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For an entire function $f$, let $F(z)=f(z)z^{-\alpha }$ with noninteger real $\alpha$, and choose the branch of the complex logarithm with $\varphi <\arg z<\varphi +2\pi$. On the cut ray $z=r{e}^{i\varphi }$, the bank whose argument approaches $\varphi$ has value $F_{+}=f(r{e}^{i\varphi })r^{-\alpha }{e}^{-i\alpha \varphi }$, while the other bank has argument approaching $\varphi +2\pi$ and value $F_{-}=e^{-2\pi i\alpha }{F}_{+}$. The signed jump is therefore $(1-e^{-2\pi i\alpha })F_{+}$ when defined as $F_{+}-F_{-}$. Stating the branch and order of subtraction removes the sign ambiguity. A keyhole contour uses this phase relation to convert its two bank integrals into a constant multiple of one real or radial integral.

On $D=\mathbb{C}\setminus [0,\infty )$ choose the continuous argument $0<\arg z<2\pi$, and putThis is holomorphic on the slit plane for every fixed $w$. The upper boundary value at $x>0$ is $x^w$, whereas the lower boundary value is $e^{2\pi iw}{x}^w$. This difference is exactly what the keyhole contour measures.

Use the branch of the complex logarithm with $0<\arg z<2\pi$, so the branch cut is the positive real axis, and take $\sqrt{z}=\exp (\mathrm{Log}z/2)$ on the same slit plane. Apply the residue theorem toon a positively oriented keyhole contour with inner radius $\rho <1$ and outer radius $R>1$. The upper bank travels from $\rho$ to $R$, the large circle counterclockwise, the lower bank from $R$ to $\rho$, and the small circle clockwise. One may first keep the banks at small nonzero angles and then take their limits.

On the upper bank, $\mathrm{Log}x=\log x$ and $\sqrt{x}$ is positive. On the lower bank, the corresponding limits are $\log x+2\pi i$ and $-\sqrt{x}$. Reversing the lower-bank orientation therefore makes the sum of the bank integrals tend toBoth are absolutely convergent: near zero compare with $x^{-1/2}(1+|\log x|)$, and near infinity with $x^{-5/2}(1+\log x)$. On the large circle, $|z^2+1|\ge R^2-1$, so its contribution is $O((\log R+1)R^{-3/2})$. On the small circle it is $O(\sqrt{\rho }(1+|\log \rho |))$. Both tend to zero.

The two simple poles are $i$ and $-i$. On the chosen branch their arguments are $\pi /2$ and $3\pi /2$, respectively. Consequently their residues areFor example, the first follows from $\mathrm{Log}i=i\pi /2$, $\sqrt{i}=e^{i\pi /4}$ and the derivative $2i$ of the denominator factor $z^2+1$. The second uses $\sqrt{-i}=e^{3\pi i/4}$, rather than the value from the principal square-root branch. Adding and applying the residue theorem yieldsAs $J$ is real, taking real parts givesThe imaginary part also gives $J=\pi /\sqrt{2}$, an independent consistency check on the orientation and branch factors.

Near zero the magnitude of the integrand is of order $t^{\mathrm{Re}z-1}$, and at infinity it is of order $t^{\mathrm{Re}z-3}$. Thus the improper integral converges exactly in the stripAt a boundary of this strip the logarithmic-variable oscillation has no limiting primitive, so conditional convergence does not enlarge it.

Use a keyhole contour about the positive real axis for $w^{z-1}/(1+w{)}^2$, taking $0<\arg w<2\pi$. The circular arcs vanish in the indicated strip. The upper and lower banks contribute $(1-e^{2\pi iz})F(z)$. The only enclosed pole is the double pole at $w=-1$, whose residue is $(z-1)e^{i\pi z}$. The residue theorem therefore givesAt $z=1$ the quotient has a removable singularity, with limit one, agreeing with ${\int }_0^{\infty }(1+t{)}^{-2}dt=1$.

Introduce the Mellin transform integral $J(s)={\int }_0^{\infty }{x}^{s-1}/(1+x^2)dx$, valid for $0<\mathrm{Re}s<2$. A keyhole contour for $z^{s-1}/(1+z^2)$ with $0<\arg z<2\pi$ has residues $i^{s-1}/(2i)$ and $(-i{)}^{s-1}/(-2i)$, with the latter argument $3\pi /2$. The two sides of the positive-axis cut give $(1-e^{2\pi i(s-1)})J(s)$. The small and large arcs vanish in this strip; simplifying the residue theorem identity yieldsDifferentiation under the integral is valid on compact substrips because the extra $\log x$ is dominated at both endpoints. ThereforeAt $s=3/2$, the requested values are

For this real integral choose $0<\arg z<2\pi$, so the upper bank of the positive real axis has $z^{1/4}=x^{1/4}>0$. Use a counterclockwise keyhole contour with inner radius $\epsilon$ and outer radius $R$, enclosing $\pm i$. The upper-bank contribution tends to $J$. On the lower bank, $z^{-1/4}=e^{-i\pi /2}{x}^{-1/4}=-i{x}^{-1/4}$, and the reversed orientation makes its contribution $+iJ$.

The large arc is $O(R^{-5/4})$ because $g=O(R^{-9/4})$, while the small arc is $O({\epsilon }^{3/4})$. Both vanish, givingThe residues on this branch areThe residue theorem yieldsThereforeThe branch cut in (a) was arbitrary; here placing it on the integration axis makes the bank contributions directly proportional to the desired positive-real integral.

Use a counterclockwise keyhole contour around the positive real branch cut, with small radius $\epsilon$ and large radius $R$. On the upper bank $\arg z=0$, so the integral tends to $I$. On the lower bank $\arg z=2\pi$, $z^{1/2}=-\sqrt{x}$, giving the negative of the upper-bank integrand. That bank is traversed from $R$ to $\epsilon$, so its contribution also tends to $I$.

The outer circle contributes $O(R^{-3/2})$, since $f=O(R^{-5/2})$ and its length is $O(R)$. The small circle contributes $O({\epsilon }^{1/2})$, since $f=O({\epsilon }^{-1/2})$. Both vanish. HenceThe contour encloses both double poles $\pm i$ but excludes the branch point.

Use the principal logarithm $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z<\pi$, whose branch cut is the negative real axis. For this choice,has three simple poles away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\mathrm{Log}1=0$, must be included.

Choose $0<\epsilon <\min \{1,a\}$ and $R>\max \{1,a\}$, and use a positively oriented keyhole contour around the negative real axis: the large circle is traversed counterclockwise, the upper bank from $-R$ to $-\epsilon$, the small circle clockwise, and the lower bank from $-\epsilon$ to $-R$. Formally, use radial segments at arguments $\pm (\pi -\eta )$, then let $\eta \downarrow 0$ with $R,\epsilon$ fixed. The function has continuous limiting values on each compact bank, which justifies this limit.

Put $L=\log a$. The residues areThe first follows from $\mathrm{Log}z=(z-1)+O((z-1{)}^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum isThese expressions are valid also for $a=1$, since $L\pm i\pi /2$ never vanish for real $L$.

On the outer circle, $|\mathrm{Log}z|\ge \log R$ for $R>1$ and $|z^2+a^2|\ge R^2-a^2$. On the inner circle, $|\mathrm{Log}z|\ge |\log \epsilon |$ and $|z^2+a^2|\ge a^2-{\epsilon }^2$. Thus the estimation lemma bounds the two arc integrals byrespectively, as $R\to \infty$ and $\epsilon \downarrow 0$.

On the upper bank, $z=-x+i0$ gives $\mathrm{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined integral isThe real improper integral converges: its integrand is bounded near zero by $1/(a^2{\pi }^2)$ and at infinity by $1/({\pi }^2{x}^2)$. Taking the arc limits in the residue theorem gives $-2\pi iI=2\pi i\Sigma$, henceThe final subtraction is exactly the contribution of the pole at $1$.