I have the following question. I know Cauchy's integral formula but I am facing an exercise where I have singularities and I can not solve it directly. The problem is to evaluate the integral $$\int_{C^{+}} \frac{\sin(z)}{z^2(z^2+1)}dz$$ with $C\equiv|z|=2$ and it is clear that $i,-i$ and $0$ are singularities so I can not apply directly the theorem. I am lost with this exercise.
Thank you very much.
On
We can partial fraction decompose the integrand since
$$\frac{1}{z^2(z^2+1)} = \frac{1}{z^2} - \frac{1}{z^2+1} = \frac{1}{z^2} - \frac{1}{2i} \frac{1}{z-i} + \frac{1}{2i} \frac{1}{z+i}$$
which means we get that the original integral is equal to
$$I = \oint_{|z|=2} \frac{\sin z}{z^2}\:dz - \frac{1}{2i}\oint_{|z|=2} \frac{\sin z}{z-i}\:dz + \frac{1}{2i}\oint_{|z|=2} \frac{\sin z}{z+i}\:dz$$
$$ = \frac{1}{2\pi i}\oint_{|z|=2} \frac{2\pi i\sin z}{z^2}\:dz - \frac{1}{2\pi i}\oint_{|z|=2} \frac{\pi\sin z}{z-i}\:dz + \frac{1}{2\pi i}\oint_{|z|=2} \frac{\pi\sin z}{z+i}\:dz$$
$$ = (2\pi i \sin z)' \Bigr|_{z=0} - \pi \sin i + \pi \sin (-i) = 2\pi i -2\pi \sin i$$
entirely by Cauchy's integral formulae, not the residue theorem.
On
You can use the Cauchy integral formula.
To evaluate a given pole rewrite the expression as $\frac {g(z)}{z-a}$ and evaluate at $a$
e.g. to evaluate at the pole $z = i$
$\oint \frac {\sin z}{z^2(z^2+1)} = \oint \frac {\frac {\sin z}{z^2(z+i)}}{(z-i)} = 2\pi i \left(\frac {\sin z}{z^2(z+i)}\ \right)|_{z=i} = 2\pi i \frac {\sin i}{-2i} = -\pi \sin i = -\pi i\sinh 1$
Do the same at the other two poles.
The function has three simple poles at $0,i$ and $-i$. The residue at a simple pole $a$ is equal to $\lim_{z \to a} (z-a)f(z)$. I will let you calculate the residues. You should get the residues as $1$, $\frac 1 2 i\sin i$ and $\frac 1 2 i\sin i$. So the given integral equals $2 \pi i(1+\sin i)$