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I want to calculate the following integral

$$f(x)=\int_0^\infty \frac{\log(x)}{(x+1)^2\sqrt{x}}dx$$

To get it, I used the Cauchy's Residue Theorem. Consider the path given in the image

enter image description here

By the Cauchy's Residue Theorem, this integral along this path is (if I didn't make any mistake)

$$f(z)=\int_\Gamma \frac{\log(z)}{(z+1)^2\sqrt{z}}dz=2\pi i Res(f,z=-1)=\pi^2i-2\pi.$$

It's easy to check that the integral along the curve paths are zero. Then, calling $\gamma_2$ the path with the arrow pointing to the left and $\gamma_4$ the path with the arrow pointing to the right, the integral comes

$$f(z)=\int_\Gamma \frac{\log(z)}{(z+1)^2\sqrt{z}}dz=\int_{\gamma_2} \frac{\log(z)}{(z+1)^2\sqrt{z}}dz+\int_{\gamma_4} \frac{\log(z)}{(z+1)^2\sqrt{z}}dz$$

or

$$f(z)= \lim_{R\to \infty, \varepsilon \to 0}\int_R^\varepsilon \frac{\log(ye^{2\pi i})}{(e^{2\pi i}y+1)^2\sqrt{ye^{2\pi i}}}dy+\int_\varepsilon^R \frac{\log(z)}{(z+1)^2\sqrt{z}}dz,$$

where I made the change $z=ye^{2\pi i}$ because the argument of $z$ increases in $2\pi$ when it rotates around the big circle.

However, when I change the first integral to have the same limits as the others, and taking in account that $e^{2\pi i}=1$ every time, they eliminated each other.

I don't know where I'm wrong. If it helps, the result must be $-\pi.$

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    What are the coordinates on the path? – 2017-01-16
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    The big circle has radius $R$ and the small $\varepsilon$. So $\gamma_1 : t \rightarrow Re^{i\theta }$ where $\theta \in [-\pi,\pi)$ and $\gamma_3 : t \rightarrow \varepsilon e^{i\theta }$ where $\theta \in [\pi,-\pi)$ – 2017-01-16
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    Note that your definition of $\log$ is $\log z=\log|z|+i\theta$ with $z=|z|\exp(i\theta)$, $0<\theta<2\pi$. So in the first integral of your last line, at the limit, you have $ \log(y\exp(2i\pi))=(\log y)+2i\pi$, and $\sqrt{y\exp(2i\pi)}=-\sqrt{y}$. – 2017-01-16
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    That's it. Thank you very much. – 2017-01-16

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