I understand the solution for $\epsilon =-1$. And I am trying the solve this question for $\epsilon =+1$. This is important for me. I want really to learn perfectly because I am continuously seein' such questions.
Please check my answer for $\epsilon =+1$
The question is that
Find the integral $$I(n,\epsilon)=\int_{-i \infty}^{+i \infty} \frac{e^{\epsilon z}}{(z+\epsilon)^n}$$
The integration is taken along the imaginary axis,
an integer $n\gt 0$ and $\epsilon \in \{-1,+1\}$
solution:
I need to consider the countour integral $$\oint_C \frac{e^{z}}{(z+1)^n}dz$$
The integral is counterclockwise (i.e closing the left)
Let's solist this integral into 2 pieces;
$$\int_{iR}^{iR} \frac{e^z}{(z+1)^n}dz +iR \int_{\pi /2}^{3\pi /2} e^{i\theta}\frac{e^{Rcos\theta} e^{iRsin\theta}}{(Re^{i\theta}+1)^n}d\theta$$
Next I need to show that the magintude of the second integral vanishes as $R\to \infty$ for $n\gt 0$
The magintude is bounded
$$R\int_{\pi/2}^{3\pi /2}\frac{e^{Rcos\theta}}{R^n}d\theta= \frac{1}{R^{n-1}}\int_{0}^{\pi}e^{-Rsin\theta}d\theta\le 2\frac{1}{R^{n-1}}\int_{0}^{\pi/2}e^{-R\theta/\pi}d\theta\le \pi/R^n$$
By the residue theorem, the contour integral equals to $2i\pi$ times the residue at the pole $z=-1$
$$\frac{1}{(n-1)!}[\frac{d^{n-1}}{dz^{n-1}}e^z]_{z=-1}=\frac{1^{n-1}}{(n-1)!}e^{-1}$$
So I $$\int_{-i\infty}^{+i\infty}\frac{e^z}{(z-1)^n}dz=2\pi i\frac{1}{(n-1)^!}e^{-1}$$
Please show my mistakes. Thank you:)