How to calculate the integral $\int\limits_0^{\infty }\frac{e^{-x}\sin(\pi x)\cos(nx)}{x}dx$?

Question

How to calculate the integral:

$$I = \int\limits_{0}^{\infty }\frac{e^{-x}\sin(\pi x)\cos(nx)}{x}dx ,$$

Using WolframAlpha, I got the following answer:

$I = \frac{1}{2}(\arctan(n + \pi) - \arctan(n- \pi))$

But I don't understand how we can get such an answer.

Answer 1

Use these two facts:

$\sin a\cos b = \frac 12 (\sin (a+b) + \sin(a-b))$

and

$I(a) = \int_0^{\infty} \frac {e^{-x}\sin ax}{x} dx\\ I' = \int_0^{\infty} e^{-x}\cos ax\ dx = \frac {e^{-x}(-\cos ax + a\sin ax)}{1+a^2}|_0^{\infty} = \frac {1}{1+a^2}\\ I(a)-I(0) =\int_0^a\frac {1}{1+a^2} \ da = \arctan a$

Answer 2

This is a straightforward integral: Hint: Split it into:

$\int\limits_0^\infty \left( {e^{-x} \over x} \right) \sin (\pi x) \cos (n x)\ dx$ and then use integration by parts to find:

$$\frac{1}{2} \left(\tan ^{-1}(n+\pi )-\tan ^{-1}(n-\pi )\right)$$

Answer 3

Valzavator, first substitute the $\sin(\pi x) \cos(nx)$ with $$ \frac{\sin((\pi + n)x) + \sin((\pi-n)x) }{2}. $$

Then also substitute $ e^{-x}$ with $$ \sum (-1)^{n} \frac{ x^{n}}{n!}$$

And now the integral can be solved by using partial integration..