i know application but want proof of following definite integration property . Question: prove
(a) If $f(x)=f(-x)$ and $f(x+\pi)=-f(x)$ then, 
$$\int_{0_{+}}^{\infty} f(x)\dfrac{\sin x}{x}dx=\int_{0_{+}}^{\dfrac{\pi}{2}}f(x) \cos x\  dx$$
Then, use this identity in proving result (b):
$$\int_{0_{+}}^{\infty}\dfrac{\tan x}{x}dx=\dfrac{\pi}{2}$$
My attempt: I easily solved part (b) as follows 
$$\int_{0_{+}}^{\infty}\dfrac{\tan x}{x}dx=\int_{0_{+}}^{\infty}\sec x\ \dfrac{\sin x}{x}dx$$
Now using identity (a) because $\sec(-x)=\ \sec x$ and $\sec(x+\pi)=-\sec x$ we get:
$$\int_{0_{+}}^{\infty}\sec x\ \dfrac{\sin x}{x}dx=\int_{0_{+}}^{\pi/2}\sec x\cos x\  dx=\dfrac{\pi}{2}$$
But I don't know how to prove identity (a) and how can I use identity to evaluate some famous definite integrals.
 A: The conditions imposed on $f$ imply its Fourier series is a linear combination of the $\cos nx$ with $n$ odd, so by linearity we need only check that case. Note for $n\ne 1$ that $$\int_0^\infty\frac{\sin x\cos nx}{x}dx=\int_0^\infty\frac{\sin (n+1)x-\sin (n-1)x}{2x}dx=0$$ and $$\int_0^{\pi/2}\cos x\cos nxdx=\frac{1}{2}\int_0^{\pi/2} (\cos(n+1)x+\cos(n-1)x)dx=0.$$By contrast, in the $n=1$ case the first integral reduces to $\int_0^\infty\frac{\sin 2x dx}{2x}=\frac{\pi}{4}$ and the second to $\int_0^{\pi/2}\cos^2 xdx=\frac{\pi}{4}$.
A: HINTS rather than a complete solution 
It appears from the question that small issues like whether the integral converges at all, domains of definition, etc., are not your concern. So employing every possible form of what a friend used to call "engineer's prerogative" (i.e., swapping integrals, taking sums inside and outside integrals, rearranging series, etc.), here's a suggestion or two: 
We have
$$
f(x)=f(-x)\\
f(x+\pi)=-f(x)
$$ 
so the value of $f$ is entirely determined by its values on the interval $[0, \pi/2]$. So let's take the big integral and break it into pieces:
\begin{align}
\int_{0_{+}}^{\infty} f(x)\dfrac{\sin x}{x}~dx
&= \int_{0}^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{x}~dx
+  \int_{\frac{\pi}{2}}^\pi f(x)\dfrac{\sin x}{x}~dx
+  \int_\pi^{\frac{3\pi}{2}} f(x)\dfrac{\sin x}{x}~dx + \ldots\\
&= \int_{0}^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{x}~dx
-  \int_{-\frac{\pi}{2}}^0 f(x)\dfrac{\sin x}{(x+\pi)}~dx
-  \int_0^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{(x+\pi)}~dx + \ldots\\
&= \int_{0}^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{x}~dx
+  \int_{\frac{\pi}{2}}^0 f(x)\dfrac{\sin (-x)}{(-x+\pi)}~dx
-  \int_0^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{(x+\pi)}~dx + \ldots\\
&= \int_{0}^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{x}~dx
-  \int_0^{\frac{\pi}{2}} f(x)\dfrac{\sin (-x)}{(-x+\pi)}~dx
-  \int_0^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{(x+\pi)}~dx + \ldots\\
&= \int_{0}^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{x}~dx
+  \int_0^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{(-x+\pi)}~dx
-  \int_0^{\frac{\pi}{2}} f(x)\dfrac{\sin x}{(x+\pi)}~dx + \ldots\\
\end{align}
by subsituting $x = u - \pi$ in the 2nd and third integrals of the first line, and then $u = -x$ in the second integral of the second line, and then swapping limits on the integral, and replacing $\sin -x$ with $-\sin x$. If you keep at this, you end up with something where all terms integrate from $0$ to $\pi/2$. 
So then you combine them all under one big integral (!), and factor out the $f(x) \sin x$ part to get something that looks like
$$
\int_0^\frac{\pi}{2} f(x) \sin x \sum \frac{1}{\pm x \pm k\pi} ~ dx
$$
where you'll now have to fiddle to get the signs and the values for $k$ right, and then observe that the stuff in the sum turns out to be $\cot x$. Indeed, maybe it's obvious from some power series that I won't know off the top of my head. 
Anyhow, that'll get you started. When you've done all the algebra, you can go back and worry about convergence and whether the stuff you've written down is well-defined or not. 
