definite integrals - How to show that $ PV int_{-infty}^{infty} frac{tan x}{x}dx = pi$

In a recent question, it was stated in a comment, without proof, that

$$ PV \int_{-\infty}^{\infty} \frac{\tan x}{x}dx = \pi$$

What is the easiest way to prove this? I was able to show that

$$
PV \int_{-\infty}^{\infty} \frac{\tan x}{x}dx = -PV\int_{-\infty}^{\infty} \frac{1}{x \tan x}dx \\
PV \int_{-\infty}^{\infty} \frac{1}{x \sin x}dx = 0 \\
\int_{-\infty}^{\infty} \frac{\sin x}{x}dx = \pi
$$

but failed to compute the original integral from this.

Answer

The
question is : What would be a right way to define the principal value
of this integral, knowing that it has infinitely many singularities at
the points $\frac{\pi}{2}+\pi\Bbb{Z}$ ? I will propose the following
$$
PV\int_0^\infty\frac{\tan x}{x}dx~\buildrel{\rm def}\over{=}~
\lim_{\lambda\to0}\int_0^{\infty}\frac{\sin
x\cos x}{x(\cos^2 x+\lambda^2)}dx
$$

Next I will show that the limit in this definition does exist and that
its value is $\frac{\pi}{2}$.

First, note that the convergence of the integral
$\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx$ is easy to prove
using integration by parts. Now
$$\eqalign{
\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx
&=\frac{1}{2}\lim_{n\to\infty}\int_{-\pi n}^{\pi (n+1)}\frac{\sin
x\cos x}{x(\cos^2 x+\lambda^2)}dx\cr

&=\frac{1}{2}\lim_{n\to\infty}\sum_{k=-n}^{n}\int_{\pi
k}^{\pi(k+1)}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx\cr
&=\frac{1}{2}\lim_{n\to\infty}\sum_{k=-n}^{n}\int_{0}^{\pi}\frac{\sin
x\cos x}{(x+ \pi k)(\cos^2 x+\lambda^2)}dx\cr
&=\frac{1}{2}\lim_{n\to\infty}\int_{0}^{ \pi}\left(\sum_{k=-n}^{n}\frac{1}{x+ \pi
k}\right)
\frac{\sin x\cos x}{ \cos^2x+\lambda^2}dx\cr
&=\frac{1}{2}\lim_{n\to\infty}\int_{0}^{ \pi}U_n(x)
\frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx\cr
}

$$
where
$$
U_n(x)=\tan(x)\left(\sum_{k=-n}^{n}\frac{1}{x+ \pi
k}\right)
$$
But using the well-known expansion of the cotangent function, it is easy to see that
$\{U_n \}_n$ converges point-wise to $1$, and that this sequence is bounded uniformely on the interval $[0,\pi]$. Thus, we can interchange the signs of integral and limit in the above formula to get
$$
\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx

=\frac{1}{2} \int_{0}^{ \pi}
\frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx
= \int_{0}^{ \pi/2}
\frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx
$$
Finally, taking the limit as $\lambda\to0$ we get
$$
PV\int_0^\infty\frac{\tan x}{x}dx~=~
\lim_{\lambda\to0}\int_0^{\infty}\frac{\sin
x\cos x}{x(\cos^2 x+\lambda^2)}dx=\frac{\pi}{2}.

$$