Solve $\int_0^{\infty} x^2 \frac{x}{\theta}e^{\frac{x^2}{2\theta}} dx $

Question

Solve $$\int_0^{\infty} x^2 \frac{x}{\theta}e^{\frac{x^2}{2\theta}} dx $$

Solution: $$\int_0^{\infty} x^2 \frac{x}{\theta}e^{\frac{x^2}{2\theta}} dx = - \int_0^\infty x^2 de^{\frac{-x^2}{2\theta}}dx^2 $$

By partial integration:

$$= \int_0^\infty e \frac{-x^2}{2\theta}dx^2$$

And taking substitution $u=\frac{x^2}{2\theta}$:

$$ = 2\theta \int_0^\infty e^{-u} du = 2\theta$$

Questions

I do not understand the solution provided, threfore I ask for a detailed breakdown which explains issues such as:

1. In $- \int_0^\infty x^2 de^{\frac{-x^2}{2\theta}}dx^2$ - what is the meaning of $de$?
2. Does $dx^2$ Mean that the whole expression is to be integrated twice?

Answer 1

Answers:

1. I think that actually means $$ \newcommand\diff{\,\mathrm d} \newcommand\e{\mathrm e} \diff \left( \e^{-x^2/2\theta} \right), $$ while $$ \diff \e^{-x^2/2 \theta} $$ does not make much sense;
2. $\diff x^2$ is a short hand for $(\diff x)^2$, while the differential of $x^2$ should be $\diff (x^2)$. I think the "solution" means the latter.

Additionally, the integral you gave does not converge, so I assume that you are actually requested to compute $$ \int_0^{+\infty} x^2 \cdot \frac x \theta \e^{-x^2/(2\theta)} \diff x, $$ for $\theta>0$ which we could compute as the following: $$ \int_0^{+\infty} x^2 \cdot \frac x \theta \e^{-x^2/(2\theta)} \diff x = \int_0^{+\infty} x^2 \e^{-x^2/(2\theta)} \diff \left(\frac {x^2}{2\theta}\right) = -\int_0^{+\infty} x^2 \diff (\e^{-x^2/(2\theta)}) = \left. x^2\e^{-x^2/(2\theta)}\right\vert_{+\infty}^0 + \int_0^{+\infty} \e^{-x^2/(2\theta)} \diff (x^2) = \cdots $$ and the rest is left to you.

P.S. This is actually a question of "improper integrals", so maybe the tag could be changed to this kind.

Answer 2

1) This doesn't make much sense, you can obtain the result using:

$\int \frac{x^2 x \left(\exp x^2\right)}{(2 \theta ) \theta } \, dx=\text{constant}+e^{\frac{x^2}{2 \theta }} \left(x^2-2 \theta \right)$

For this, you will use the same substitution $u = \frac{x^2}{2 \theta}$ and integrate it by parts.

Then you evaluate it using the fundamental theorem of calculus. The infinity limit is evaluated taking the limit of $x\rightarrow \infty$.

$ \int_0^{\infty } \frac{x^2 x \left(\exp x^2\right)}{(2 \theta ) \theta } \, dx=2 \theta $

Which is only true if $\text{Re}(\theta)<0$. Otherwise the integral diverges.

2) Means that the integral is respectively to $x^2$. Just like:

$\int \sin(x)d\sin(x)=\frac{\sin^2(x)}{2}+C$