Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 – 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
up vote 4 down vote accepted

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...

share|cite|improve this answer
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! – Sangchul Lee 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.

share|cite|improve this answer
The OP added PV in front of the integral. – 1015 Apr 3 '13 at 11:48
And I do not know what PV means with multiple singularities. He still needs to explain. – GEdgar Apr 3 '13 at 11:49
Me neither actually...I was just trying to give this question a chance. Maybe with contour integration. – 1015 Apr 3 '13 at 11:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.