Evaluating an integral of the form $\int_a^{\infty}x^{2}e^{-x^{2}}dx$

Question

How do you evaluate an integral of the form $$\int_a^{\infty}x^{2}e^{-x^{2}}dx ?$$ This integral is very similar to the error function integral $$\int_a^{\infty}e^{-x^{2}}dx =\frac{\sqrt{\pi}}{2}\big(1-\text{erf}(a)\big).$$

And so I thought that you could integrate by parts the first integral somehow and end up writing it in terms of the error function, however I don't know how to go about it (how do you evaluate the boundary terms?).

Answer 1

Hint: Try this for integration by parts.

$\int u dv$

Where

$u = x$

$dv = xe^{-x^2}$

Answer 2

$$I=\int_a^{\infty}x^{2}e^{-x^{2}}dx $$ $$I=-\frac 12\int_a^{\infty}x (-2xe^{-x^{2}})dx $$ Integrate by part now $$I=-\frac 12\int_a^{\infty}x (e^{-x^{2}})'dx $$ $$I=-\frac 12( \left. (x e^{-x^{2}})\right |_a^{\infty} -\int_a^{\infty}e^{-x^{2}}dx)$$

Answer 3

You can insert a general parameter and use Feynman's Trick. Let's denote $a$ as the new inserted parameter and $b$ as the lower limit for the integral$$I(a)=\int\limits_b^{\infty}dx\, x^2e^{-ax^2}=-\frac {\partial}{\partial a}\int\limits_b^{\infty}dx\, e^{-ax^2}=-\frac {\partial}{\partial a}\left\{\sqrt{\frac {\pi}{4a}}\operatorname{erfci}\left(b\sqrt a\right)\right\}$$Expanding the parenthesis and differentiating with respect to $a$ gives us$$I(a)=\frac 1{4a}\sqrt{\frac {\pi}a}+\frac {be^{-ab^2}}{4a}$$Let $a=1$ to retrieve the desired integral$$I(1)\color{blue}{=\frac {\sqrt{\pi}}4+\frac {be^{-b^2}}4}$$