# Finding convolution of two functions

A common engineering notational convention is: wikipedia

$${\displaystyle f(x)*g(x)\,:=\underbrace {\int_{0}^{x}f(\tau )g(x-\tau )\,d\tau } _{(f*g)(x)}.}$$

I want to write the following expression as the convolution of two functions. In the other words; what is the function $$g(x)$$ in the definition for the integral below:

$$$$\frac{(\rho+1)^{1-\alpha}}{\Gamma(\alpha)} \int_{a}^{x}\left(x^{\rho+1}-\tau^{\rho+1}\right)^{\alpha-1} \tau^{\rho} f(\tau) \,d \tau$$$$ where $$\alpha$$ and $$\rho \neq-1$$ are real numbers and $$x > a$$, $$\Gamma$$ is gamma function.

My Try: Let

\begin{align} k&=\tau^{\rho+1}\\ dk&=(\rho+1)\tau^{\rho}\,d \tau \implies \frac{dk}{(\rho+1)}=\tau^{\rho}\,d \tau. \end{align}

Substituting the last equality to the integral in the question, we have

$$$$\frac{(\rho+1)^{-\alpha}}{\Gamma(\alpha)} \int_{a^{\rho}}^{x^{\rho}}\left(x^{\rho+1}-k\right)^{\alpha-1} f(k^{1/(\rho+1)})\, dk$$$$

And then?

P.S. If the last expression was

$$$$\frac{(\rho+1)^{-\alpha}}{\Gamma(\alpha)} \int_{0}^{x}\left(x-k\right)^{\alpha-1} f(k)\, dk$$$$ $$g(x)$$ would be as follows:

$$g(x)=\frac{(\rho+1)^{-\alpha}}{\Gamma(\alpha)} x^{\alpha-1}.$$

• It seems some of your \rho became p, I've fixed it, but please verify that it's correct. Jan 16, 2021 at 19:07

Let $$z:=x^{\rho+1}$$ and $$\sigma:=\tau^{\rho+1}$$.
$$\int(x^{\rho+1}-\tau^{\rho+1})^{\alpha-1}\tau^\rho f(\tau)d\tau=\frac1{\rho+1}\int(z-\sigma)^{\alpha-1}f(\sigma^{1/(\rho+1)})d\sigma$$
$$g(t)\equiv t^{\alpha-1}$$ and $$f(t)\equiv f(t^{1/(\rho+1)}).$$