I have got the following question in my homework at which I am stuck.

Use Cauchy's Integral Formula to evaluate the following integral:
Now, here $z=1$ is a point on the boundary of the given curve $|z|=1$, at which the integrand has a singularity. And Cauchy's integral formula is applicable for points in the interior of the curve. But the question specifies to use Cauchy's integral formula.

I am really confused as to how solve this and whether Cauchy's Theorem can be used here which states that:
If f(z) is holomorphic in a simply connected open region R and $\gamma$ is a rectifiable closed curve contained in R then, $\int_{\gamma} f(z)dz$ = 0

  • $\begingroup$ The integral does not even exist in the Lebsgue sense. $\endgroup$ Commented Jan 15, 2021 at 7:48
  • $\begingroup$ I do not have any idea about that $\endgroup$
    – Esha
    Commented Jan 15, 2021 at 7:49
  • $\begingroup$ Are you sure you have the right contour? $\endgroup$ Commented Jan 15, 2021 at 9:11
  • $\begingroup$ @Tito Eliatron Yes $\endgroup$
    – Esha
    Commented Jan 15, 2021 at 11:25

1 Answer 1


The integral on a path that passes through a pole will likely not exist in the normal sense. However, one convention is instead to calculate its principal value. This amounts to calculating the integral on the curve excluding a small symmetrical segment on either side of the pole and then letting the size of the excluded portion approach zero. If the limit exists then that is the principal value.

It can be evaluated using the "Indentation Lemma" which says that for $f$ holomorphic except for a simple pole at $z_0$, integration along an incomplete circular shaped arc around $z_0$, given by $\gamma= z_0+\varepsilon e^{i\theta}$, $\theta_1 \leqslant \theta \leqslant\theta_2$ has limiting value $i \cdot(\theta_2-\theta_1)\cdot\text{res}(f, z_0)$ as $\varepsilon \to 0$.

Then, to proceed in your case, evaluate the integral around $|z|=1$ except take a small (nearly) semi-circular detour around the outside of $z=1$. By Cauchy's residue theorem this integral is equal to $2\pi $ times the sum of the residues inside the curve, which includes that at $z=1$ by virtue of the detour. You can use the Indentation lemma to calculate how much the detour contributes in the limit, an by subtraction you obtain the limiting value of the integral along the curve $|z| = 1$ excluding a small segment.

The result will be $2\pi i \cdot \sum \text{res}(f, |z|<1)$, zero in this case, plus, $\pi i\cdot \text{res}(f,z=1)$


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