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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.

The tangent half-angle substitution $u=\tan (t/2)$ maps $0\le t<\pi$ to $0\le u<\infty$, withBecause $a>1$, both $a+1$ and $a-1$ are positive. ThusEvaluating the improper integral givesAlthough the change of variable has an infinite endpoint, the original integrand is continuous on the closed interval, and the transformed integrand decays as $u^{-2}$, so the limiting substitution is justified.

Inside the unit circle, use the geometric series for $1/(1-z)$:Outside the unit circle, expand in powers of $1/z$ instead:The different Laurent series reflect different annuli separated by the pole at one. Both follow from absolutely convergent geometric series in their stated regions.

The Laurent theorem states that a holomorphic function on the annulus $R<|z-a|<S$ has a unique Laurent seriesThe integration circle is counterclockwise. The series converges absolutely at every point of the annulus and uniformly on every compact subannulus. The coefficient integrals are independent of the choice of $\rho$ by the Cauchy integral theorem applied between two such circles. These statements include $R=0$ or $S=\infty$ when the indicated annulus has those endpoints.