I was trying to solve the integral $$ A=\int_0^{\infty}\frac{x}{e^x + 1}\, dx. $$
Firstly, I was trying to solve the indefinite one $$ I(x)=\int\frac{x}{e^x + 1}\, dx. $$ By subtitution $t = e^x$, fraction decomposition, and lastly partial integration for the second integrand i get :
$\displaystyle I(x)=\frac{x^2}{2} - x\cdot\ln(e^x + 1) - \text{Li}_2(-e^x)$ (ignoring the constant)
Where $\text{Li}_2(-e^x) = \sum\limits_{k\geq 1} \dfrac{(-e^x)^k}{k}$
so then of course :
$A = I(x\to \infty) - I(0)$
The result that i get on calculator is $A = \dfrac{\pi^2}{12}$,
which makes sense considering $\dfrac{x}{e^x+1} \to 0$ as $x\to \infty$ (thus it has finite area).
But when i tried to do it manually by hand (with fourier series to compute $I(0)$ and Convergence test for $I(x\to \infty)$)
i get : $A = \text{divergent} - \dfrac{\pi^2}{12}$
edit : for $I(x\to\infty)$ i do L'hopital and Convergence test
$I(x\to \infty) = N-K$ where $\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-2\ln(e^x+1))}{2}$ and $\displaystyle K=\lim_{x\to\infty} \sum\limits_{k\geq1} (-1)^k \cdot \dfrac{e^{kx}}{k}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-2\ln(e^x+1))}{2} = \lim_{x\to\infty} \dfrac{x-\ln(e^x+1)}{\cfrac{2}{x}}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-\ln(e^x+1))}{2} = \lim_{x\to\infty} \dfrac{1-\dfrac{2e^x}{e^x+1}}{-\cfrac{2}{x^2}}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-\ln(e^x+1))}{2} = \dfrac{1}{2}\lim_{x\to\infty} \dfrac{\dfrac{e^x(e^x+1)-e^{2x}}{(e^x+1)^2}}{\cfrac{1}{x^3}}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-\ln(e^x+1))}{2} = \dfrac{1}{2}\lim_{x\to\infty} \dfrac{\dfrac{e^x}{(e^x+1)^2}}{\cfrac{1}{x^3}}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-\ln(e^x+1))}{2} = \dfrac{1}{2}\lim_{x\to\infty} \dfrac{x^3e^x}{(e^x+1)^2}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-\ln(e^x+1))}{2} = \dfrac{1}{2}\lim_{x\to\infty} \dfrac{(3x^2+x^3)e^x}{2(e^x+1)\cdot e^x}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-\ln(e^x+1))}{2} = \dfrac{1}{4}\lim_{x\to\infty} \dfrac{3x^2+x^3}{e^x+1}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-\ln(e^x+1))}{2} = \dfrac{1}{4}\lim_{x\to\infty} \dfrac{6x+3x^2}{e^x}$
$\displaystyle N=\lim_{x\to\infty} \dfrac{x(x-\ln(e^x+1))}{2} = \dfrac{3}{2}\lim_{x\to\infty} \dfrac{x+1}{e^x} \to 0$
and for K :
$\displaystyle K=\lim_{x\to\infty} \sum\limits_{k\geq1} (-1)^k \cdot \dfrac{e^{kx}}{k}$
$\displaystyle K= \sum\limits_{k\geq1} \dfrac{(-1)^k}{k} \cdot \lim_{x\to\infty}e^{kx} $
$ K= \sum\limits_{k\geq1} \dfrac{(-1)^k}{k} \cdot\infty = \infty$
other things i tried : using series representation of $\dfrac{1}{x+1}$, change $x$ into $e^x$, multiply the series by $x$ and then integrate it, i still get Divergent-$\dfrac{\pi^2}{12}$
Anyone can point out where it went wrong done manually ?
Edit : Resolved
So all i need to do is just turn $\dfrac{x}{e^x+1}$ into $\dfrac{xe^{-x}}{e^{-x}+1}$ then use maclaurin series for $\dfrac{1}{x+1}$