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experimentX
 3 years ago
Evaluate:
\[ \displaystyle \int_0^{\infty} \dfrac{(\log x)^2}{x^2 + 1} dx \]
experimentX
 3 years ago
Evaluate: \[ \displaystyle \int_0^{\infty} \dfrac{(\log x)^2}{x^2 + 1} dx \]

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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0the problem hasn't been solved http://math.stackexchange.com/questions/191736/helpwithintegratingdisplaystyleint0inftydfraclogx2x21

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0going with \[t=\log x\]\[\displaystyle \int_{\infty}^{\infty} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t=\int_{\infty}^{0} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t+\int_{0}^{\infty} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t\]first one for example\[\int_{\infty}^{0} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t=\int_{\infty}^{0} t^2e^t \sum_{n=0}^{\infty}(1)^ne^{2nt} \text{d}t= \sum_{n=0}^{\infty} (1)^n\int_{\infty}^{0} t^2e^te^{2nt} \text{d}t\\=\sum_{n=0}^{\infty} (1)^n \frac{2}{(2n+1)^3}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but i'd like to know how can we go with complex integration

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0the series looks like Fourier expansion.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0hold on .. is that 3?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this one is reachable by fourier series i think

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0you made things a bit more complicated http://www.wolframalpha.com/input/?i=sum [1%2F%282n%2B1%29^3%2C+{n%2C+0%2C+Infinity}]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0but this seems nice http://www.wolframalpha.com/input/?i=sum [%281%29^n%2F%282n%2B1%29^3%2C+{n%2C+0%2C+Infinity}]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0No not really ,,, this is nice ... and interesting http://www.wolframalpha.com/input/?i=sum [%281%29^n%2F%282n%2B1%29^3%2C+{n%2C+1%2C+Infnity}]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahh yeah so this is reachable man

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0probably some nasty Fourier analysis.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hae you tried using the residue theorem?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0tried ... but stuck. I need picture of contour .... that t^2 term is bugging me badly http://openstudy.com/users/experimentx#/updates/50476f64e4b0c3bb09860ba6

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0See http://en.wikipedia.org/wiki/Methods_of_contour_integration Example (V)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0thanks ... i this is helpful ...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0i'll try it ... if it get answer i'll post solution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1347643651001:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0man this contour works

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0oh great ... but I have QM exam six days later.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\oint \frac{z^2 e^z}{1+e^{2z}} dz=a_{1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0man try it after exams

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0sure ... 14 days to go.
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