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

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

Mathematics
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http://tinyurl.com/9csdea3
the problem hasn't been solved http://math.stackexchange.com/questions/191736/help-with-integrating-displaystyle-int-0-infty-dfrac-log-x2x2-1
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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\[\int_{-\infty}^{0} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t=\int_{-\infty}^{0} t^2e^t \sum_{n=0}^{\infty}(-1)^ne^{2nt} \text{d}t= \sum_{n=0}^{\infty} (-1)^n\int_{-\infty}^{0} t^2e^te^{2nt} \text{d}t\\=\sum_{n=0}^{\infty} (-1)^n \frac{2}{(2n+1)^3}\]
but i'd like to know how can we go with complex integration
the series looks like Fourier expansion.
yeah
hold on .. is that 3?
lol .. yes
this one is reachable by fourier series i think
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}]
but this seems nice http://www.wolframalpha.com/input/?i=sum[%28-1%29^n%2F%282n%2B1%29^3%2C+{n%2C+0%2C+Infinity}]
i made it worser
No not really ,,, this is nice ... and interesting http://www.wolframalpha.com/input/?i=sum[%28-1%29^n%2F%282n%2B1%29^3%2C+{n%2C+1%2C+Infnity}]
ahh yeah so this is reachable man
probably some nasty Fourier analysis.
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Hae you tried using the residue theorem?
tried ... but stuck. I need picture of contour .... that t^2 term is bugging me badly http://openstudy.com/users/experimentx#/updates/50476f64e4b0c3bb09860ba6
See http://en.wikipedia.org/wiki/Methods_of_contour_integration Example (V)
thanks ... i this is helpful ...
*think
i'll try it ... if it get answer i'll post solution.
|dw:1347643651001:dw|
man this contour works
did you try it?
yes
oh great ... but I have QM exam six days later.
\[\oint \frac{z^2 e^z}{1+e^{2z}} dz=a_{-1}\]
man try it after exams
sure ... 14 days to go.

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