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How might I evaluate this integral: $$\int_0^\infty \frac{\log\ x}{1+x^2}dx$$

I tried substituing $x = \tan\ v$ but that didn't get me anywhere. Any suggestions as to a thought process for this one?

Mikasa
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kloptuy554
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1 Answers1

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Break up the integral into two pieces:

$$\int_0^1 dx \frac{\log{x}}{1+x^2} + \int_1^{\infty} dx \frac{\log{x}}{1+x^2}$$

Sub $x=1/y$ in the second integral and see that the sum of the above integrals, and hence the original integral, is zero.

Ron Gordon
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    thats really clever! is there is a reasoning behind why you split the integral up? – kloptuy554 Jan 29 '14 at 16:05
  • @kloptuy554: always a useful transformation when there is a quadratic in the denominator. Really even more useful when there is a log. – Ron Gordon Jan 29 '14 at 16:05
  • @kloptuy554: Please register. It 'll help you to trace your works here better. :-) and +1 for my friend Ron. – Mikasa Jan 29 '14 at 16:12