My question is regarding a integral that´s giving me a huge headache.
I want to show $$\int_{0}^{\infty}y^2e^{-\frac{y^2}{2}}dy=\sqrt{\frac{\pi}{2}}$$
I'm studying for an exam. I'm suppose to find the integral of $(x^2/c^2)\, e^{-x^2/(2c^2)}$ between $0$ and $\infty$.
So first i substituted $x/c = y$ which makes the integral simpler. Then I'm stuck, I'm guessing that the equation that appears when substituting like this, is well know(a standard integral used a lot) which is equal to integral of pi/2.
$\endgroup$ 63 Answers
$\begingroup$Hint:
note that: $$ \int x^2e^{-x^2}dx=\int x\cdot\left(xe^{-x^2}\right)dx=\int x d\left( -\frac{e^{-x^2}}{2}\right) $$
and integrate by parts using the fact that $$ \int e^{-x^2}dx=\frac{\sqrt{\pi}}{2}\mbox{erf}(x)+C $$ where $\mbox{erf}(x)$ is the Error function.
$\endgroup$ 2 $\begingroup$To expand upon one of the earlier comments, you can use differentiation under the integral sign. Notice
$$\int_{0}^{\infty}y^2{e^{\frac{-y^2}{2}}}dy=\int_{0}^{\infty}-2\lim_{t \rightarrow 1} \frac{\partial}{\partial t}{e^{\frac{-ty^2}{2}}}dy=-2\lim_{t \rightarrow 1} \frac{\partial}{\partial t}\int_{0}^{\infty}{e^{\frac{-ty^2}{2}}}dy.$$
To solve for $$\int_{0}^{\infty}{e^{\frac{-ty^2}{2}}}dy,$$ you can square the integral:
$$\int_{0}^{\infty}\int_{0}^{\infty}{e^{\frac{-t(x^2+y^2)}{2}}}dxdy,$$ and convert to polar coordinates:
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}r{e^{\frac{-tr^2}{2}}}drd\theta=\int_{0}^{\frac{\pi}{2}}\frac{1}{t}d\theta=\frac{\pi}{2t}.$$
This means $$\int_{0}^{\infty}{e^{\frac{-ty^2}{2}}}dy=\sqrt{\frac{\pi}{2t}}.$$
Now compute $$-2\lim_{t \rightarrow 1} \frac{\partial}{\partial t}\sqrt{\frac{\pi}{2t}}=\frac{{}{\sqrt{2\pi}}}{2}=\sqrt{\frac{\pi}{2}}$$ which is your answer.
$\endgroup$ $\begingroup$Let $u=e^{-y^2/2}$, then $\mathrm{d}u=-ye^{-y^2/2}$
$$
\begin{align}
\int_0^\infty y^2e^{-y^2/2}\,\mathrm{d}y
&=-\int_0^\infty y\,\mathrm{d}u\tag{1}\\
&=\int_0^\infty u\,\mathrm{d}y\tag{2}\\
&=\int_0^\infty e^{-y^2/2}\,\mathrm{d}y\tag{3}\\
\end{align}
$$
Explanation:
$(1)$: substitute $u=e^{-y^2/2}$
$(2)$: integration by parts
$(3)$: undo the substitution
Now if we have $A=\int_0^\infty e^{-y^2/2}\,\mathrm{d}y$, we can square to get
$$
\begin{align}
A^2
&=\int_0^\infty e^{-x^2/2}\,\mathrm{d}x\int_0^\infty e^{-y^2/2}\,\mathrm{d}y\tag{4}\\
&=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)/2}\,\mathrm{d}x\,\mathrm{d}y\tag{5}\\
&=\int_0^\infty\int_0^{\pi/2}e^{-r^2/2}r\,\mathrm{d}\theta\,\mathrm{d}r\tag{6}\\
&=\frac\pi2\int_0^\infty e^{-r^2/2}r\,\mathrm{d}r\tag{7}\\
&=\frac\pi2\left[-e^{-r^2/2}\right]_0^\infty\tag{8}\\
&=\frac\pi2\tag{9}
\end{align}
$$
Explanation:
$(4)$: square the integral
$(5)$: combine the two integrals (since they are absolutely convergent)
$(6)$: convert to polar coordinates
$(7)$: integrate in $\theta$
$(8)$: integrate in $r$
$(9)$: evaluate
Combining the two results we get above yields $$ \int_0^\infty y^2e^{-y^2/2}\,\mathrm{d}y=\sqrt{\frac\pi2}\tag{10} $$
$\endgroup$ 4
