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

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 $a \neq b$.

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I would use the fact that cosine is the real part of the exponential function, then you would have a complex part, and with the inverse polynomial, use its roots to split it, that are complex, so you would have a more comfortable integral to work with. –  Sebastian Griotberg Apr 3 '13 at 11:27
Then, check out the exponential integral in wikipedia en.wikipedia.org/wiki/Exponential_integral –  Sebastian Griotberg Apr 3 '13 at 11:28
First I would explain what to do with the poles in the integrand (where the denominator is zero). At least when $a=b$ the zeros all cancel, but surely your answer is wrong in that case. –  GEdgar Apr 3 '13 at 11:33
@GEdgar: You are right. The answer is wrong for $a=b$. –  Anthony Apr 3 '13 at 11:36
Should this be understood as Cauchy principal value? –  1015 Apr 3 '13 at 11:43
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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...

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Your result is a lot more plausible. –  Ron Gordon Apr 3 '13 at 13:08
+1, for the same reason as Ron Gordon. I also obtained the same conclusion, but your technique $(*)$ is much more appealing! –  sos440 Apr 3 '13 at 13:20
Could you explain how you define PV of this integral and why this coincides with the limit (*)? You seem to be using an idiosyncratic definition of principal value, but maybe it is equivalent to the usual one and at the moment I fail to see the equivalence. –  Did Apr 3 '13 at 13:21
@Did, $$PV \int_{\mathbb{R}} = \lim_{\delta\to 0}\int_{\mathbb{R} \setminus \cup_{k\in\mathbb{Z}}(\frac{(2k-1)\pi}{2b} - \delta,\frac{(2k-1)\pi}{2b} + \delta)}$$ i.e. the PV is the $\delta \to 0$ limit of the integral where points within a distance $\delta$ from the poles are excluded. The contour appear in $\lim_{\epsilon\to 0+}\int_{-\infty+i\epsilon}^{\infty+i\epsilon}$ can be deformed to this plus a bunch of half circles of radius $\delta$ running clockwisely around the poles. –  achille hui Apr 3 '13 at 13:38
There seems to be a double limit here, when $\epsilon\to0^+$ and when $R\to+\infty$, of the integral from $-R+i\epsilon$ to $+R+i\epsilon$ but I guess that one can show the limit $R\to+\infty$ does not trouble things. –  Did Apr 3 '13 at 14:16
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.