Question

Is it to do with \(\color{green}{\textsf{Gamma Function}}\) ??

\[\color{blue}{\int\limits_{-\infty}^{\infty}(e^{-x^2} \cdot x^2) dx}\]

How to solve it??

Answer 1

\[\color{red}{\Gamma(n) = \int\limits_0^{\infty} e^{-x} \cdot x^{(n-1)} \cdot dx}\]

Answer 2

Is this right for Gamma Function??

Answer 3

yes @waterineyes

Answer 4

I am here only.. :)

Answer 5

If in my question I put : \(x^2 = t\), then \(2x \cdot dx = dt\)

But, then what will be the limits for t ??

Answer 6

I get limits from : Infinity from Infinity, and here I got stuck....

Answer 7

The function is even, so you can do
\[ \begin{align} \int_{-\infty}^{\infty}e^{-x^2}x^2dx&=2\int_{0}^{\infty}e^{-x^2}x^2dx \end{align}\]

Answer 8

then using your substitution will be okay :) it will still be bounds of 0 to \(\infty\) and you can use the gamma function

Answer 9

Wow..!!

Answer 10

One more doubt, I have. I have further problem in solving it..

Answer 11

The function is even, so you can do
\[ \begin{align} \int_{-\infty}^{\infty}e^{-x^2}x^2dx&=2\int_{0}^{\infty}e^{-x^2}x\cdot dx \\&=2\int_{0}^{\infty}\end{align}\]
\(u = x^2\)
\(du = 2x dx \implies \dfrac{du}{2}=xdx\)

Answer 12

By substitution, I will get:

\[2 \int\limits_{0}^{\infty} e^{-t} \cdot \sqrt{t} \cdot dt = \Gamma(\frac{3}{2})\]

Is this right or not?

Answer 13

oops well ipressed enter too soon :)

Answer 14

yeah I get Gamma(3/2)

Answer 15

Then??

\[\Gamma (\frac{3}{2}) = \frac{3}{2} \Gamma(\frac{1}{2})\] ??

Answer 16

\[\implies \frac{3 \sqrt{\pi}}{2}\]

Answer 17

Or I have done it too fast, there must be a mistake.. :(

Answer 18

\(\Gamma \left( \dfrac{3}{2}\right)=\dfrac{1}{2}\Gamma \left( \dfrac{1}{2}\right)=\dfrac{\sqrt{\pi}}{2}\)
since
\(\Gamma(n)=(n-1)\Gamma(n-1)\)

Answer 19

\[\Gamma(n+1) = n \Gamma(n)\]

Answer 20

Sorry, Okay Okay.. I got it..

\[\Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2}\]

Answer 21

n is half there not 3/2..

Answer 22

\(\Large\checkmark \)

Answer 23

\[\color{green}{\checkmark}\]

Answer 24

@AccessDenied you want to add something??

Answer 25

I had a different route I was writing up. Just checking it over. :P

Answer 26

I did not check for even or odd, that's why it was looking typical to me.. :)

Answer 27

I won't mind, if we can go by that route now.. :)

Answer 28

This one involves the Gaussian integral with a parameter. Basically using the method for differentiation under the integral.
\( \displaystyle
\begin{align} \int_{-\infty}^{\infty} e^{-tx^2} \cdot x^2 \ dx &= -\int_{-\infty}^{\infty} \frac{\partial}{\partial t} \left( e^{-tx^2} \right) \ dx \\
&= \dfrac{d}{dt} \left(-\int_{-\infty}^{\infty} e^{-tx^2} \ dx \right) \\
&= \dfrac{d}{dt} \left(-\sqrt{\dfrac{\pi}{t}}\right) \\
&= \dfrac{d}{dt} \left(-\sqrt{\pi} \cdot t^{-1/2} \right) \\
&= \sqrt{\pi} \dfrac{1}{2} t^{-3/2} \\
\end{align} \)
From there we can substitute t = 1 for the original integral, which gives back sqrt(pi)/2 again
It was actually the first thing I thought of before I saw the above.

Answer 29

^nice method =]

Answer 30

Sorry @kirbykirby you content with the 3, this is for @AccessDenied ..

Answer 31

I will keep it in my notes.. :) Thank you both @kirbykirby and @AccessDenied ..

Answer 32

Glad to share! I've always liked that method after reading this online document and seeing some of its real power:
http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf

Answer 33

I will definitely favourite that page