For $f(x)=\sum\limits_{n=1}^{\infty}nxe^{-nx^2}$, evaluate $\int\limits_{-1}^{1}f(x)\, dx$

Question

For $f(x)=\sum\limits_{n=1}^{\infty}nxe^{-nx^2}$, evaluate $\int\limits_{-1}^{1}f(x)\, dx$

What I have done:

for every $x \in \mathbb{R}, $ series is uniformly convergent. So we have $$\int\limits_{-1}^{1}f(x)\, dx=\int\limits_{-1}^{1}\sum\limits_{n=1}^{\infty}nxe^{-nx^2}=\sum\limits_{n=1}^{\infty}\int\limits_{-1}^{1}nxe^{-nx^2}=0$$

If this approach is not OK please let know. For the alternative way, please give me a hint.

Answer 1

You can see immediately from the definition of $f(x)$ that it is an odd function, i.e. $f(-x)=-f(x)$. The same result follows from the fact that integrating an odd function over a symmetrical interval will result in $0$.

Your methodology is completely fine aswell.