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

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

Mathematics
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Evaluate \[\large \int\limits\limits_0^{\infty} \dfrac{\cos x}{x^2+1}\,dx\]
I found a nice solution few days back while attempting SithsAndGiggle's latest problem It's really nice, doesn't use complex analysis/series...
Lol I was thinking series

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But the one I have is rather lenthy I have posted this here to see alternative ways... series might work nicely too..
Can it be like a power series?
This looks fun and interesting, I really wanna work on this one and find a pretty answer
it sure feels like multiple approaches are possible...
Well 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.
\(|\cos x|\le 1\) so replacing \(\cos x\) by \(1\) is it ?
thats 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
Yeah, it's just a guess though haha
Oh wow, that e is unexpected, but \(e \approx 2\) hahaha
Can't you use Euler's equation otherwise this seems tedious xD
I'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.
For 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.
\[\int\limits_{0}^{\infty}\frac{ cosx }{ x^2+1 }dx\] 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: \[\lim_{b \rightarrow \infty}\int\limits_{0}^{b}\frac{ cosx }{ x^2+1 }dx\]
\[\int_0^{\infty}\frac{1}{x^2+1}dx \cdot \int_0^{\infty}\cos(x)dx\] Would this make it more complicated? I have feeling Kai already wrote this in a different form...lol.
\[\mathcal F \left\{ \frac{1}{t^2+1} = \pi \ e^{-2 \pi |f|}\right\}\] 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?!
oops \[\mathcal F \left\{ \frac{1}{t^2+1} \right\} = \pi \ e^{-2 \pi |f|}\]
We can use l'hopital's rule , it ll be much quicker right ?
u=x^2+1 du=2xdx dx=du/2x x=sqrt(u-1) integral (cos x)/x . 1/2u du hmmm how this is related to cosine integral
Ohhhh... integration by parts :P

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