Here's the question you clicked on:
KingGeorge
[EDIT: Now that I've managed to solve this on my own, this is now a (very difficult) challenge problem] Evaluate \[\large \int_{-\infty}^\infty \frac{\cos(z)}{z^2+1} dz\]
I doubt that many people here (myself included) can help you with a problem of this level. You're better off posting this at some site like this: http://www.physicsforums.com where many of the members have Ph.Ds in physics, math, etc...
Not at all. At least, not in any obvious way that I can see.
Since it's an even function, you could just double the integral from 0 to infinity.
You could. But I don't know how to integrate it from 0 to infinity before finding the integral from -infinity to infinity. The way I know how to do this only really works from -infinity to infinity.
\[\int\limits_{- \infty}^{\infty}\frac{ e^{ix} }{ x^2+ 1}\] i hope you can do it now, by using integration by parts ,this is what i can think of so far , its been ages since i have done this :(
This is the method that I used. Anyone can feel welcome to describe why we can substitute \[\cos(z)=e^{ix}.\]
Why does cos z = e^(iz) ?
\[e^{iz}=\cos(z)+i\sin(x)\neq \cos(z)\]
Ooops, nevermind that's an inverse trig function... it's gonna make life harder.
So you made the substitution cos z = e ^(ix) hm..
Since one is an odd function, the result is 0?
@KingGeorge \(\sin(x)\) is odd so \[ \int_{-a}^a i\sin(x)dx = 0 \]??
hmm, but there is something in the denominator so I guess a little more thought needs to be put into it.
BTW isn't \(\large \cos(z) = \frac{e^{iz} + e^{-iz}}{2}\)
Then, if you want to factor the denominator z^2+1 = (z+i)(z-i)
@cruffo yes. Also, the substitution made by ghazi is valid. I'm just waiting on a good explanation of why it's valid.
To me, it looks like what you are doing is changing the real integral into a complex one. Your taking an integral defined on the real line, and changing it to one defined on the complex plane. In the solution, did you have a path for the complex integral?
Half circle sitting on the real axis. Then let the diameter go to infinity.
ah ok, so you claim that:\[\int\limits_{-r}^{r}\frac{\cos(z)}{z^2+1}dz=\int\limits_{C}\frac{e^{iz}}{z^2+1}dz\]where C is the half circle or radius r sitting on the x axis
note the integral on the left is a real integral, while the one on the right a complex one.
which is why you don't use the imaginary part of e^iz. Got it.
Almost. There is one more integral needed on the right.I claim that\[\int\limits_{-r}^{r}\frac{\cos(z)}{z^2+1}dz=\int\limits_{C}\frac{e^{iz}}{z^2+1}dz-\int\limits_{\gamma}\frac{e^{iz}}{z^2+1}{dz}\]where \(\gamma\) is the upper part of the semi-circle.
my guess is that the upper part of the semi circle tends to 0 as r goes to infinity.
er, i mean the value of the integral on the upper part of the semi circle.
However, when we work it out, that extra integral does go to 0 as the radius increases. But that's still something that needs to be shown.
and the part on the real line matches perfectly with the original integral since:\[real(e^{iz})=\cos (z)\]
you can consider that as a corollary of Liouville theorem that says if \[fe^g\] has an elementary antiderivative, where f and g are rational functions provided that "g" is not constant, then it has an antiderivative of the form \[he^g\] , h is a rational function. For this to be an antiderivative of \[fe^g\], we need this condition to be satisfied h′+hg′=f. Now lets say , \[f= \frac{ 1 }{ 1+z^2 }\]and g=iz, the condition is h′+ih=\[\frac{ 1 }{ 1+z^2 }\]. The right side has a pole of order 1 at z=i. In order for the left side to have a pole there, h must have a pole there, but wherever h has a pole of order k, h′ has a pole of order k+1, so the left side can never have a pole of order 1. @KingGeorge hope this complex answer helps you lol i cant go further , i have forgotten most of it :( :(
try to figure this out, i'll be back :) hope this helps you
I dont think my associate's degree has prepared me for this o_O I wish I could give best responses to multiple people.
Liouville's theorem is actually not needed to finish the problem. That being said, you still need to know how to work out Laurent series, and know how to find values of integrals using these.
this is what is an explanation of the substitution made for cos z , another alternate is to use expansion of cos z ,ignore for higher order
my complex variables class was pretty lackluster =/ so im auditing the course next semester with a better professor.
For real? (bad pun)
My professor wasn't the greatest either. As long as we came to class and made some sort of attempt on the homework it was an A. If anyone an finish it, feel free. If no one does, I'll post up a solution sometime in the (hopefully) near future.
i learned what i needed for the Subject GRE and called it a day lol. Hopefully next semester i'll gain a better understanding.
by the way answer is around pi cos z ?? is it?
The solution is \(\pi/e\). However, finding this is not too easy.
What an interesting solution..
^^ my thoughts exactly. Before I knew about the substitution for \(e^{iz}\), I was stunned when I saw \(e\) in the solution.
hmm , alright, it'll be done in a while or may be morrow
I saw something similar to this in a document about Differentiation under the Integral by Keith Conrad. --> http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf page 13, on 11. \( \displaystyle \int_{\mathbb{R}} \frac{\cos(xt)}{1+x^2} \; \text{d}x \) This integral involves an extra variable, t. He is able to figure out the answer with some change of variables and differentiation under the integral work. This problem was really interesting, using a lot of different techniques. (: With the solution, we may let t=1 to match this integral here.
going with jordan's lemma\[ \int_{-\infty}^{\infty} \frac{\cos x}{1+x^2} \; \text{d}x=\text{Re} \left( \int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^2} \; \text{d}x \right)=2 \pi i \frac{e^{-1}}{2i}=\frac{\pi}{e}\]
this is what i was expecting when i saw @mukushla here, nice :)