How does $\frac{1}{2\pi i}\int_{|z|=3} \frac{e^{zt}}{z^2+1}dz=sin(t)$ when $t>0$?
My class is working on the Cauchy Integral Formula currently. Here is my work:
The zeros are z=1,-1 so both lie within the circle we're integrating over.
$\int_{|z|=3} \frac{e^{zt}}{(z+1)(z-1)}dz$
Then using partial fractions: $\int_{|z|=3} \frac{\frac{e^{t}}{2}}{(z-1)}dz$-$\int_{|z|=3} \frac{\frac{e^{-t}}{2}}{(z+1)}dz$
=$2\pi i f(1)$-$2\pi i f(-1)$
=$2\pi i \frac{e^t}{2}$-$2\pi i \frac{e^{-t}}{2}$
=$\pi i e^t$-$\pi ie^{-t}$
I think I'm missing some relationship between $e^t$ and $sin(t)$, but I'm not sure.
The roots of the denominator are $\pm\mathrm i$, not $\pm1$. The partial fraction decomposition should then come out as $\frac{\mathrm i\mathrm e^{zt}}{2}\left(\frac{1}{z+\mathrm i}-\frac{1}{z-\mathrm i}\right)$. In the end, you should get
$2\pi\mathrm i\frac{\mathrm i}{2}\left(\mathrm e^{-\mathrm it}-\mathrm e^{\mathrm it}\right)$.
With Euler's formula, you also get that $\sin t=-\frac{\mathrm i}{2}\left(\mathrm e^{\mathrm it}-\mathrm e^{-\mathrm it}\right)$.