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I have the following integral to compute: $$ \int_{0}^{\infty}\frac{\log x}{1 + x^2}\text{d}x.$$ Following is my attempt:
$$ \int_{0}^{\infty}\frac{\log x}{1 + x^2}\text{d}x = \int_{0}^{1}\frac{\log x}{1 + x^2}\text{d}x + \int_{1}^{\infty}\frac{\log x}{1 + x^2}\text{d}x .$$ But using the substitution $x=1/u$, we get: $$\int_{0}^{1}\frac{\log x}{1 + x^2}\text{d}x= -\int_{\infty}^{1}\frac{1}{u^2}\cdot\frac{\log (1/u)}{1 + (1/u)^2}\text{d}u = -\int_{1}^{\infty}\frac{\log u}{1 + u^2}\text{d}u.$$ Hence $$ \int_{0}^{\infty}\frac{\log x}{1 + x^2}\text{d}x = -\int_{1}^{\infty}\frac{\log u}{1 + u^2}\text{d}u + \int_{1}^{\infty}\frac{\log x}{1 + x^2}\text{d}x = 0$$ since $u$ is a dummy variable.

What I'd like to do now is to compute the same integral using the method of residues(I have no experience with it) and I'd gladly appreciate any kind of help.
Thanks.

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    I believe you meant to have $\frac{1}{u^2}$ inside the middle integral in line 3?2011-10-29
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    Even as an enthusiast of residue calculations I wouldn't change a word of your solution. But if you want some problems and methods to cut your teeth on you can look at Markushevich's "Theory of functions of a complex variable" (the second volume in particular). Then there is Krantz's book, which contains a lot of exercises, but really any book on complex analysis (really, any one at all!) will have a chapter on these things.2011-10-29
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    @Gunnar: Thanks, but could you show me how to use the method to evaluate the above integral?2011-10-29

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