A hard integral gives a simple closed form, π/(4a)^3

[Tony1]

Proposed:

How can we prove $(1)?$

$$\int_{0}^{\infty}\mathrm dx{\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}={\pi\over (4a)^3}\tag1$$

Spoiler

We can start to decompose $(1)$ to...

$$\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-\int_{0}^{\infty}\mathrm dx{\cos^2\left({a\over x}\right)\over (4a^2+x^2)^2}\tag2$$

$$\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}\tag3$$

$${1\over 2}\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}\tag4$$

So far...

 

[GJA]

Hi, Tony.

Spoiler

Your work looks good so far. Next, you can use a $u$-substitution to show that
$$\int_{0}^{\infty}\frac{\cos\left(\frac{2a}{x}\right)}{(4a^{2}+x^{2})^{2}}dx=\frac{1}{(2a)^{3}}\int_{0}^{\infty}\frac{x^{2}\cos (x)}{(x^{2}+1)^{2}}dx.$$
The idea is to now notice that the two integrals in (4) --i.e.,
$$\int_{0}^{\infty}\frac{1}{(4a^{2}+x^{2})^{2}}dx\qquad\text{and}\qquad\int_{0}^{\infty}\frac{x^{2}\cos (x)}{(x^{2}+1)^{2}}dx$$
-- are even functions of $x$, so can be expressed as
$$\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{(4a^{2}+x^{2})^{2}}dx\qquad\text{and}\qquad\frac{1}{2}\int_{-\infty}^{\infty}\frac{x^{2}\cos (x)}{(x^{2}+1)^{2}}dx.$$
Now use the method of contour integration on these integrals to obtain the desired result.