\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx = \frac{\pi \sqrt{a}}{2} J_{1} \left( 2 \sqrt{a} \right) where J_{1} is the Bessel function of the first kind of order 1.

Some calculations I did already

\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx= \int_{0}^{\infty} \sum_{k=0}^{\infty }(-1)^{k}\frac{x^{2k+1}}{2k+1!} \cdot  \sum_{l=0}^{\infty }(-1)^{l}\frac{a^{2l+1}x^{-2l-1}}{2l+1!} \ dx

=?????? \int_{0}^{\infty} \sum_{l=0}^{\infty } \sum_{k=0}^{\infty }(-1)^{k+l}\frac{x^{2(k-l)}}{(2k+1)!(2l+1)!} a^{2l+1} \ dx


\frac{\pi \sqrt{a}}{2}J_{1}(2\sqrt{a})=\frac{\pi \sqrt{a}}{2} \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{2l+1}l!(1+l)!}   2^{l+\frac{1}{2}}a^{l+\frac{1}{2}}
=\pi \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{l+\frac{3}{2}}l!(1+l)!} a^{l+1}

I put ??? because I think this step is not allowed because of the singularity of \sin \left(\frac{a}{x}\right) at x=0. Can someone confirm if this equality is true?