Trigonometric mixed with exponential integral

Let $a>0$. Prove that:

$$\int_{0}^{\infty}e^{-2ax} \left(\frac{\sin(x)}{x}\right)^{2}\, {\rm d}x=a\log\left(\frac{a}{\sqrt{1+a^{2}}}\right)+\arccot a$$

Solution

For $a>0$ we define the function:

$$f(a)= \int_{0}^{\infty}e^{-2ax}\left ( \frac{\sin x}{x} \right )^2 \, {\rm d}x$$

We differentiate $f$ twice , hence:

$$ f''(a)=4\int_{0}^{\infty}e^{-2ax}\sin^2 x \, {\rm d}x = \frac{4}{4a^3+4a}= \frac{1}{a^3+a}$$

since the Laplace transform of $\sin^2 x$ is quite known. It can be found in tables or someone can invoke the formula $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$.

Integrating back twice (and at the same time evaluating the constants) we have that:

$$ f(a)=\int_{0}^{\infty}e^{-2ax} \left ( \frac{\sin x}{x} \right )^2 \, {\rm d}x= a \log a - \frac{1}{2}a \log \left ( a^2+1 \right )+\arccot a$$

ending the exercise.

The exercise can also be found in mathimatikoi.org