Integration by parts of a Gaussian function.

Question

I've attempted the following question which involves integration by parts. I've only recently learnt come across this skill, which means there is a very large chance that I'm doing something wrong.

Find $$\int_{-∞}^{∞} Ax^2e^{-x^2/2σ^2} \,dx$$ given that $$\int_{-∞}^{∞} Ae^{-x^2/2σ^2} \,dx =1$$

Using the general form for integration by parts, $$\int_{-∞}^{∞} f(x)g'(x) \,dx=f(x)g(x)-\int_{-∞}^{∞} g(x)f'(x) \,dx$$, I let $f(x)=x^2$, and $g'(x)=Ae^{-x^2/2σ^2}$. Therefore, $f'(x)=2x$ and $g(x)=1$. Substituting these in to the formula, I get $$\int_{-∞}^{∞} Ax^2e^{-x^2/2σ^2} \,dx=x^2-\int_{-∞}^{∞}2x \, dx=x^2-(x^2+c)=-c$$. However, this is incorrect (I know this because I have to submit the answer and the website, which is Isaac Physics, will tell me whether it is correct or not with no explanation). I don't know where I've gone wrong.

Answer 1

Integration by parts for this improper integral reads $$ \int_{-\infty}^\infty f(x)g'(x)~dx = \lim_{b\to\infty}\lim_{a\to-\infty}f(b)g(b)-f(a)g(a) - \int_{-\infty}^\infty f'(x)g(x)~dx. $$ The middle term corresponds to the limit as the endpoints go to $\pm\infty$ of the difference $f(b)g(b)-f(a)g(a)$. The integral you start with basically tells you that $\lim_{b\to\infty}\lim_{a\to-\infty} g(b)-g(a) = 1$. If you apply integration by parts with the choices of $f$ and $g$ you started with, the middle term reads $$ \lim_{b\to\infty}\lim_{a\to-\infty}b^2g(b) - a^2g(a), $$ which does not exist. However, if you choose $f(x) = x$ and $g'(x) = Axe^{-x^2/2\sigma^2}$, you will get limits that exist and can proceed as in this answer.