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  • $\begingroup$ It is very mysterious. By the way, I obtained the 6.58887 value by first changing to your form 2*Pi*Integrate[r*Exp[-r^2]/RealAbs[r^2 - 1]^(3/5), {r, 0, Infinity}], and then further analytically performing the substitution $r^2=u$. In this case, regardless of setting the integral range as {u,0,Infinity} or {u,0,1,Infinity}, the value is 6.58887. $\endgroup$ Commented Oct 10, 2021 at 11:38
  • $\begingroup$ @eigenvalu: Which version do you use? In 12.3.1 on Windows 10 N[Pi*Integrate[Exp[-x]/RealAbs[x - 1]^(3/5), {x, 0, 1, Infinity}]] still results in 2.56359 and N[Pi*Integrate[Exp[-x]/RealAbs[x - 1]^(3/5), {x, 0, Infinity}]] still produces 6.58887 - 5.13246*10^-16 I. $\endgroup$ Commented Oct 10, 2021 at 14:12
  • $\begingroup$ I use version 12.1 on Window 10. $\endgroup$ Commented Oct 11, 2021 at 1:15
  • $\begingroup$ @eigenvalue: It's strange: the documentation to Integrate says the command was updated in 12.0 latest time. $\endgroup$ Commented Oct 11, 2021 at 5:33