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:

\begin{equation} \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 \end{equation} 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

\begin{equation} \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 \end{equation}

And then?

P.S. If the last expression was

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

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

  • 1
    $\begingroup$ It seems some of your \rho became p, I've fixed it, but please verify that it's correct. $\endgroup$
    – DMcMor
    Jan 16, 2021 at 19:07

1 Answer 1


Let $z:=x^{\rho+1}$ and $\sigma:=\tau^{\rho+1}$.

$$\int(x^{\rho+1}-\tau^{\rho+1})^{\alpha-1}\tau^\rho f(\tau)d\tau=\frac1{\rho+1}\int(z-\sigma)^{\alpha-1}f(\sigma^{1/(\rho+1)})d\sigma$$

and you could be tempted to conclude

$$g(t)\equiv t^{\alpha-1}$$ and $$f(t)\equiv f(t^{1/(\rho+1)}).$$

Unfortunately, the bounds are not those of a convolution. What you are asking is not possible.


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