Evaluate $\int\limits_0^\infty \frac{\cos(ax)}{\cos(bx)}\frac{1}{1+x^2}dx$

Question

I would like to show that

$$\text{PV}\int_0^\infty \frac{\cos(ax)}{\cos(bx)}\frac{1}{1+x^2}dx = \frac{\pi}{2}\mathrm{sech}(b)$$

using complex analysis. $a$ and $b$ are real numbers and $a \neq b$.

Please give some hints.

Answer 1

What the question asking for cannot be right!

At least for $0 < a < b$, we have:

$$\begin{align}\operatorname{PV} \int_0^{\infty} \frac{\cos a x}{\cos b x} \frac{dx}{1+x^2}&= \frac12 \operatorname{PV} \int_{-\infty}^{\infty} \frac{\cos a x}{\cos b x}\frac{dx}{1+x^2} \\&= \frac12 \lim_{\epsilon\to 0+} \Re\left[\int_{-\infty+i\epsilon}^{\infty+i\epsilon} \frac{\cos a z}{\cos b z}\frac{dz}{1+z^2}\right]\tag{*} \end{align}$$ The last equality is true because at the poles $\pm \frac{(2k-1)\pi}{2 b}, k = 1, 2,\ldots$ of the integrand $\frac{\cos a z}{\cos b z}\frac{1}{1+z^2}$, the residues are all real. Their contribution to the integral is $-\pi i$ times the residues and hence is imaginary.

We can evaluate the integral $(*)$ by completing the contour in upper half plane.

Notice when the $y$ in $z = x + iy$ becomes big, $\frac{\cos a z}{\cos b z} \sim e^{-(b-a)(y - ix)} \to 0$. The upper half circle at infinity contributes nothing to the contour integral and we have:

$$\lim_{\epsilon\to 0+}\int_{-\infty+i\epsilon}^{\infty+i\epsilon} \frac{\cos a z}{\cos b z}\frac{dz}{1+z^2} = 2 \pi i \operatorname{Res}( \frac{\cos a z}{\cos b z}\frac{1}{1+z^2}; z = i ) = 2 \pi i \frac{\cos a i}{\cos b i}\frac{1}{2i} = \pi \frac{\cosh a}{\cosh b}$$

From this, we get:

$$\operatorname{PV} \int_0^{\infty} \frac{\cos a x}{\cos b x} \frac{dx}{1+x^2} = \frac{\pi}{2} \frac{\cosh a}{\cosh b}$$

This is not what the OP asking to show...

Answer 2

So, you need to explain what you want. Take this one $$ \int_0^\infty \frac{\cos x}{\cos(2x)}\;\frac{dx}{1+x^2} $$ The first spot where it is improper is $x=\pi/4$, and $$ \int_0^{\pi/4} \frac{\cos x}{\cos(2x)}\;\frac{dx}{1+x^2} = +\infty . $$ So unless you provide some explanation, your request is impossible.