The following result is well-known (ie; I read it in a book) $$ \int_0^\infty d\phi\ \cos(x\phi)\cos(y\phi) = \frac{\pi}{2} \delta( x + y ) + \frac{\pi}{2} \delta( x - y ) $$
I read it in the appendix of a paper once, wrote it down, and can't find the paper since then. I don't understand how this is derived. My questions are:
How is the above result derived? I can see that setting $x = y$ in the original integral gives $\infty$ and everything else gives $0$, but I don't understand.
Can a similar kind of result be stated for the integral $\int_0^\infty d\phi\ \cos(x\phi)\sin(y\phi)$?
From trig,
$\cos((x+y)\phi) = \cos(x\phi)\cos(y\phi)-\sin(x\phi)\sin(y\phi) \\ \cos((x-y)\phi) = \cos(x\phi)\cos(y\phi)+\sin(x\phi)\sin(y\phi) $
Therefore,
$\cos((x+y)\phi) + \cos((x-y)\phi) = 2\cos(x\phi)\cos(y\phi)$
I figure you can take it from there.
As for your part 2, $\sin((x+y)\phi)+\sin((x-y)\phi)=2\sin(x\phi)\sin(y\phi)$