Integration by rational substitution

Question

I was taking a look at a proof of the Riemann-Liouville Integral of consecutive integrations, and at some point I reached a step where it shows the following substitution:

$$_{a}I_{x}^\alpha(_{a}I_{x}^\beta f(x))=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_{a}^x \int_\zeta^x(x-t)^{\alpha-1}(t-\zeta)^{\beta-1}f(\zeta)\mathop{dt}\mathop{d\zeta}$$

using the substitution $u=\frac{t-\zeta}{x-\zeta}$ we finally obtain

$$_{a}I_{x}^\alpha(_{a}I_{x}^\beta f(x))=\frac{B(\beta,\alpha)}{\Gamma(\alpha)\Gamma(\beta)}\int_{a}^x (x-\zeta)^{\alpha+\beta-1}f(\zeta)\mathop{d\zeta}=\,_{a}I_{x}^{\alpha+\beta} f(x)$$

My question is, is there any motivation to make such a substitution as $u=\frac{t-\zeta}{x-\zeta}$? I mean, is some kind of special rational substitution? it reminds me to Möbius transformation somewhow but i can't connect the ideas.

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Edit:
After reading Markus Scheuer reply, I've found the next document:
Aygören, Aysel - Fractional Derivative and Integral [2014], pp 27-29

Answer 1

We can factor out $f(\zeta)$ in the first line and consider the inner integral in (1): \begin{align*} &_{a}I_{x}^\alpha(_{a}I_{x}^\beta f(x))\\ &\qquad=\frac{1}{\Gamma(\alpha)\Gamma(\beta)} \int_{a}^x f(\zeta)\left(\color{blue}{\int_\zeta^x(x-t)^{\alpha-1}(t-\zeta)^{\beta-1}\mathop{dt}}\right)\mathop{d\zeta}\tag{1} \end{align*}

The main theme here is to separate the integration variables $\zeta$ and $t$ in order to be able to separate the inner and outer integral. This can be achieved by the convenient substitution \begin{align*} t&=\zeta+(x-\zeta)u\tag{2}\\ dt&=(x-\zeta)du\tag{3} \end{align*} Since now we obtain from (2) \begin{align*} t-\zeta&=(x-\zeta)u\tag{4}\\ x-t&=x-\zeta-(x-\zeta)u=(x-\zeta)(1-u)\tag{5} \end{align*}

Using (3) to (5) we can write the inner integral in (1) as \begin{align*} \color{blue}{\int_\zeta^x}&\color{blue}{(x-t)^{\alpha-1}(t-\zeta)^{\beta-1}\mathop{dt}}\\ &=\int_0^1(x-\zeta)^{\alpha-1}(1-u)^{\alpha-1}(x-\zeta)^{\beta-1}u^{\beta-1}(x-\zeta)du\\ &=(x-\zeta)^{\alpha+\beta-1}\int_0^1(1-u)^{\alpha-1}u^{\beta-1}du\\ &\,\,\color{blue}{=B(\alpha,\beta)(x-\zeta)^{\alpha+\beta-1}}\tag{6} \end{align*} and we have now when putting (6) into (1) the integral $\int_{a}^x f(\zeta)(x-\zeta)^{\alpha+\beta-1}d\zeta$ separated from $B(\alpha,\beta)$.

The nice property \begin{align*} \color{blue}{_{a}I_{x}^\alpha(_{a}I_{x}^\beta f(x))=\;_{a}I_{x}^{\alpha+\beta} f(x)} \end{align*} of the Riemann-Liouville Integral is called a semigroup property of fractional integration. The substitution in (2) which when written as \begin{align*} u=\frac{t-\zeta}{x-\zeta} \end{align*} might remind us on Möbius transformation is not that strongly coupled with it, since the Möbius transformations does not share such a semigroup property when iteratively applied to the argument of a function $f$.