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ganeshie8

  • one year ago

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

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

  2. ganeshie8
    • one year ago
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    I found a nice solution few days back while attempting SithsAndGiggle's latest problem It's really nice, doesn't use complex analysis/series...

  3. Astrophysics
    • one year ago
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    Lol I was thinking series

  4. ganeshie8
    • one year ago
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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..

  5. Jhannybean
    • one year ago
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    Can it be like a power series?

  6. Empty
    • one year ago
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    This looks fun and interesting, I really wanna work on this one and find a pretty answer

  7. ganeshie8
    • one year ago
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    it sure feels like multiple approaches are possible...

  8. Empty
    • one year ago
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    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.

  9. ganeshie8
    • one year ago
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    \(|\cos x|\le 1\) so replacing \(\cos x\) by \(1\) is it ?

  10. ganeshie8
    • one year ago
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    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

  11. Empty
    • one year ago
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    Yeah, it's just a guess though haha

  12. Empty
    • one year ago
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    Oh wow, that e is unexpected, but \(e \approx 2\) hahaha

  13. Astrophysics
    • one year ago
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    Can't you use Euler's equation otherwise this seems tedious xD

  14. Empty
    • one year ago
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    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.

  15. Empty
    • one year ago
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    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.

  16. Owlcoffee
    • one year ago
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    \[\int\limits_{0}^{\infty}\frac{ cosx }{ x^2+1 }dx\] 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: \[\lim_{b \rightarrow \infty}\int\limits_{0}^{b}\frac{ cosx }{ x^2+1 }dx\]

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

  18. IrishBoy123
    • one year ago
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    \[\mathcal F \left\{ \frac{1}{t^2+1} = \pi \ e^{-2 \pi |f|}\right\}\] 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?!

  19. IrishBoy123
    • one year ago
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    oops \[\mathcal F \left\{ \frac{1}{t^2+1} \right\} = \pi \ e^{-2 \pi |f|}\]

  20. anonymous
    • one year ago
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    We can use l'hopital's rule , it ll be much quicker right ?

  21. anonymous
    • one year ago
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    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

  22. Jhannybean
    • one year ago
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    Ohhhh... integration by parts :P

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