Series expansion of integral (ln(x))^2/(1+x^2) dx from 0 to infinity

  • Thread starter newmike
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  • #1
newmike
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Hi everyone, once I again I turn to all of your expertise in complex analysis.

Homework Statement


Evaluate

[tex]\int\frac{(ln(x))^{2}}{1+x^{2}}dx[/tex]

from 0 to +infinity by appropriate series expansion of the integrand to obtain

4[tex]\sum(-1)^{n}(2n+1)^{-3}[/tex]

where the sum goes from n=0 to +infinity


Homework Equations



Laurant series, anything in complex analysis

The Attempt at a Solution



At this point, I'm looking for how to even start this. We never covered this in class, and the professor is not really helpful with our (the other students too) questions. I started with the substitution

[tex] z=e^{t} [/tex]

and came up with:

[tex]\frac{1}{2}\int t^{2}sech(t) dt[/tex]

but I think the limits of integration change from -infinity to +infinity with this substitution. either way I am stuck here. I think I have to create a series and integrate term by term. I guess the cleanest way to get to the answer is to leave the series in terms of n and integrate that.

So I guess my question is: should I abandon the substution I have and try to come up with a series of ln about some point other than zero and substitute back to zero at the end? I'm just not sure that's legit.

Either way, just looking for tips. I'm sure it is easy, and it's just a matter of us not learning the material and I can't seem to find this covered in the books I looked at.

Thanks in advance!
 

Answers and Replies

  • #2
vela
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You're on the right track. Note that the integrand is even, so you can say
$$\int_{-\infty}^\infty \frac{t^2}{e^t+e^{-t}}\,dt = 2\int_0^\infty \frac{t^2}{e^t + e^{-t}}\,dt = 2\int_0^\infty \frac{t^2 e^{-t}}{1 + e^{-2t}}\,dt.$$ Then expand in powers of ##e^{-2t}##. The resulting integrals can be done by parts.
 

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