how would I show that this indefinite integral equals 1/2?

Question

richyw · Answer

$$\lim_{r\rightarrow \infty}\int^r_0p^3e^{-p^2}=\frac{1}{2} $$

anonymous · Answer

@mukushla

anonymous · Answer

integration by parts will work here...i think...lets work on that

richyw · Answer

tried that

richyw · Answer

after three times, to get eliminate p, it just seems to get uglier

turingtest · Answer

$$u=e^{-p^2}\implies du=-2pe^{-p^2}dp$$$$\ln u=-p^2$$$$p^3e^{-p^2}dp=-\frac12p^2du=\frac12\ln udu$$maybe?

turingtest · Answer

just trying things...

turingtest · Answer

now you could integrate by parts I think

anonymous · Answer

hmm...

$\frac{1}{2}(u ln u -1) \ from\ \  1 \ to\  0=1/2 $

nice work...

turingtest · Answer

$$\frac12\int\ln udu=\lim_{r\to\infty}\left.\frac12u(\ln u-1)\right]_0^r$$thanks, means a lot from you :D

turingtest · Answer

oh yeah after the sub it's from 1 to 0, right...

anonymous · Answer

$u(\ln u-1)$  or  $u \ln u-1$

anonymous · Answer

second is right ha?

anonymous · Answer

im wondering $$\lim_{u \rightarrow 0} u \ln u =0$$

anonymous · Answer

Integration by parts

$u=p^2$  and  $dv=pe^{-p^2}$  so  $v=\frac{1}{2} e^{-p^2}$ and we have $$I=\frac{p^2}{2} e^{-p^2}]_0^\infty-\int\limits_{0}^{\infty} p e^{-p^2} dp=1/2$$

turingtest · Answer

so it works by parts after directly all, nice :)
I've done integrals very much like this before, I should've seen that
at least my solution was novel though :D

anonymous · Answer

:)