Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Analysis / 5 / c / Solution

Use the branch from part (b) and close the real interval $[-R,R]$ with the upper semicircle, oriented counterclockwise. For $\mathrm{Im}z\ge 0$, the number $z+i\alpha$ has positive imaginary part, so the numerator is a holomorphic logarithm throughout a neighbourhood of the contour and its interior. The branch point $-i\alpha$ lies strictly below the contour. The only enclosed pole ofis the simple pole at $i\beta$, with residueTo justify discarding the arc, take $R>\max \{\beta ,\alpha +1\}$. On it, $|z+i\alpha |$ lies between $R-\alpha$ and $R+\alpha$, and the argument has bounded absolute value. HenceThe real-axis integral is absolutely convergent, since its integrand is $O(\log |x|/x^2)$ at infinity and has no real singularities. The residue theorem therefore givesFinally, $\mathrm{Re}\mathrm{Log}(x+i\alpha )=\frac{1}{2}\log (x^2+{\alpha }^2)$. Taking twice the real part yieldsThe branch choice affects the first integral's imaginary part; the second integral is real and independent of that choice.