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How do I solve the Integral $$ \frac{1}{2\pi j} \oint \frac{b^{ - s} \Gamma[2 + i - s] \Gamma[s] \Gamma[-1 - i + s]}{ (2 + i - s) \Gamma[3 + i - s]} \:\mathrm{d}s$$

This integral is an inverse Mellin transform. Therefore, the contour extends from $l+j\infty$ to $l-j\infty$, where $l\in\mathbb{R}$.

$j=\sqrt{-1}$

$b \in\mathbb{R},\quad b>0$

$i\in\mathbb{R}, \quad i\ge 0$

$\Gamma(.) $ is the gamma function. Does it make any difference when $i$ becomes an integer?

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    $\begingroup$ Why not use $\Gamma[3+i-s]=(2+i-s)\Gamma[2+i-s]$ to cancel a Gamma function in the numerator and denominator? $\endgroup$
    – Stopple
    Mar 22 '13 at 22:45
  • $\begingroup$ I agree with you . I can substitute $$\frac{1}{2+i-s} = \frac{\Gamma(2+i-s)}{\Gamma(3+i-s)} $$, but would this result in a Meijer G-function? because two of the poles will be coinciding and have a difference of 1. $\endgroup$
    – Remy
    Mar 23 '13 at 16:36

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