Question

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

Answer 1

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

Answer 2

I found a nice solution few days back while attempting SithsAndGiggle's latest problem
It's really nice, doesn't use complex analysis/series...

Answer 3

Lol I was thinking series

Answer 4

But the one I have is rather lenthy
I have posted this here to see alternative ways... series might work nicely too..

Answer 5

Can it be like a power series?

Answer 6

This looks fun and interesting, I really wanna work on this one and find a pretty answer

Answer 7

it sure feels like multiple approaches are possible...

Answer 8

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.

Answer 9

\(|\cos x|\le 1\) so replacing \(\cos x\) by \(1\) is it ?

Answer 10

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

Answer 11

Yeah, it's just a guess though haha

Answer 12

Oh wow, that e is unexpected, but \(e \approx 2\) hahaha

Answer 13

Can't you use Euler's equation otherwise this seems tedious xD

Answer 14

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.

Answer 15

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.

Answer 16

\[\int\limits_{0}^{\infty}\frac{ cosx }{ x^2+1 }dx\]
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:
\[\lim_{b \rightarrow \infty}\int\limits_{0}^{b}\frac{ cosx }{ x^2+1 }dx\]

Answer 17

\[\int_0^{\infty}\frac{1}{x^2+1}dx \cdot \int_0^{\infty}\cos(x)dx\] Would this make it more complicated? I have feeling Kai already wrote this in a different form...lol.

Answer 18

\[\mathcal F \left\{ \frac{1}{t^2+1} = \pi \ e^{-2 \pi |f|}\right\}\]

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?!

Answer 19

oops

\[\mathcal F \left\{ \frac{1}{t^2+1} \right\} = \pi \ e^{-2 \pi |f|}\]

Answer 20

We can use l'hopital's rule , it ll be much quicker right ?

Answer 21

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

Answer 22

Ohhhh... integration by parts :P