Compute complex integral inside an open curve

Question

I need to compute this complex integral:

$$ \int_\gamma \frac{1}{(z-i)(z-2i)} dz $$

$\gamma$ is defined as:

$$ \gamma (t) = t + i(3e^t\cos^2(t)) $$

The parameter t belongs to the following interval: $[\frac{-\pi}{2},\frac{\pi}{2}] $

My approach to this question was to use the Residue's Theorem. However, the theorem requires a closed curve.

The problem above does not portray a closed curve. It is an open one.

I know that open curves can be manipulated as closed ones. You treat it artificially as a closed one and, by the end, you take out (or add) something.

How can I approach this integral?

The partial result of the residues is:

$Res(f,i)= i$ and $Res(f,2i)=-i$

Hence:

$2\pi i(+i-i)=0 $

P.S.: I also tried a different approach with the formula:

$$ \int f(\gamma(t))\gamma'(t)dt $$

Nonetheless, this approach results in a massive and ugly integral. I think the correct approach is the manipulation described above.

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UPDATE

Professor Robert Israel commented bellow and suggested using a third approach with $F(\gamma(\pi/2)) - F(\gamma(-\pi/2))$. In order to do that, I needed to find the primitives. I got them with a little help from Wolfram Alpha. As you can see bellow, the primitive is not really friendly. I am still curious on how to solve this problem treating the curve as a closed one and manipulating it:

Answer 1

Here is a brute force approach. First, decompose the integrand:

$\int_\gamma \frac{1}{(z-i)(z-2i)} dz=i\int_\gamma \frac{1}{z-i} dz-i\int_\gamma \frac{1}{z-2i} dz.$

Now, treat each one separately:

For the first integral, complete $\gamma $ to a closed curve $\Gamma$ by taking $\gamma_1$ to be the circular arc centered at $z=i$ of radius $|i-\pi/2|$ going from $x=-\pi/2$ to $x=\pi/2$ and find that the angle $\theta $ subtended by this arc is given by $\cos \frac{1}{2}\theta=\frac{1}{|i-\pi/2|}.$ Then, $i\int_{\gamma_1} \frac{1}{z-i} dz=-\theta.$

Now, the argument principle applies to show that $i\int_{\Gamma} \frac{1}{z-i} dz=i(2\pi i)(1)=-2\pi.$

Therefore, $i\int_{\Gamma} \frac{1}{z-i} dz=i\int_{\gamma_1} \frac{1}{z-i} dz-i\int_{\gamma} \frac{1}{z-i} dz\Rightarrow \int_{\gamma} \frac{1}{z-i} dz=i\theta-2\pi i$.

The second integral is treated in the same way, this time with a circle centered at $2i$.

Answer 2

Hint: find an antiderivative $F$ of the integrand that is analytic in a region containing the curve $\gamma$. For this it suffices that any branch cuts are on the imaginary axis below $2i$. Then the integral is $F(\gamma(\pi/2)) - F(\gamma(-\pi/2))$.