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KingGeorge
 3 years ago
[EDIT: Now that I've managed to solve this on my own, this is now a (very difficult) challenge problem]
Evaluate \[\large \int_{\infty}^\infty \frac{\cos(z)}{z^2+1} dz\]
KingGeorge
 3 years ago
[EDIT: Now that I've managed to solve this on my own, this is now a (very difficult) challenge problem] Evaluate \[\large \int_{\infty}^\infty \frac{\cos(z)}{z^2+1} dz\]

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JasonDeRulo
 3 years ago
Best ResponseYou've already chosen the best response.0I 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...

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Not at all. At least, not in any obvious way that I can see.

cruffo
 2 years ago
Best ResponseYou've already chosen the best response.0Since it's an even function, you could just double the integral from 0 to infinity.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2You 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.

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{ \infty}^{\infty}\frac{ e^{ix} }{ x^2+ 1}\] 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 :(

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2This is the method that I used. Anyone can feel welcome to describe why we can substitute \[\cos(z)=e^{ix}.\]

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0Why does cos z = e^(iz) ?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2\[e^{iz}=\cos(z)+i\sin(x)\neq \cos(z)\]

wio
 2 years ago
Best ResponseYou've already chosen the best response.0Ooops, nevermind that's an inverse trig function... it's gonna make life harder.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0So you made the substitution cos z = e ^(ix) hm..

wio
 2 years ago
Best ResponseYou've already chosen the best response.0Since one is an odd function, the result is 0?

wio
 2 years ago
Best ResponseYou've already chosen the best response.0@KingGeorge \(\sin(x)\) is odd so \[ \int_{a}^a i\sin(x)dx = 0 \]??

wio
 2 years ago
Best ResponseYou've already chosen the best response.0hmm, but there is something in the denominator so I guess a little more thought needs to be put into it.

cruffo
 2 years ago
Best ResponseYou've already chosen the best response.0BTW isn't \(\large \cos(z) = \frac{e^{iz} + e^{iz}}{2}\)

cruffo
 2 years ago
Best ResponseYou've already chosen the best response.0Then, if you want to factor the denominator z^2+1 = (z+i)(zi)

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2@cruffo yes. Also, the substitution made by ghazi is valid. I'm just waiting on a good explanation of why it's valid.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0To 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?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Half circle sitting on the real axis. Then let the diameter go to infinity.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0ah 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

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0note the integral on the left is a real integral, while the one on the right a complex one.

cruffo
 2 years ago
Best ResponseYou've already chosen the best response.0which is why you don't use the imaginary part of e^iz. Got it.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Almost. 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 semicircle.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0my guess is that the upper part of the semi circle tends to 0 as r goes to infinity.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0er, i mean the value of the integral on the upper part of the semi circle.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2However, 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.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0and the part on the real line matches perfectly with the original integral since:\[real(e^{iz})=\cos (z)\]

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0you 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 :( :(

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0try to figure this out, i'll be back :) hope this helps you

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0I dont think my associate's degree has prepared me for this o_O I wish I could give best responses to multiple people.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2Liouville'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.

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0this 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

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0my complex variables class was pretty lackluster =/ so im auditing the course next semester with a better professor.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0For real? (bad pun)

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2My 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.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0i learned what i needed for the Subject GRE and called it a day lol. Hopefully next semester i'll gain a better understanding.

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0by the way answer is around pi cos z ?? is it?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2The solution is \(\pi/e\). However, finding this is not too easy.

LogicalApple
 2 years ago
Best ResponseYou've already chosen the best response.0What an interesting solution..

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.2^^ my thoughts exactly. Before I knew about the substitution for \(e^{iz}\), I was stunned when I saw \(e\) in the solution.

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0hmm , alright, it'll be done in a while or may be morrow

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2going with jordan's lemma\[ \int_{\infty}^{\infty} \frac{\cos x}{1+x^2} \; \text{d}x=\text{Re} \left( \int_{\infty}^{\infty} \frac{e^{ix}}{1+x^2} \; \text{d}x \right)=2 \pi i \frac{e^{1}}{2i}=\frac{\pi}{e}\]

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0this is what i was expecting when i saw @mukushla here, nice :)
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