How do you write $\int_0^\infty d\phi\ \cos(x\phi)\sin(y\phi)$ as a sum of Dirac deltas?

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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:

  1. 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.

  2. Can a similar kind of result be stated for the integral $\int_0^\infty d\phi\ \cos(x\phi)\sin(y\phi)$?

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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, $\sin((x+y)\phi)+\sin((x-y)\phi)=2\sin(x\phi)\sin(y\phi)$

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First note that $$ \cos(x\phi)\cos(y\phi) =\cos((x+y)\phi)+\cos((x-y)\phi) $$ In any case, the integral $$ \int_0^\infty\cos(u\phi)\,\mathrm{d}\phi $$ doesn't converge. However, we can write $$ \begin{align} \int_0^\infty e^{-\lambda\phi^2}\cos(u\phi)\,\mathrm{d}\phi &=\frac12\int_{-\infty}^\infty e^{-\lambda\phi^2}\cos(u\phi)\,\mathrm{d}\phi\tag1\\ &=\frac12\int_{-\infty}^\infty e^{-\lambda\phi^2}e^{iu\phi}\,\mathrm{d}\phi\tag2\\ &=\frac12e^{-\frac{u^2}{4\lambda}}\int_{-\infty}^\infty e^{-\lambda\left(\phi-i\frac{u}{2\lambda}\right)^2}\,\mathrm{d}\phi\tag3\\ &=\frac12e^{-\frac{u^2}{4\lambda}}\int_{-\infty}^\infty e^{-\lambda\phi^2}\,\mathrm{d}\phi\tag4\\ &=\sqrt{\frac\pi{4\lambda}}e^{-\frac{u^2}{4\lambda}}\tag5 \end{align} $$ Explanation:
$(1)$: symmetry
$(2)$: $e^{-\lambda u^2}\sin(u\phi)$ is odd
$(3)$: complete the square
$(4)$: no singularities, so we can move the contour
$(5)$: $\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u=\sqrt\pi$

$(5)$ is $\pi$ times an approximation of the Dirac-delta function.