$\int _\gamma \frac {sinz} {z^2+1}dz$ for circle of radius 2

Question

My task is to compute $\int _\gamma \frac {sinz} {z^2+1}dz$, where $\gamma$ is the positvely oriented circle centered at the origin of radius 2.

I tried an approach similar to Jacky Chong's solution for a previous question I asked. I wrote $\int _\gamma \frac {sinz} {z^2+1}dz=\int _0 ^{2π} \frac {sin(2e^{it})} {4e^{i2t}+1}2ie^{it}dt=\int _0 ^{2π}f(t)dt+i\int _0 ^{2π}g(t)dt$ for some real functions $f$ and $g$.

But these functions are not pretty. We have $f(t)= \frac {6sint[e^{-2sint}sin(2cost)+e^{2sint}sin(2cost)]-10cost[e^{2sint}cos(2cost)-e^{-2sint}cos(2cost)} {36cos^2t+18}$

One could, in theory, compute this ugly integral, although I am having trouble doing that. But looking at the graph of $f(t)$, we should have $\int _0 ^{2π}f(t)dt=0$.

Now, to finish the problem, one could compute $g(t)$ (another ugly function) and integrate it.

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Anyways, I suspect that there is an easier way to solve the problem. (Hopefully without involving the Cauchy Residue Theorem, since my class hasn't learned that yet.)

An acceptable answer will be one of the following:

• Show me how to integrate $f$ and $g$
• Come up with a different way of solving the problem

Answer 1

HINT: use the expansion of the sine function, that is, $\sin z=\sum_{k=0}^\infty(-1)^k\frac{z^{2k+1}}{(2k+1)!}$. It "reduces" to a series of integrals of the kind

$$\int_\gamma\frac{z^{2k+1}}{z^2+1}\, dz$$

that probably you can handle using the ideas of the answer of Jacky Chong that you had linked. However, the expression of the series as a simple value could be difficult.

Answer 2

Cauchy's Integral formula is the way to go. However, I think you misunderstood the example; there's no need to do partial fractions. Note that

$$ \int_\gamma \frac{\sin z}{z^2+1} dz = \int_{\gamma_1} \frac{\sin z}{z^2+1} dz + \int_{\gamma_2} \frac{\sin z}{z^2+1} dz $$

where $\gamma_1$ is a small circle centered on $z_1=i$ and $\gamma_2$ is a small circle centered on $z_2=-i$

Let $A(z) = \dfrac{\sin z}{z+i}$, then $A(z)$ is analytic in $\gamma_1$, therefore

$$ \int_{\gamma_1} \frac{\sin z}{(z+i)(z-i)} dz = \int_{\gamma_1} \frac{A(z)}{z-i}dz = 2\pi i A(i) = 2\pi i \cdot \frac{\sin i}{2i} = \pi \sin i $$

The second integral follows the same process.