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we have the integral : $\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds$

which diverges for every value of n except $n=0$ if we perform the change of variables :

$s\rightarrow \frac{1}{s}$

then : $\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds=\int_{-i}^{i}\frac{(1-s)^{n}}{s^{n+1}}ds$

which converges . am i missing something here , or is this correct !?

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    @MohammadAlJamal All this is not that relevant.2012-03-30

2 Answers 2

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I assume that the original path of integration is a line between the endpoints. If so, the limiting path gets mapped to the circle below which loops from the origin, through $\frac12$, and back to the origin again. Note that $ -\int_C\frac{(1-s)^{n}}{s^{n+1}}ds $ diverges near the origin.

$\hspace{5cm}$complex paths

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For integrals in the complex plane you must specify not just the endpoints, but the path involved. I'm assuming your original integral is $ J_T = \int_{\Gamma_T} \frac{(s-1)^n}{s} \ ds $ where $\Gamma_T$ is the straight line from $2-iT$ to $2+iT$. The substitution $z=1/s$ transforms this to $\int_{C_T} \frac{(1-z)^n}{z^{n+1}}\ dz$ where $C_T$ goes from $1/(2+iT)$ to $1/(2-iT)$ on the arc not containing $0$ of a circle of radius $1/4$ centred at $1/4$. There's no reason to expect convergence.