Can Tabular Integration Be Used for Integrating Functions That Go to Zero?

[b2386]

How do I integrate this? It is not of any form with which I am familiar.

$$\int_{-\infty}^{\infty}x^2e^{-b^2x^2}dx$$

 

[courtrigrad]

$$\int_{-\infty}^{\infty}x^2e^{-b^2x^2}dx = 2\int_{0}^{\infty} x^2e^{-b^2x^2}dx$$ because the function is even.

Then use integration by parts.

 

[quasar987]

There is a swifter method. See

https://www.physicsforums.com/showthread.php?t=139196

 

[b2386]

Thanks for the replies.

hmmm...

Quasar, could you please explain the steps in your link. I looked over it but do not fully understand what is going on. Let's start here: $$\int_{-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2}} = \sqrt{2 \pi \sigma^2}$$

How do we know this?

 

[quasar987]

By parts it's u=x, dv=xexp{...}dx

Of course if you do not know a priori that

$$\int_{-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2}} = \sqrt{2 \pi \sigma^2}$$,

this complicates things, as you won't be able to do the integral by parts either. Is this for a QM course? In this case, it is probably assumed that you know the result of the above integral. If you don't and want to know how to do it, here it is roughly. It's a beautiful and ingenious trick, watch. If I remember correctly, it goes like...

$$\int_{-\infty}^{+\infty} e^{-b^2 x^2}dx=\sqrt{\left( \int_{-\infty}^{+\infty} e^{-b^2 x^2}dx \right)^2} = \sqrt{\left( \int_{-\infty}^{+\infty} e^{-b^2 x^2}dx \right)\left( \int_{-\infty}^{+\infty} e^{-b^2 y^2}dy \right)}=\sqrt{ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{-b^2( x^2+y^2)}dx dy }$$

Now switch to polar coordinates: r²=x²+y², $dxdy=rdrd\theta$ r going from 0 to infinity, theta going from 0 to 2pi [in order to cover the whole R² plane]:

$$\int_{-\infty}^{+\infty} e^{-b^2 x^2}dx = \sqrt{\int_{0}^{+\infty}\int_{0}^{+2\pi} re^{-b^2( r^2)}dr d\theta }=\sqrt{2\pi\int_{0}^{+\infty}re^{-b^2( r^2)}dr}$$

Set u=b²r² ==> du=2rb²dr ==>rdr=du/2b²:

$$\int_{-\infty}^{+\infty} e^{-b^2 x^2}dx=\sqrt{\frac{\pi}{b^2}\int_{0}^{+\infty}e^{-u}du} = \sqrt{\frac{\pi}{b^2}\left[-e^{-u}\right]_0^{+\infty}}= \sqrt{\frac{\pi}{b^2}\left[0-(-1)\right]} = \sqrt{\frac{\pi}{b^2}}$$

Phew that was longer than I tought it would be!

 

[SNOOTCHIEBOOCHEE]

Tabular integration is the easiest method here. when you think your gunna use integration by parts, if one of the functions (when differentiated many times) goes to zero, then tabular integration will work. just look it up. its easy enough to understand