I am working for an investment institution and we need to use upper partial moments. To evaluate them, I need to integrate
$$ \int_a^\infty x^2 e^{-ax^2} dx $$
with $a>0$. I found this formula online :
$$ \int x^2 e^{-ax^2} dx = \frac{1}{4} \sqrt{\frac{\pi}{a^3}}erf \left(x\sqrt{a} \right) - \frac{x}{2a}e^{-ax^2} $$
I was hoping someone can help me prove it so I can be convinced before I use it. Thanks!
We have:
$$\int_0^ue^{-t^2}\mathrm dt=\frac{\sqrt\pi}2\mbox{erf}(u)$$
Letting $t=x\sqrt a$, $u=w\sqrt a$:
$$\int_0^we^{-ax^2}\mathrm dx=\frac{\sqrt\pi}{2\sqrt a}\mbox{erf}\left(\frac{w}{\sqrt a}\right)$$
Also:
$\displaystyle\quad\int\mbox{erf}(x)\ \mathrm dx$
$\displaystyle=x\mbox{erf}(x)-\int x\ \mathrm d\left(\mbox{erf}(x)\right)$
$\displaystyle=x\mbox{erf}(x)-\frac2{\sqrt\pi}\int xe^{-x^2}\ \mathrm dx$
$\displaystyle=x\mbox{erf}(x)+\frac1{\sqrt\pi}\int e^{-x^2}\ \mathrm d(-x^2)$
$\displaystyle=x\mbox{erf}(x)+\frac1{\sqrt\pi}e^{-x^2}+C$
The rest of the proof is left to the reader as an exercise.
This is the most common asked question. If you google it, you will find many links explain different ways to solve such integrals.
Let us try ourselves,
\begin{align}
\int{x^2 e^{(-a x^2)}}dx=\int{x (x e^{(-a x^2)})}dx=x\int{x e^{(-a x^2)}dx}-\int{\int{x e^{(-a x^2)}dx}}dx
\end{align}
Now we know that $\int{x e^{(-a x^2)}}dx=-\frac{1}{2}\,{\frac {{{\rm e}^{-a{x}^{2}}}}{a}}$. using this, we will get
\begin{align}
\int{x^2 e^{(-a x^2)}}dx=-\frac{1}{2}\,{\frac {x{{\rm e}^{-a{x}^{2}}}}{a}}+\frac{1}{4} \frac{\sqrt{\pi} erf({\sqrt{a}x})}{a^{3/2}}
\end{align}
We present first a direct approach here that applied Leibniz's Rule for Differentiating Under the Integral.
Let $I(a)$ be the function given by
$$I(a)=\int_a^\infty e^{-ax^2}\,dx \tag 1$$
Enforcing the substitution $x \to x/\sqrt{a}$ yields
$$\begin{align} I(a)&=\frac{1}{\sqrt{a}}\int_{a^{3/2}}^\infty e^{-x^2}\,dx\\\\ &=\frac12 \sqrt{\frac{\pi}{a}}\left(1-\text{erf}(a^{3/2})\right) \tag 2 \end{align}$$
where the error function $\text{erf}(x)$ is given by
$$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt$$
Differentiating the right-hand sides of $(1)$ and $(2)$ and equating results reveals
$$\begin{align} -\int_a^\infty x^2e^{-ax^2}\,dx-e^{-a^3}&=-\frac14\sqrt{\frac{\pi}{a^3}}\left(1-\text{erf}(a^{3/2})\right)-\frac34 \sqrt{\pi}\text{erf}'(a^{3/2})\\\\ &=-\frac14\sqrt{\frac{\pi}{a^3}}\left(1-\text{erf}(a^{3/2})\right)-\frac32 e^{-a^3}\tag 3 \end{align}$$
Finally, solving $(3)$ for the term of interest, we obtain
$$\bbox[5px,border:2px solid #C0A000]{\int_a^\infty x^2e^{-ax^2}\,dx=\frac14\sqrt{\frac{\pi}{a^3}}\left(1-\text{erf}(a^{3/2})\right)+\frac12 e^{-a^3}}$$
as was to be shown.
We can derive the formula in question in a similar way. Note that we can write
$$\begin{align} J(a)&=\int e^{-ax^2}\,dx\\\\ &=\frac12\sqrt{\frac{\pi}{a}}\text{erf}(\sqrt{a}\,x)+C(a) \end{align}$$
where $C(a)$ is an integration constant. Now, upon differentiating with respect to $a$, we find
$$\begin{align} -\int x^2 e^{-ax^2}\,dx&=-\frac14 \sqrt{\frac{\pi}{a^3}}\text{erf}(\sqrt{a}\,x)+\frac{x}{2a}e^{-ax^2}+C'(a) \tag 4\\\\ \end{align}$$
whereupon solving $(4)$ for the term of interest yields
$$\int x^2 e^{-ax^2}\,dx=\frac14 \sqrt{\frac{\pi}{a^3}}\text{erf}(\sqrt{a}\,x)-\frac{x}{2a}e^{-ax^2}-C'(a)$$
When evaluating the indefinite integral between lower and upper limits of integration, the contribution from $C'(a)$ is zero. Therefore, we can drop the integration constant and simply write
$$\bbox[5px,border:2px solid #C0A000]{\int x^2 e^{-ax^2}\,dx=\frac14 \sqrt{\frac{\pi}{a^3}}\text{erf}(\sqrt{a}\,x)-\frac{x}{2a}e^{-ax^2}}$$
