Question

How would you integrate\[\int\limits_{0}^{\infty}\Phi \sin \phi \Phi''d \phi\]

Answer 1

what is \[\Phi \]

Answer 2

Sorry,\[\Phi(\phi)\]

Uppercase phi is a function of lowercase phi

Answer 3

@mahmit2012

Answer 4

can you state the original problem ?

Answer 5

Sure

Answer 6

Given\[\Phi''+\frac{\cos \phi}{\sin \phi}\Phi'+(\nu-\frac{n^2}{\sin^2 \phi})\Phi=0\]Notice that this is a bessel function of order v-n^2/sin^2(phi)

Show that\[\nu>0\]We have to start by multiplying both sides by\[\Phi \sin \phi\]And integrate from 0 -> infinity

Answer 7

I meant of order n^2 >.<

Answer 8

lol...i got it...

Answer 9

i think we can rearrange to the standard form with a variable change or something like this :\[t=\sqrt{\nu} \sin \phi\]

Answer 10

Hm might play around with that, the professor literally just multiplied everything by\[\Phi \sin \phi\] and integrated.