# Integrate this singularity?

• Jan 6th 2013, 04:39 PM
NowIsForever
Integrate this singularity?
I found this problem recently on Yahoo Answers! It is gone now because it went unanswered after four days, and it is the policy of YA! to delete such questions.

Given:

$\displaystyle {I(n)=\int^{\pi/2}_{-\pi/2}\frac{\cos(2nx)}{(1+e^x)\cos(2x)}dx}$

Prove:

$\displaystyle I(n + 2) = -I(n)$, for all natural numbers $\displaystyle n$. (Since "natural number" is ambiguous, let's say for non negative—or possibly positive integers.)

I couldn't get passed the singularities at $\displaystyle -\pi/4$ and $\displaystyle \pi/4$ (Headbang). WolframAlpha couldn't provide an answer either, even for (e.g.), $\displaystyle n=2$.

I have my doubts about this being true; however, if it is I would like to see a proof.
• Jan 6th 2013, 11:58 PM
chiro
Re: Integrate this singularity?
Hey NowIsForever.

Use the fact that cos(a+b) = cos(a)cos(b) - sin(a)sin(b) where a = 2nx and b = 2x. This gives cos(2nx)cos(2x) - sin(2nx)sin(2x) for which you get some cancellation and simplification for the cos and sin of 2x.
• Jan 7th 2013, 12:43 AM
JJacquelin
Re: Integrate this singularity?
The integral has to be understood as a Cauchy integral (i.e. as Principal Value) since it is not convergent except for n=1.
On this condition, I(n+2) = -I(n) , as shown in attachment and I(1) = pi/2
• Jan 7th 2013, 04:44 PM
NowIsForever
Re: Integrate this singularity?
Quote:

Originally Posted by JJacquelin
The integral has to be understood as a Cauchy integral...

Ah, yes, that does it; however, you have (n + 2) in the last four integrals where it should be (n + 1).

You side stepped the singularity by clever use of trigonometric identities, and the identity $\displaystyle \frac{1}{1+e^{-x}}=1-\frac{1}{1+e^x}$.

I thought it would require something like the Cauchy principal value calculated with limits, but you've made that unnecessary, thanks.

Nice—and elegant.
• Jan 8th 2013, 01:28 AM
JJacquelin
Re: Integrate this singularity?
Hi !
We saw that I(1)=pi/2 which leads to I(n)=((-1)^((n-1)/2))*(Pi/2) in case of odd n.
In case of even n, we have to compute I(0) which requires to follow the computation rules for the principal value of a Cauchy integral. The result is I(0)=0.
As a consequence I(n)=0 in case of even n
.
Quote:

Originally Posted by NowIsForever
you have (n + 2) in the last four integrals where it should be (n + 1).

Thanks you for bringing the typo to attention. Of course, it doesn't change the result.