I was solving the residues of $f(z)e^{zt} = e^{zt}\frac{\ln(z)}{z^2+1}$ as follows:
$$\operatorname{Res}(f(z)e^{zt}, i) = \lim_{z\to i} (z-i)\frac{e^{zt}\ln(z)}{(z-i)(z+i)} = \frac{e^{it}\ln(i)}{2i}$$
and
$$\operatorname{Res}(f(z)e^{zt}, -i) = \lim_{z\to -i} (z+i)\frac{e^{zt}\ln(z)}{(z-i)(z+i)} = \frac{e^{-it}(\ln(i)+i\pi)}{-2i}$$
$$\sum \textrm{Residues} = \frac{e^{it}(\ln(i)+i\pi)}{-2i}+\frac{e^{it}\ln(i)}{2i} = \frac{-\pi e^{-it}}{2}$$
My book says the answer is $\frac{\pi}{2}\cos(t)$ is it possible to simplify $\frac{-\pi e^{-it}}{2}$ to $\frac{\pi}{2}\cos(t)$.
Thanks!
Assuming "$\ln$" is the principal branch of the complex logarithm: $$ \operatorname{Res}\limits_{z=i}\left( \frac{e^{zt}\ln z}{z^2+1} \right) = \frac{e^{it}\ln i}{2i} = \frac{e^{it} i\pi}{4i} = \frac{\pi e^{it}}{4} $$ and $$ \operatorname{Res}\limits_{z=-i}\left( \frac{e^{zt}\ln z}{z^2+1} \right) = \frac{e^{-it}\ln(-i)}{-2i} = \frac{e^{-it} i\pi}{4i} = \frac{\pi e^{-it}}{4}. $$ Finally: $$ \frac{\pi e^{it}}{4} + \frac{\pi e^{-it}}{4} = \frac{\pi}{2} \cdot \frac{e^{it}+e^{-it}}{2} = \frac{\pi}{2}\cos t. $$
Note that $\ln(i) = \ln|i| + i\arg(i) = i\pi/2$ and $\ln(-i) = \ln|-i| + i\arg(-i) = -i\pi/2$. Here $\arg$ is the principal argument, i.e. the argument between $-\pi$ and $\pi$.