# Integration by rational substitution

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.

Edit:
Aygören, Aysel - Fractional Derivative and Integral [2014], pp 27-29

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 $$f$$.

• Thanks @Markus, althought I managed to come up to the result stated, my doubt is about the substitution itself. It works, but to me it still looks like something pulled out from under the sleeve, or not something intuitive to do. Maybe my integration methods are rusty after all, lol. Anyway, thanks again for your reply. :) Jul 10, 2021 at 23:47
• @ppmbb: You're welcome! I was also curious about any connections with the Möbius function and checked my books for it. But there was no indication. Jul 11, 2021 at 8:40