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

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experimentXBest ResponseYou've already chosen the best response.0
the problem hasn't been solved http://math.stackexchange.com/questions/191736/helpwithintegratingdisplaystyleint0inftydfraclogx2x21
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
going 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\[\int_{\infty}^{0} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t=\int_{\infty}^{0} t^2e^t \sum_{n=0}^{\infty}(1)^ne^{2nt} \text{d}t= \sum_{n=0}^{\infty} (1)^n\int_{\infty}^{0} t^2e^te^{2nt} \text{d}t\\=\sum_{n=0}^{\infty} (1)^n \frac{2}{(2n+1)^3}\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
but i'd like to know how can we go with complex integration
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
the series looks like Fourier expansion.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
hold on .. is that 3?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
this one is reachable by fourier series i think
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
you 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}]
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.0
No 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}]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
ahh yeah so this is reachable man
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
probably some nasty Fourier analysis.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
Hae you tried using the residue theorem?
 one year ago

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

eliassaabBest ResponseYou've already chosen the best response.1
See http://en.wikipedia.org/wiki/Methods_of_contour_integration Example (V)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
thanks ... i this is helpful ...
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
i'll try it ... if it get answer i'll post solution.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
dw:1347643651001:dw
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
man this contour works
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
oh great ... but I have QM exam six days later.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
\[\oint \frac{z^2 e^z}{1+e^{2z}} dz=a_{1}\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
man try it after exams
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
sure ... 14 days to go.
 one year ago
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