I want to ask a practical integration and I hope the answer could be a line by line form to for me to understand.
Consider complex integration. Assume the integration is consist of a vertical line $K: [c-i\infty,c+i\infty]$, and enclose it by large radius circle $L$ at the infinity. Then complex analysis is teaching us $$ \frac{1}{2\pi i } \oint_{L+K} f(z) dz =\frac{1}{2\pi i } \int_{c-i\infty}^{c+i\infty} f(z) dz + \frac{1}{2\pi i } \int_{L} f(z) dz $$ On the left hand side, one can evaluate it by residue. On the RHS people usually evaluate the integration over large radius $K$ and find it either vanishing or some finite answer (See Chapter 2 of E. Elizalde, S. D. Odintsov, A. Romeo, Andrei A. Bytsenko, S. Zerbini - Zeta regularization techniques with applications-World Scientific (1994) for an example of finite answer for the infinity integration).
In short, I feel I lack the knowledge of how to do integration on the the vertial line $K$. I only know to do it by computing the residue and subtracting the integration from the infinity. But I assume some computation should be straightforwardly doable for the vertical straightline. This is very common in Mellin transformation.
For example: Cahen–Mellin integral (I do not find the resources for doing this line by line) $$ e^{-z} = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s) z^{-s} ds $$ I want to see the integration on the RHS can be evaluated like take $s=c+i t$. Then $$ \frac{1}{2\pi i} \int_{-\infty}^{\infty} \Gamma(c+ it) z^{-c - it} i(dt) = \text{Re}() + \text{Im} () $$ and compute for each of them. Show me this is indeed $e^{-z}$ and independent of $c$. Is this possible? I can also accept other examples.
Thank you