Question

Prove: \[\int_{-∞}^∞ e^{-\lambda x^2}\text{d}x =\sqrt{ \frac{π}{\lambda}}\]

Answer 1

what kind of analysis are you doing here?

Answer 2

I mean, is this from some section where you are doing integrals like this in polar coordinates? 'cause I don't know how to do that...

Answer 3

hmm well it is a Gaussian,
this is just one step in a question i am trying to solve

Answer 4

This question,
i have all the other working i just cannot see how these two terms are equivalent

Answer 5

Try the new-ish tag feature:
@Zarkon @across @JamesJ
darn, none of them are online, but they are your best bet for this type of thing
you should address your questions at them more often, due to the caliber of you problems

Answer 6

this is a pain, because you will not find an anti derivative. you have to go into another dimension if i recall.
proof is standard, so a google search will probably do it

Answer 7

i dont know what i would enter into google

Answer 8

write as a double integral and switch to polar coordinates

Answer 9

http://en.wikipedia.org/wiki/Gaussian_integral

Answer 10

@Zarkon I knew it!
gotta learn how to do those kinds of integrals

Answer 11

so first step is to recognise it as a even function , then to make in terms of the gamma function,
@Zarkon, i will try that also

Answer 12

\[\int_{-∞}^∞e^{-\lambda x^2}\text{d}x\]\[=2\int_{0}^∞ e^{-\lambda x^2} \text{d} x\] let \[\qquad \lambda x^2=t\]\[\qquad x=\sqrt{t/\lambda} \]\[\qquad \text{d}x=\frac{t^{-1/2}}{2\sqrt\lambda}\text{d}t\]\[={1 \over \sqrt{\lambda}}\int_{0}^∞ t^{-1/2}e^{-t} \text{d} t\]\[= \frac {1} { \sqrt{ \lambda} } \Gamma(1/2) \]\[=\sqrt{\frac {π}{\lambda}}\]