Assume we are working on the interval $[a,b]$, we have that $f \in L^1 [a,b]$. Let $\gamma \in (0,1)$. We then have that
$$\int_a^uf(t)(u-t)^{1-\gamma}dt, u \in [a,b]$$
is well-defined. Because $(u-t)^{1-\gamma}$ is continuous on $[a,u]$, and hence bounded.
A book I am reading states that:
$$\frac{d}{du }\int_a^uf(t)(u-t)^{1-\gamma}dt=(1-\gamma)\int_a^uf(t)(u-t)^{-\gamma}dt.$$
But why is this the case? I tried using the leibniz integral rule. But from what I see I can not use it because the derivative $(u-t)^{-\gamma}$ does not behave well when $t=u$.
Another problem I have is that by Fubini/Tonelli we may have that $(1-\gamma)\int_a^uf(t)(u-t)^{-\gamma}dt$ is well defined only a.e. on $[a,b]$? But do you see how we can prove that identity if $(1-\gamma)\int_a^uf(t)(u-t)^{-\gamma}dt$ is well-defined?
I tried writing the definition of the derivative, and then taking limits etc.. But I do not get that the fraction is bounded(so I can use the dominated convergence theorem), because $(u-t)^{-\gamma}$ is not bounded. If I am supposed to solve it using this I get:
$\frac{\int_a^{u+\Delta u}f(t)(u+\Delta u-t)^{1-\gamma}dt-\int_a^{u}f(t)(u-t)^{1-\gamma}dt}{\Delta u}=\int_a^u\frac{f(t)[(u+\Delta u-t)^{1-\gamma}-(u-t)^{1-\gamma}]dt}{\Delta u}+\int_u^{u+\Delta u}\frac{f(t)(u+\Delta u-t)^{1-\gamma}dt}{\Delta u}.$
Do you see what happens with the last terms when $\Delta u$ goes to zero? (If what I wrote in the start is correct, the last term is supposed to go to zero. And in the second last term we are supposed to be able to write the derivative inside, but how do I argue for this using DCT?)
ANSWER
For those who are interested I think I found the answer. The book I was reading was a book containg articles, and I think the article my quesiton is based on was badly written. I looked at the source for the article and it seemed better. Note first if we can show that
$\int_a^u f(t)(u-t)^{1-\gamma}dt=(1-\gamma)\int_a^u[\int_a^t\frac{f(s)ds}{(t-s)^\gamma}]dt , H(t)=\int_a^t\frac{f(s)ds}{(t-s)^\gamma}\in L^1[a,b]$, we are done, because then the result follows from the fundamental theorem of calculus.
First we show that $H$ is well defined, and simultaneously show that we can use Fubini to interchange limits.:
$\int_a^b\int_a^t\frac{|f(s)|ds}{(t-s)^\gamma}dt=\int_a^b \int_s^b|f(s)|(t-s)^{-\gamma} dtds=\int_a^b |f(s)|(b-s)^{1-\gamma} \cdot1/(1-\gamma)ds<\infty$.
And now the same calculuation with $u$ instead of b, and $f(s)$ instead of $|f(s)|$ tells us that $\int_a^u f(t)(u-t)^{1-\gamma}dt=(1-\gamma)\int_a^u[\int_a^t\frac{f(s)ds}{(t-s)^\gamma}]dt$.