integral resulting in Bessel function Prove that
$$\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx = \frac{\pi \sqrt{a}}{2} J_{1} \left( 2 \sqrt{a} \right)$$ where $J_{1}$ is the Bessel function of the first kind of order 1.
Some calculations I have done
$$\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx= \int_{0}^{\infty} \sum_{k=0}^{\infty }(-1)^{k}\frac{x^{2k+1}}{2k+1!} \cdot  \sum_{l=0}^{\infty }(-1)^{l}\frac{a^{2l+1}x^{-2l-1}}{2l+1!} \ dx$$
$$= \int_{0}^{\infty} \sum_{l=0}^{\infty } \sum_{k=0}^{\infty }(-1)^{k+l}\frac{x^{2(k-l)}}{(2k+1)!(2l+1)!} a^{2l+1} \ dx$$

$$\frac{\pi \sqrt{a}}{2}J_{1}(2sqrt{a})=\frac{\pi \sqrt{a}}{2} \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{2l+1}l!(1+l)!}   2^{l+\frac{1}{2}}a^{l+\frac{1}{2}}$$
$$=\pi \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{l+\frac{3}{2}}l!(1+l)!} a^{l+1}$$
 A: Let $f(a) = \int_0^\infty \sin(x) \sin\left(\frac{a}{x}\right) \mathrm{d} x$, where $a\in \mathbb{R}$. Without loss of generality we can assume $a > 0$.
Observe that
$$ \begin{eqnarray}
    a f^{\prime\prime}(a) = \int_0^\infty \sin(x) \left(-\sin\left(\frac{a}{x} \right) \right) \frac{a}{x} \frac{\mathrm{d} x}{x} \stackrel{x \to a/y}{=} \int_0^\infty \sin\left( \frac{a}{y} \right) (-\sin(y)) \mathrm{d} y = -f(a)
\end{eqnarray}
$$
The differential equation so obtained, $a f^{\prime\prime}(a) + f(a) = 0$, reduces to Bessel differential equation, with general solution 
$$
    f(a) = c_1 \cdot \sqrt{a} J_1(2\sqrt{a}) + c_2 \cdot \sqrt{a} Y_1(2\sqrt{a})
$$
where $J_1$ and $Y_1$ are Bessel functions of the first and the second kind. Since $f(0)=0$ and $\lim_{a \downarrow 0^+} \sqrt{a} Y_1(2 \sqrt{a} ) = -\frac{1}{\pi}$, we get $c_2 = 0$.
By splitting the integration range into $(0,1)$ and $(1,\infty)$ and changing variables in 
the first one to $x \to 1/x$ we get
$$
  f(a) = \int_{1}^{\infty} \sin(1/x) \sin(a x) \frac{\mathrm{d} x}{x^2}  + \int_1^\infty \sin(x) \sin(a/x) \mathrm{d} x
$$
From here we see that small $a$ expansion first term is 
$$
   f(a) = a \left( \int_{1}^\infty \frac{1}{x} \sin(x) \mathrm{d} x + \int_{1}^\infty \sin(1/x) \frac{\mathrm{d} x }{x} \right) + \mathcal{o}(a) = \frac{\pi}{2} a + \mathcal{o}(a)
$$
This fixes $c_1 = \frac{\pi}{2}$, proving the requested equality.
