UnkleRhaukus
  • UnkleRhaukus
Prove: \[\int_{-∞}^∞ e^{-\lambda x^2}\text{d}x =\sqrt{ \frac{π}{\lambda}}\]
Mathematics
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chestercat
  • chestercat
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TuringTest
  • TuringTest
what kind of analysis are you doing here?
TuringTest
  • TuringTest
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...
UnkleRhaukus
  • UnkleRhaukus
hmm well it is a Gaussian, this is just one step in a question i am trying to solve

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UnkleRhaukus
  • UnkleRhaukus
This question, i have all the other working i just cannot see how these two terms are equivalent
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TuringTest
  • TuringTest
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
anonymous
  • anonymous
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
UnkleRhaukus
  • UnkleRhaukus
i dont know what i would enter into google
Zarkon
  • Zarkon
write as a double integral and switch to polar coordinates
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Gaussian_integral
TuringTest
  • TuringTest
@Zarkon I knew it! gotta learn how to do those kinds of integrals
UnkleRhaukus
  • UnkleRhaukus
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
UnkleRhaukus
  • UnkleRhaukus
\[\int_{-∞}^∞e^{-\lambda x^2}\text{d}x\]\[=2\int_{0}^∞ e^{-\lambda x^2} \text{d} x\] let \[\qquad \lambda x^2=t\]\[\qquad x=\sqrt{t/\lambda} \]\[\qquad \text{d}x=\frac{t^{-1/2}}{2\sqrt\lambda}\text{d}t\]\[={1 \over \sqrt{\lambda}}\int_{0}^∞ t^{-1/2}e^{-t} \text{d} t\]\[= \frac {1} { \sqrt{ \lambda} } \Gamma(1/2) \]\[=\sqrt{\frac {π}{\lambda}}\]

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