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

1. experimentX
2. experimentX

the problem hasn't been solved http://math.stackexchange.com/questions/191736/help-with-integrating-displaystyle-int-0-infty-dfrac-log-x2x2-1

3. mukushla

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4. mukushla

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}$

5. mukushla

but i'd like to know how can we go with complex integration

6. experimentX

the series looks like Fourier expansion.

7. mukushla

yeah

8. experimentX

hold on .. is that 3?

9. mukushla

lol .. yes

10. mukushla

this one is reachable by fourier series i think

11. experimentX

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}]

12. experimentX

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}]

13. mukushla

i made it worser

14. experimentX

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}]

15. mukushla

ahh yeah so this is reachable man

16. experimentX

probably some nasty Fourier analysis.

17. eliassaab

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18. eliassaab

Hae you tried using the residue theorem?

19. experimentX

tried ... but stuck. I need picture of contour .... that t^2 term is bugging me badly http://openstudy.com/users/experimentx#/updates/50476f64e4b0c3bb09860ba6

20. eliassaab
21. experimentX

thanks ... i this is helpful ...

22. experimentX

*think

23. experimentX

i'll try it ... if it get answer i'll post solution.

24. mukushla

|dw:1347643651001:dw|

25. mukushla

man this contour works

26. experimentX

did you try it?

27. mukushla

yes

28. experimentX

oh great ... but I have QM exam six days later.

29. mukushla

$\oint \frac{z^2 e^z}{1+e^{2z}} dz=a_{-1}$

30. mukushla

man try it after exams

31. experimentX

sure ... 14 days to go.