How should I pick the contour to compute the integral
$$\int_{-a}^a\frac{1-z}{\sqrt{(z-a)(z+a)}}\mathrm d z\,, $$
where $a$ is a real number?
My problem is that when I choose a keyhole contour around the cut $(-a,a)$, the big circle with radius $R$ going to infinity diverges. The integrand goes as
$$\frac{1-z}{\sqrt{(z-a)(z+a)}} \sim i-\frac{i}{z}+i\frac{ a^2}{2 z^2}+O\left(\frac{1}{z^3}\right)\,,$$
for $|z|\rightarrow \infty$ and thus
$$ \lim_{R\rightarrow \infty}\int_R\frac{1-z}{\sqrt{(z-a)(z+a)}}\mathrm d z < \lim_{R\rightarrow\infty}( i\times (2\pi R))=\infty\,.$$
But I know that the answer is finite as
$$\int_{-a}^a\frac{1-z}{\sqrt{(z-a)(z+a)}}\mathrm d z =\pi\,. $$
Where am I doing something wrong?
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