Can the integral be made to converge by changing the variable?

[mmzaj]

greetings . 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 !?

 

[jackmell]

greetings . 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 :

Can you solve it for n=2 using antiderivatives? That is, what is:

\lim_{T\to\infty}\left(-2s+\frac{s^2}{2}+\log(s)\right)_{2-iT}^{2+iT}

 

[mathman]

greetings . 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 !?

You need to split the original integration range into two parts at s = 0. Now when you change s to 1/s, you will be able to get the correct integration limits. Note also that the original T becomes 1/T.