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I want to evaluate $$I = \int_0^\infty \frac{\ln z ~dz}{z \sqrt {1+z}}.$$ So I consider integral $$I^* = \int_0^\infty \frac{\ln ^2 z ~dz}{z \sqrt {1+z}}$$ in the contour:

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By the residue theorem $$I^* + \oint_{\gamma_\varepsilon} + \oint_{\Gamma_R} - \int_0^\infty \frac{(\ln + 2 \pi i)^2 z ~dz}{z \sqrt {1+z}} = res_{z = -1} f(z)$$ $$I^* + \oint_{\gamma_\varepsilon} + \oint_{\Gamma_R} - I^* +4\pi\int_0^\infty \frac{dz}{z \sqrt {1+z}} - 4\pi i \int_0^\infty \frac{\ln z ~dz}{z \sqrt {1+z}}= res_{z = -1} f(z)$$ $$\oint_{\gamma_\varepsilon} + \oint_{\Gamma_R} +4\pi\int_0^\infty \frac{dz}{z \sqrt {1+z}} - 4\pi i \int_0^\infty \frac{\ln z ~dz}{z \sqrt {1+z}}= res_{z = -1} f(z)$$ Integral $$4\pi\int_0^\infty \frac{dz}{z \sqrt {1+z}}$$ diverges, but I think I'll manage to separate imaginary and real part of the whole expression in the end. What I can't do is to calculate the residue. The series I get is so ugly I don't believe I'm right.

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  • $\begingroup$ $\frac{\log z}{z}$ is not an integrable function over $(0,1)$, so you are just trying to compute a divergent integral. $\endgroup$
    – Jack D'Aurizio
    Sep 12, 2015 at 13:46
  • $\begingroup$ Further to @Jack's comment, even if there were no nonintegrable singularity at $z=0$, there is no pole at $z=-1$ and thus the residue theorem does not apply. In this case, you would construct a contour in the complex plane that avoids the branch point there and traverses either side of a branch cut emanating from that branch point. Typically, Cauchy's theorem produces a new integral in terms of the original integral and if you are lucky, you can evaluate the new integral much more easily. $\endgroup$
    – Ron Gordon
    Sep 14, 2015 at 13:53

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