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.
Best Answer
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 $\int_0^{\pi/2}\cos^2 xdx=\frac{\pi}{4}$.