Integrate under assumptions

Question

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 adding 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.

Answer 1

In V12.2, we get the expected answer:

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

(*  1/(c sigma)  *)

I assume they fixed the bug.