Hi can i get a proof check please. Thanks Define $|z|=3 \implies S$ ie denote the circle of radius 3 by S \begin{equation} \frac{1}{2\pi i}\int_{S} \frac{e^{zt}}{z^2(z^2+2z+2)} dz = \\ \frac{1}{2\pi i}\int_{S} \frac{e^{zt}}{z^2(z+1+i)(z+1-i)} dz = \\ W(S,0)Res(f,0)+W(S,-1+i)Res(f,-1+i)+W(S,-1-i)Res(f,-1-i) \end{equation}
We have three poles, inside the circle of radius 3, 1 pole of order 2, and 2 simple poles. with the contour winding around each pole only once.
Calculating these i have
- $z^2 = 0;$ A pole of Order 2: then let $\psi(z) = (z-0)^{2}\frac{e^{zt}}{z^2(z^2+2z+2)} = \frac{e^{zt}}{z^2+2z+2}$ then $$Res(f,0) = \frac{1}{(2-1)!}\lim_{z \longrightarrow 0}\frac{d^{(2-1)}}{dz^{(2-1)}}(\psi(z))$$ $$Res(f,0) = \lim_{z \longrightarrow 0}\psi'(z) = \lim_{z \longrightarrow 0}\frac{e^{zt}\left(tz^2+(2t-2)z+2t-2\right)}{(z^2+2z+2)^{2}} = \frac{t-1}{2}$$
- $z=-1+i;$ A pole of Order 1: it's a simple pole then $$Res(f,-1+i) = \lim_{z \longrightarrow (-1+i)} \frac{(z+1-i)e^{zt}}{z^2(z+1+i)(z+1-i)}$$ $$= \lim_{z \longrightarrow (-1+i)} \frac{e^{zt}}{z^2(z+1+i)} = \frac{e^{(i-1)t}}{4}$$
- $z=-1-i;$ A pole of Order 1: it's a simple pole then $$Res(f,-1-i) = \lim_{z \longrightarrow (-1-i)} \frac{(z+1+i)e^{zt}}{z^2(z+1+i)(z+1-i)}$$ $$= \lim_{z \longrightarrow (-1+i)} \frac{e^{zt}}{z^2(z+1-i)} = \frac{e^{-(i+1)t}}{4}$$ \begin{equation} \frac{1}{2\pi i}\int_{S} \frac{e^{zt}}{z^2(z+1+i)(z+1-i)} dz = \\ Res(f,0)+Res(f,-1+i)+Res(f,-1-i) = \\ \frac{t-1}{2}+\frac{e^{(i-1)t}}{4}+\frac{e^{-(i+1)t}}{4} \implies \\ \frac{1}{2\pi i}\int_{S} \frac{e^{zt}}{z^2(z+1+i)(z+1-i)} dz = \frac{2(t-1)+e^{(i-1)t}+e^{-(i+1)t}}{4} \end{equation}