Existence and calculation of a fractional integral

Let $b >0$ be arbitrary but fixed and let $\alpha >0$ if $$g_{\alpha}(x)=\frac{1}{\Gamma(\alpha)}\frac{1}{x^{1-\alpha}} \chi_{]0,b]}(x)$$ If $f:[0,b] \rightarrow \mathbb{R}$ is continous in [0,b] and $F=f\chi_{[0,b]}$ show that for every $x \in [0,b]$ exists $$I_{0}^{\alpha}[f](x)=F*g_{\alpha}(x)$$ The function $I_{0}^{\alpha}[f](x)$ is called the fractional integral of order $\alpha$ of $f$ in $[0,b]$.
Now if $f(x)=x, \forall x\in [0,1]$ and $$I_{0}^{\frac{1}{2}}[f](x)=\frac{4}{3\sqrt{\pi}} x^{\frac{3}{2}},\hspace{.5cm}\forall x \in [0,1]$$ compute $$I_{0}^{\frac{1}{2}}\left[I_{0}^{\frac{1}{2}}[f]\right](x)$$

I have already tried the first part, using some properties about the convolution, but the calculation of $$I_{0}^{\frac{1}{2}}\left[I_{0}^{\frac{1}{2}}[f]\right](x)$$ has been complicated, I have been given the suggestion to use the Beta function and its relation with the Gamma function but I can not see the connection, any help, suggestion to finish the problem would be very grateful.

Solution 1:

What you want to calculate is

$$ I_0^\alpha\left(I_0^\alpha f\right)(x) = \left(g_\alpha\ast g_\alpha \ast f\right)(x). $$

Now consider \begin{align*} g_\alpha \ast g_\alpha(t) &= \int_0^t g_\alpha(t-s)g_\alpha(s)\mathrm{d}s \\ &= \frac{1}{\Gamma(\alpha)^2} \int_0^t (t-s)^{\alpha-1}s^{\alpha-1}\mathrm{d}s\\ &= \frac{t^{2\alpha-1}}{\Gamma(\alpha)^2}\int_0^1 (1-u)^{\alpha-1} u^{\alpha-1}\mathrm{d}u\\ &= \frac{B(\alpha,\alpha)}{\Gamma(\alpha)^2} t^{2\alpha-1} = \frac{t^{2\alpha-1}}{\Gamma(2\alpha)} \end{align*} where we used the substitution $u = s/t$ and the relation between the beta function $B$ and Gamma function $\Gamma.$

In case $\alpha=1/2,$ the function $g_{1/2}\ast g_{1/2}$ reduces to the constant function $1.$ Can you take it from here?