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?

  • $\begingroup$ By keyhole contour do you mean dogbone contour? $\endgroup$
    – B. Mehta
    May 26, 2017 at 11:24
  • 1
    $\begingroup$ You might like to read about the residue at infinity $\endgroup$
    – B. Mehta
    May 26, 2017 at 11:25

1 Answer 1


I think that taking into account the helpful comment asking me to consider the residue at infinity, the right contour is the following figure

Call $C_{\epsilon,\pm}$ the small circles with radius $\epsilon$ around $-a$ and $a$ respectively, $J_{\pm}$ the part of the path just above and below the cut connecting $a$ and $-a$ and the big circle going to infinity has a radius $R$.

As the only pole when circling the branch cut joining $a$ to $-a$

$$2\pi i\text{ Res}(f(z),z=\infty)=\left(\int_{C_{\epsilon,+}}+\int_{C_{\epsilon,-}}+\int_{J_+}+\int_{J_-}\right)f(z)\mathrm d z\,, $$ where $$f(z)=\frac{(1-z)}{\sqrt{(z-a)(z+a)}}\,. $$


$$\lim_{\epsilon\rightarrow 0}\int_{C_{\epsilon,\pm}}\frac{(1-z)}{\sqrt{(z-a)(z+a)}} \mathrm d z \rightarrow 0$$

the equality reduces to

$$\text{Res}(f(z),z=\infty)=\left(\int_{J_+}+\int_{J_-}\right)f(z)\mathrm d z\,, $$

The two integrals along the straight lines $J_{\pm}$ add up as one can remark that $$\sqrt{(z-a)(z+a)}=\begin{cases} \sqrt{x^2-a^2} && a<x<\infty\\ -\sqrt{x^2-a^2} && -\infty<x<-a\\ i\sqrt{a^2-x^2} && -a<x<a &y\rightarrow 0^+\\ -i\sqrt{a^2-x^2} && -a<x<a &y\rightarrow 0^-\end{cases} $$ and so that $$\int_{J_+} \mathrm d z \, f(z +i\epsilon)=\int_{J_-} \mathrm d z \, f(z-i\epsilon)= -iI $$ with $\epsilon$ an infinitessimally positive small real number and $I$ the integral we are looking for.

Now Res$(f(z),z=\infty)=-1$ such that $$I= \pi. $$


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