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ganeshie8
 one year ago
Evaluate \[\large \int\limits_0^{\infty} \dfrac{\cos x}{x^2+1}\,dx\]
ganeshie8
 one year ago
Evaluate \[\large \int\limits_0^{\infty} \dfrac{\cos x}{x^2+1}\,dx\]

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Evaluate \[\large \int\limits\limits_0^{\infty} \dfrac{\cos x}{x^2+1}\,dx\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I found a nice solution few days back while attempting SithsAndGiggle's latest problem It's really nice, doesn't use complex analysis/series...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Lol I was thinking series

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0But the one I have is rather lenthy I have posted this here to see alternative ways... series might work nicely too..

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Can it be like a power series?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0This looks fun and interesting, I really wanna work on this one and find a pretty answer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0it sure feels like multiple approaches are possible...

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Well it's less than \(\pi/2\) I guess that's the best guess, maybe like half of that, like \(\pi/4\) would be a fun guess.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(\cos x\le 1\) so replacing \(\cos x\) by \(1\) is it ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0thats not too crude it seems, http://www.wolframalpha.com/input/?i=%5Cint%5Climits_0%5E%7B%5Cinfty%7D+cos%28x%29%2F%28x%5E2%2B1%29

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, it's just a guess though haha

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Oh wow, that e is unexpected, but \(e \approx 2\) hahaha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Can't you use Euler's equation otherwise this seems tedious xD

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to see some kind of way to do differentiation under the integral sign or somehow back out into a double integral and flip the order of integration but I can't really think of anything.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0For what it's worth maybe someone can do something with this: \[I(y) = \int_0^\infty \frac{ \arctan(xy)\cos x}{x} dx \] \[I'(1) = \int_0^\infty \frac{\cos x dx}{1+x^2}\] I think this is true, not completely sure.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\infty}\frac{ cosx }{ x^2+1 }dx\] What you can do, is assign a constant that tends to infinity on the superior number of the integral, this way, you can create a finite integral, solvable with Barrows rule: \[\lim_{b \rightarrow \infty}\int\limits_{0}^{b}\frac{ cosx }{ x^2+1 }dx\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_0^{\infty}\frac{1}{x^2+1}dx \cdot \int_0^{\infty}\cos(x)dx\] Would this make it more complicated? I have feeling Kai already wrote this in a different form...lol.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\[\mathcal F \left\{ \frac{1}{t^2+1} = \pi \ e^{2 \pi f}\right\}\] this is an even function, \(\hat f = \hat f_c\) but we only want \( \hat f_c/2\) so that's \[\frac{1}{2} \pi \ e^{(2 \ \pi (\frac{1}{2 \pi})) }\] pretty crap way to do it, eh?!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0oops \[\mathcal F \left\{ \frac{1}{t^2+1} \right\} = \pi \ e^{2 \pi f}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We can use l'hopital's rule , it ll be much quicker right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u=x^2+1 du=2xdx dx=du/2x x=sqrt(u1) integral (cos x)/x . 1/2u du hmmm how this is related to cosine integral

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Ohhhh... integration by parts :P
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