Lecture Notes

This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong.

====== September 27, Wednesday ======
Gaussian integral is of the form (modulo constant)
$$I = \int_{-\infty}^\infty e^{- x^2} dx $$
How to evaluate this? here residue theorem cannot help you. you need to use a little trick. 

Consider the double integral 
$$I^2 = \int_{-\infty}^\infty\int_{-\infty}^\infty  e^{- x^2 -y^2 } dx dy $$
then switch to radial coordinate
$$ I^2 = \int_{r=0}^{\infty}\int_{\theta=0}^{2\pi} e^{-r^2} r drd\theta = 2 \pi \int_{r=0}^{\infty} e^{-r^2} r dr$$
subsitatue $r^2=u$, we get $2rdr = du$, and
$$ I^2 = \pi \int_0^\infty e^{-u} du = \pi $$
ok, we get 
$$ I = \sqrt{\pi}. $$

OK, not bad. How about more general case? For $a>0$, consider
$$I_a = \int_{-\infty}^\infty e^{- a x^2} dx $$
we can change variable, let $u = \sqrt{a} x$, then $dx = (1/\sqrt{a}) du$, 
$$I_a = (1/\sqrt{a}) \int_{-\infty}^\infty e^{-  u^2} du = (1/\sqrt{a}) I = (1/\sqrt{a}) \sqrt{\pi}. $$

How about $a = r e^{i \theta}$? and $0<\theta<<1$? and then making $\theta$ larger, and larger?

Lecture Notes

User Tools

Site Tools

Page Tools