I ran into a set of four double integrals that I would like to find analytic solutions for. They are:
$$I_1=\int_0^\infty dy \int_0^\infty dx f(x) \cos(xy)\cos(yz)$$ $$I_2=\int_0^\infty dy \int_0^\infty dx f(x) \sin(xy)\cos(yz)$$ $$I_3=\int_0^\infty dy \int_0^\infty dx g(x) \cos(xy)\sin(yz)$$ $$I_4=\int_0^\infty dy \int_0^\infty dx g(x) \sin(xy)\sin(yz),$$ where $\int_{-\infty}^\infty dx |f(x)|$ and $\int_{-\infty}^\infty dx |g(x)|$ both converge, $f(x)=f(-x)$ is symmetric, and $g(x)=-g(-x)$ is antisymmetric.
So far, I've found that $I_1$ and $I_4$ can be solved by first using the fact that $f(x)\cos(xy)$ and $g(x)\sin(xy)$ are symmetric with respect to $x=0$ to extend the integral over $x$ to $-\infty$. Then, by applying the identities $$\cos(a)\cos(b)=\frac{1}{2}\cos(a-b)+\frac{1}{2}\cos(a+b)$$ $$\sin(a)\sin(b)=\frac{1}{2}\cos(a-b)-\frac{1}{2}\cos(a+b)$$ and using the Fourier integral theorem to recognize that $$f(z) = \frac{1}{\pi}\int_0^\infty dy\int_{-\infty}^\infty dx f(x)\cos(y(x-z))$$ we can get $$I_1=\frac{\pi}{4}(f(z)+f(-z))=\frac{\pi}{2}f(z)$$ $$I_4=\frac{\pi}{4}(g(z)-g(-z))=\frac{\pi}{2}g(z).$$
Unfortunately, this approach does not seem to work to solve $I_2$ and $I_3$. I've tried using the similar identities for $\sin(a)\cos(b)$ but have had no luck yet because there is no analog of the Fourier transform formula for $\sin(y(x-z))$ that I can find.
My question is: do the remaining two integrals not have analytic solutions without making further assumptions on $f(x)$ and $g(x)$, or am I missing something simple?