \frac{1}{1+(k+1)^2} $$ by partial fration composition, with $i=\sqrt{-1}$, the imaginary unit $$ \sum_{k\ge 0} (-1)^k \frac{1}{2+2k+k^2} = \sum_{k\ge 0} (-1)^k \frac{1}{(k+1-i)(k+1+i)} = \frac{i}{2}\sum_{k\ge 0} (-1)^k [\frac{1}{k+1+i}-\frac{1}{k+1-i}] $$ $$ = \frac{i}{2}[\frac12 \psi(1+i/2)-\frac12 \psi(1/2+i/2)- \{\frac12 \psi(1-i/2)-\frac12 \psi(1/2-i/2)\}] $$ $$ = -\frac{1}{2}\Im [\frac12 \psi(1+i/2)-\frac12 \psi(1/2+i/2)- \frac12 \psi(1-i/2)+\frac12 \psi(1/2-i/2)] $$ $$ = -\frac{1}{2}[\Im \frac12 \psi(1+i/2)-\frac12 \frac{\pi}{2}\tanh\frac{\pi}{2}- \Im \frac12 \psi(1-i/2)+\frac12 \frac{\pi}{2}\tanh\frac{-\pi}{2}] $$ $$ = -\frac{1}{2}[\Im \frac12 \psi(1+i/2)-\frac{\pi}{2}\tanh\frac{\pi}{2}- \Im \frac12 \psi(1-i/2)] $$ $$ = -\frac{1}{4}[\Im \psi(1+i/2)-\pi\tanh\frac{\pi}{2}- \Im \psi(1-i/2)] $$ $$ = -\frac{1}{4}\{\Im [\psi(i/2)+\frac{1}{i/2}]-\pi\tanh\frac{\pi}{2}- \Im [\psi(-i/2)+\frac{-1}{i/2}]\} $$ $$ = -\frac{1}{4}\{\Im \psi(i/2)-2-\pi\tanh\frac{\pi}{2}- \Im \psi(-i/2)-2\} $$ $$ = -\frac{1}{4}\{\frac{1}{2\times 1/2}+\frac{\pi}{2}\coth\frac{\pi}{2}-2-\pi\tanh\frac{\pi}{2}- (\frac{1}{2\times (-1/2)}+\frac{\pi}{2}\coth\frac{-\pi}{2}) -2\} $$ $$ = -\frac{1}{4}[-2-\pi\tanh\frac{\pi}{2}+\pi\coth\frac{\pi}{2}]. $$ $$ \Omega_2=\frac{1}{4\pi}[2+\pi\tanh \frac{\pi}{2}-\pi\coth \frac{\pi}{2}] \approx 0.11586017\ldots $$ For $\sum_{n\ge 0} (-1)^n/(1+n^2)$ see e.g. the Kotesovec formula of 2015 in https://oeis.org/A002522 .

Answer 2

Both of these integrals can be solved together pretty easily with geometric series. Consider

$$\frac{x^2+\sin\pi x}{1+e^{\pi x}} = \frac{(x^2+\sin\pi x)e^{-\pi x}}{1+e^{-\pi x}}= \sum_{n=1}^\infty (-1)^{n+1}e^{-n\pi x}(x^2+\sin\pi x)$$

The first integral is similar to how you computed it, just simpler without the intermediary substitution

$$\Omega_1=\sum_{n=1}^\infty(-1)^{n+1}\int_0^\infty x^2e^{-n\pi x}dx = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{\pi^3n^3}\int_0^\infty t^2e^{-t}dt$$

$$=\frac{2}{\pi^3}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^3} = \frac{2}{\pi^3}\left(-\zeta(3)+\frac{1}{4}\zeta(3)\right)=-\frac{3\zeta(3)}{2\pi^3}$$

by recognizing the $2!$ and subtracting off twice the sum of the even terms. The second piece then is

$$\Omega_2 = \sum_{n=1}^\infty(-1)^{n+1}\int_0^\infty\operatorname{Im}\{e^{(-n+i)\pi x}\}dx$$ $$ = \sum_{n=1}^\infty \operatorname{Im}\left\{\frac{(-1)^{n+1}}{n-i}\right\} =\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2+1}$$