How do I integrate this ?
$$\int^{ \infty }_0\frac{x^4}{e^{x^2}+1}dx$$
I just brute force integrated it , but didn't get a valid answer . I'm sure it must be done by some properties of definite integral , I'm not sure what properties though.
My attempt : https://ibb.co/b422TH
In that attempt , $1+e^{-t}=m\implies -e^{-t}dt=dm$ Minus sign was missing , but even after I corrected it and then integrated my final expression by parts , I'm still getting answer as not defined $(\infty)$
$$A=\int_{0}^{\infty}\frac{x^4}{e^{x^2}+1} dx = \int_{0}^{\infty} \frac{x^4 e^{-x^2}}{1+e^{-x^2}} dx\\ =\int_{0}^{\infty}\sum_{n=0}^{\infty}{x^4 (-1)^ne^{-(n+1)x^2}} dx$$
In the interval, $[0, \infty]$, For all $n \in \mathbb{N}$, $|(-1)^n x^4 e^{- (n+1) x^2}| \le x^4 e^{- (n+1) x^2}$ and $\sum_{n=0}^{\infty}{x^4 e^{- (n+1) x^2}}=\frac{x^4}{e^{x^2}-1}$ which is bounded function and$\sum_{n=0}^{\infty}\int_{0}^{\infty}{x^4 e^{- (n+1) x^2}}dx = \frac{3 \sqrt{\pi}}{8}\zeta(\frac{5}{2}) < + \infty$.
Therefore, we could interchange integration and summation.
$$A=\sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty}{x^4 e^{-(n+1)x^2}} dx\\ = \sum_{n=0}^{\infty} (-1)^n\int_{0}^{\infty}{\frac{1}{2} t^{3/2} e^{-(n+1)t}} dt\\ = \sum_{n=0}^{\infty} (-1)^n \frac{3 \sqrt{\pi}}{8}\frac{1}{(1+n)^{5/2}}\\ = \frac{3 \sqrt{\pi}}{8}\sum_{n=0}^{\infty} (-1)^n \frac{1}{(1+n)^{5/2}}\\ = \frac{3 \sqrt{\pi}}{8}\eta(\frac{5}{2})\\ = \frac{3 \sqrt{\pi}}{8}(1-2^{-3/2})\zeta(\frac{5}{2}) $$