Question

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

Answer 1

http://tinyurl.com/9csdea3

Answer 2

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

Answer 3

*

Answer 4

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\[\int_{-\infty}^{0} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t=\int_{-\infty}^{0} t^2e^t \sum_{n=0}^{\infty}(-1)^ne^{2nt} \text{d}t= \sum_{n=0}^{\infty} (-1)^n\int_{-\infty}^{0} t^2e^te^{2nt} \text{d}t\\=\sum_{n=0}^{\infty} (-1)^n \frac{2}{(2n+1)^3}\]

Answer 5

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

Answer 6

the series looks like Fourier expansion.

Answer 7

yeah

Answer 8

hold on .. is that 3?

Answer 9

lol .. yes

Answer 10

this one is reachable by fourier series i think

Answer 11

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}]

Answer 12

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}]

Answer 13

i made it worser

Answer 14

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}]

Answer 15

ahh yeah so this is reachable man

Answer 16

probably some nasty Fourier analysis.

Answer 17

*

Answer 18

Hae you tried using the residue theorem?

Answer 19

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

Answer 20

See http://en.wikipedia.org/wiki/Methods_of_contour_integration
Example (V)

Answer 21

thanks ... i this is helpful ...

Answer 22

*think

Answer 23

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

Answer 24

man this contour works

Answer 25

did you try it?

Answer 26

yes

Answer 27

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

Answer 28

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

Answer 29

man try it after exams

Answer 30

sure ... 14 days to go.