1

I'm working on stats problems and am a bit rusty with calculus as I haven't worked with it in 3 years. My prof gave us an equation $\int_{-\infty}^\infty ke^{\frac{-x^2}{2}} ~dx =1$ where XER and f(x) is a PDF to solve for k.

Could someone please explain to me how he arrived at $k = \frac{1}{\sqrt{2\pi}}$

Curious Cat
  • 13
  • 6
    This comes from the beautiful gaussian result: http://math.stackexchange.com/questions/9286/proving-int-0-infty-mathrme-x2-dx-dfrac-sqrt-pi2 – Olivier Oloa Aug 04 '16 at 17:33
  • $\int_{-\infty}^{\infty} k \exp(-\frac{x^2}{2}) , dx = k \int_{-\infty}^{\infty} \exp(-\frac{x^2}{2}) , dx$, so if you believe the latter integral evaluates to $\sqrt{2\pi}$, you're done. There are a lot of good proofs for this. – David Kraemer Aug 04 '16 at 17:35
  • Substitute $\frac{x^2}{2}=t$ and go from there. – StubbornAtom Aug 04 '16 at 17:38
  • @G.Sassatelli I think the integrand becomes $e^{-t}\frac{1}{\sqrt {2t}}$ actually. – StubbornAtom Aug 04 '16 at 18:04
  • @StubbornAtom Yes: it wouldn't have changed my point, though. However, I found a method of solution in the linked question which used it, and I therefore decided to put on some duct tape. :D –  Aug 04 '16 at 18:08
  • It's basically http://math.stackexchange.com/q/1343672/321264 – StubbornAtom Aug 04 '16 at 18:10

0 Answers0