How do I evaluate $\frac{1}{2 \pi i} \oint_C \frac{e^{zt}}{(z^2 + 1)^2} \,dz$

Question

How do I evaluate this integral when t > 0 and C is the Circle $\mid z \mid$ = 3 using Cauchy's Integral Formula

Answer 1

To evaluate the integral $\displaystyle \frac{1}{2 \pi i} \oint_C \frac{e^{zt}}{(z^2 + 1)^2} \,dz$ we see $(z^2+1)^2=(z+i)^2(z-i)^2$. The Cauchy Integral formula says $$f'(z_0)=\dfrac{1}{2\pi i}\int_{|z-z_0|=\rho}\frac{f(z)}{(z-z_0)^2}\,dz$$ Since $C$ is the circle $|z|=3$ oriented counter-clockwise, we divide it to $C_1: |z-i|=\frac12$ and $C_2: |z+i|=\frac12$ both oriented counter-clockwise. So here $z_0=\pm i$ and \begin{eqnarray} \dfrac{1}{2\pi i}\oint_C \frac{e^{zt}}{(z^2 + 1)^2}\,dz &=& \dfrac{1}{2\pi i}\oint_{C_1} \frac{\frac{e^{zt}}{(z+i)^2}}{(z-i)^2}\,dz + \dfrac{1}{2\pi i}\oint_{C_2}\frac{\frac{e^{zt}}{(z-i)^2}}{(z+i)^2}\,dz\\ &=& \left(\frac{e^{zt}}{(z+i)^2}\right)'\Big|_{z=i} + \left(\frac{e^{zt}}{(z-i)^2}\right)'\Big|_{z=-i}\\ &=& \frac{e^{+it}(2it-2)}{-8i} + \frac{e^{-it}(-2it-2)}{8i}\\ &=& \color{blue}{\dfrac{\sin t-t\cos t}{2}} \end{eqnarray}