The <residue theorem> states that a <meromorphic function> with finitely many poles inside a positively oriented simple closed contour and none on it satisfies $\oint_C F(z)\,dz=2\pi i\sum\operatorname{Res}(F;a)$. The function must be holomorphic on a neighbourhood of the contour and its interior away from those poles.

Write $J=\int_0^\infty x^{1/2}\log x/(1+x^2)\,dx$ and $I=\int_0^\infty x^{1/2}/(1+x^2)\,dx$. Both integrals converge absolutely. Use the <principal complex logarithm> in the upper half-plane, with $z^{1/2}=e^{\log z/2}$, and an upper semicircle of radius $R>1$ indented above zero by a clockwise semicircle of radius $\epsilon<1$. The branch values on the negative real side are upper limits; equivalently use contours arbitrarily slightly above that side before taking a limit.

For $F(z)=z^{1/2}\log z/(1+z^2)$, the large arc is $O(R^{-1/2}\log R)$ and the small arc is $O(\epsilon^{3/2}|\log\epsilon|)$, so both vanish. On $z=-x$ approached from above, $z^{1/2}=i\sqrt x$ and $\log z=\log x+i\pi$. The oriented negative segment therefore contributes $iJ-\pi I$, and the positive segment contributes $J$. The only enclosed pole is $i$, with <residue>
$$
\operatorname{Res}(F;i)=\frac{e^{i\pi/4}(i\pi/2)}{2i}=\frac\pi4e^{i\pi/4}.
$$
Consequently
$$
(1+i)J-\pi I=2\pi i\operatorname{Res}(F;i)
=\frac{\pi^2}{2\sqrt2}(-1+i).
$$
Comparing imaginary and real parts yields the <upper-half-plane contour for square-root logarithmic integrals> evaluation:
$$
\boxed{\int_0^\infty\frac{x^{1/2}\log x}{1+x^2}\,dx=\frac{\pi^2}{2\sqrt2},\qquad
\int_0^\infty\frac{x^{1/2}}{1+x^2}\,dx=\frac\pi{\sqrt2}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 2 / 13A / Solution

{title2=$I=\pi/\sqrt2,\quad J=\pi^2/(2\sqrt2)$}

Set $I=\int_0^\infty x^{1/2}/(1+x^2)\,dx$ and $J=\int_0^\infty x^{1/2}\log x/(1+x^2)\,dx$. In an indented upper semicircle use the <principal complex logarithm> and $z^{1/2}=e^{\log z/2}$. The negative real boundary contributes $iJ-\pi I$, the positive boundary contributes $J$, and both circular arcs vanish. The pole at $i$ has <residue> $\pi e^{i\pi/4}/4$. Thus $(1+i)J-\pi I=(\pi^2/(2\sqrt2))(-1+i)$; comparing real and imaginary parts yields the displayed integrals.

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