Originally Posted by

**blorpinbloo** $\displaystyle

\frac{1}{2pi} \int k^2\frac{\sin \left(kr\right)}{kr} \frac{\sin \left(kr'\right)}{kr'} {dk}

$

(Couldn't figure out how to make the pi symbol, but that is a pi in the denominator at the beginning. Also the limit is from 0 to inf.). I'm supposed to show this integral equals 1/(4*pi*r*r') times the delta function of (r-r').

I solved the integral and ended up with

$\displaystyle

\frac{1}{4(pi)^2rr'} \left(\frac{sin\left(k\left(r-r'\right)\right)}{r-r'} - \frac{sin\left(k\left(r+r'\right)\right)}{r+r'}\ri ght)

$

With k ranging from 0 to infinity. Now my problem is, I'm not sure exactly how to get a delta function out of this result. I can sort of see how it might emerge on the right sided term, but I'm confused as to how to express that mathematically. Also there is apparently a pi term that comes out as well when I compare it to the expected answer. Can anyone help me?