Is the Dirichlet integral a shortcut for solving this difficult integral?

[Mr Davis 97]

I have the integral $\displaystyle \int_{- \infty}^{\infty} \frac{\cos x}{x^2+1} dx$. We are going to use differentiation under the integral sign, so we let $\displaystyle I(t) = \int_{- \infty}^{\infty} \frac{\cos tx}{x^2+1} dx$, and then, after manipulation, $\displaystyle I'(t) = \int_{- \infty}^{\infty} \frac{\sin tx}{x(x^2+1)} dx - \int_{- \infty}^{\infty} \frac{\sin tx}{x} dx$. My question lies in the rightmost integral. In a solution I've seen, the rightmost integral is linked to the Dirichlet integral: https://en.wikipedia.org/wiki/Dirichlet_integral. And so $\pi$ is simply substituted for this expression. What I don't understand is how can it be linked to this known integral when there is that $t$ in the argument of $\sin$?

 

[kuruman]

Can you not change variables? Let $u=tx$, etc. etc.

 

[Mr Davis 97]

Can you not change variables? Let $u=tx$, etc. etc.

Okay, I see how that could work... But I don't know the sign of t, right? As such I can't tell whether the upper bound on the integral goes to $+ \infty$ or $- \infty$

 

[kuruman]

The final answer should not depend on the sign of $t$ because $I(t)=I(-t)$. You have an even function in $x$ and $t$. That helps you to link to the Dirchlet integral by putting a $2$ up front and taking the limits from zero to infinity.

 

[jostpuur]

Is there a reason for why you don't deal with the original integral via calculus of residues?