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I know the proof for the fractional calculus power rule using the definition of the Riemann-Liouville Fractional Integral, $_{c}D_x^{-\nu}f(x) = \frac{1}{\Gamma({\nu})}\int_{c}^{x}(x-t)^{\nu - 1}f(t)dt$, but I don't understand one part in the process.

We start with,

$D^{- \nu}x^{\mu} = \frac{1}{\Gamma({\nu})}\int_{0}^{x}(x-t)^{\nu - 1}t^{\mu}dt$

$= \frac{1}{\Gamma({\nu})}\int_{0}^{x}(1-\frac{t}{x})^{\nu - 1}x^{\nu - 1}t^{\mu}dt$

which equals, $ \frac{1}{\Gamma({\nu})}\int_{0}^{1}(1-u)^{\nu - 1}x^{\nu -1}{(xu)}^{\mu}xdu$, $(u = \frac{t}{x})$.

At the last step I don't understand why you change from integrating from 0 to x to integrate from 0 to 1, everything else makes sense. I kind of see how it works, but I need an in depth explanation for why it works.

Carpenter
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1 Answers1

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When making the substitution $t=xu$ we incorporate the integral bounds.

Since $0 \leq t \leq x$ we obtain $0 \leq xu \leq x$. Dividing by $x$ yields $0 \leq u \leq 1$, which gives the new integral bounds.

William T.
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