# Integrate under assumptions

I am trying to solve the following integral: $$\int_0^\infty u^{\sigma-1} \exp\left[-c u^\sigma\right] \mathrm{d}u$$ under the assumptions $$\sigma \in (0,1)$$ and $$c>0$$. I know the result is $$\dfrac{1}{c \sigma}.$$ However, using the following code in Mathematica

 Integrate[ Power[u, sigma - 1] * Exp[-c*Power[u, sigma]] , {u, 0, Infinity},  Assumptions -> {c > 0 && 0 < sigma && sigma <1 && Element[c, Reals] && Element[sigma, Reals]}]


I obtain

Integrate Integral of e^(-c u^sigma) u^(-1+sigma) does not converge on {0,[Infinity]}.

Surprisingly, deleting just the assumption $$\sigma <1$$, it works as expected, that is running

Integrate[Power[u, sigma - 1] * Exp[-c*Power[u, sigma]] , {u, 0, Infinity}, Assumptions -> {c > 0 && 0 < sigma && Element[c, Reals] &&Element[sigma, Reals]}]


I obtain

1/(c sigma).

I do not understand why to add an additional a restriction changes the general result, but I do not find my mistake.

Edit: In the version 12.2 of Mathematica both the expressions work correctly. I assume they fixed the bug.

• It is a weird behavior indeed. This: Integrate[u^(s - 1)*Exp[-c*u^s], {u, 0, Infinity}, Assumptions -> {c > 0, s > 0}]returns the expected result. But this: Integrate[u^(s - 1)*Exp[-c*u^s], {u, 0, Infinity}, Assumptions -> {c > 0, s > 0, s < 1}] brings the error message you mentioned. I would report it to Wolfram. Nov 7 '18 at 13:00
• As best I can tell there is a faulty convergence test in play. Treating it as a bug (belatedly). Apr 23 '19 at 21:24
• It would be better if you put your finding as an answer. I can delete mine, if you want to post it instead. Apr 3 '21 at 1:39
– Gio
Apr 3 '21 at 18:51

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