Consider the expression

fun = 1/(y^2-1)^a;

If I do the Mellin Transform explicitly, I get a result that involves a few conditions on the parameters a,s

Integrate[(y)^(s - 1)fun, {y, 0, Infinity}]

ConditionalExpression[ 1/2 Gamma[ 1 - a] (Gamma[a - s/2]/Gamma[1 - s/2] + ((-1)^-a Gamma[s/2])/ Gamma[1 - a + s/2]) , 2 a > Re[s] && Re[s] > 0 && a < 1]

However, if I just use the Mellin transform function, no restrictions on the parameters a,s appear:

MellinTransform[fun, y, s]

1/2 Gamma[ 1 - a] (Gamma[a - s/2]/Gamma[1 - s/2] + ((-1)^-a Gamma[s/2])/ Gamma[1 - a + s/2])

Which result should I trust? Are the restrictions really necessary, or is there maybe some analytic continuation going on which the MellinTransform function knows about while Integrate does not?


From the docs: enter image description here

Your case:

MellinTransform[1/(y^2 - 1)^a, y, s, GenerateConditions -> True]
  1/2 Gamma[1 - a] (Gamma[a - s/2]/Gamma[1 - s/2] + ((-1)^-a Gamma[s/2])/Gamma[1 - a + s/2]), 
  2 a > Re[s] && Re[s] > 0 && a < 1
  • $\begingroup$ Wonder why this option is off by default. If no conditions are given, the result is wrong. There is no instance I can think of, in which not giving the conditions would be useful. $\endgroup$ – Kagaratsch Mar 17 '17 at 14:58
  • 3
    $\begingroup$ @Kagaratsch To be honest, I don't know. For computing formal results, it's useful. It's also less verbose. I found this answer though: mathematica.stackexchange.com/a/46492/4346 $\endgroup$ – Chip Hurst Mar 17 '17 at 15:06

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