This question arose on a very complicated context, but it is very simple:
I have for $0<a<1$ the integral $\int_0^t (t-x)^{a-1} dx$ and 'd like to derive on $t$. The imediate result is $t^{a-1} $, which is a positive function for $t$ positive.
Suppose that I'd like to use Leibniz rule, so $\dfrac{\partial}{\partial t} \displaystyle\int_0^t (t-x)^{a-1} dx=(t-t)^{a-1}+ \displaystyle\int_0^t \dfrac{\partial}{\partial t}(t-x)^{a-1} dx=\displaystyle\int_0^t (a-1) (t-x)^{a-2} dx$
But this is a positive function? Because $(a-1)<0$ and $(t-x)^{a-2} \geq0$, so the integrandum is everywhere nonpositive...!
Thank you so much
The problem lies in the fact that the function $g(x) = (t - x)^{a - 1}$ is not defined at $x = t$. You are treating it as if $g(t) = (t - t)^{a - 1} = 0$, but this does not actually make sense; the graph of $g$ has vertical asymptote as $x$ approaches $t$ from the left.
If you treat your problem carefully by instead writing $\int_0^{t - \delta} (t - x)^{a - 1} \, dx$, you can get the proper calculation.