I encountered behaviour I wasn't expecting with Integrate and thought I'd ask here before reporting a bug in case I'm just missing something obvious. A simple example is the following: Computing

Assuming[n >= 1 && n \[Element] Integers && p < \[Mu] && p > 0 && \[Mu] > 0,
  Integrate[1/((p0 - I*\[Mu])^2 + p^2)^n, {p0, -Infinity, Infinity}]]

gives a general answer, which eg. in the case $n=1$ reduces to $\pi/p$. However, this is the answer in the case $p>\mu$, which is the exact opposite of the assumption given to Mathematica. In the region $p<\mu$ the integral should vanish, which is easily confirmed by specifying the value of $n$ before integrating:

Assuming[p < \[Mu] && p > 0 && \[Mu] > 0, 
 Integrate[1/((p0 - I*\[Mu])^2 + p^2), {p0, -Infinity, Infinity}]]

Which outputs zero, as it should.

Search gave some at least vaguely similar bugs, but all of those were supposed to have been fixed around version 10, whereas this is present in 11.3.

  • $\begingroup$ Seems to continue in 12, I also get (p^(1 - 2 n) Sqrt[\[Pi]] Gamma[-(1/2) + n])/Gamma[n], which for n=1 reduces to your $\pi/p$. $\endgroup$ – Mauricio Fernández Apr 18 '19 at 8:54
  • $\begingroup$ Please do not use the bugs tag when posting new questions. See the tag description for why. $\endgroup$ – Szabolcs Apr 24 '19 at 17:56

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