Question # a) If displaystyle f{{left({t}right)}}={t}^{m}{quadtext{and}quad} g{{left({t}right)}}={t}^{n}, where m and n are positive integers. show that displays

Integrals
ANSWERED a) If $$\displaystyle f{{\left({t}\right)}}={t}^{m}{\quad\text{and}\quad} g{{\left({t}\right)}}={t}^{n}$$, where m and n are positive integers. show that $$\displaystyle{f}\ast{g}={t}^{{{m}+{n}+{1}}}{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}$$
b) Use the convolution theorem to show that
$$\displaystyle{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}=\frac{{{m}!{n}!}}{{{\left({m}+{n}+{1}\right)}!}}$$
c) Extend the result of part b to the case where m and n are positive numbers but not necessarily integers. 2021-01-31
(a)
By definition,
$$\displaystyle{f}\cdot{g}={\int_{{0}}^{{t}}} f{{\left(\tau\right)}} g{{\left({t}-\tau\right)}}{d}\tau={\int_{{0}}^{{t}}}\tau^{m}{\left({t}-\tau\right)}^{n}{d}\tau$$
Since we must get $$\displaystyle{u}^{m}{\left({1}-{u}\right)}^{n}$$, we can try with the substitution $$\displaystyle\tau={u}{t}$$:
$$\displaystyle{\int_{{0}}^{{t}}}\tau^{m}{\left({t}-\tau\right)}^{n}{d}\tau={\left\lbrace{\left(\tau={u}{t}\ {0}\mapsto{0}\right)},{\left({d}\tau={t}{d}{u}\ {t}\mapsto{1}\right)}\right\rbrace}$$
$$\displaystyle={\int_{{0}}^{{1}}}{\left({u}{t}\right)}^{m}{\left({t}-{u}{t}\right)}^{n}{t}{d}{u}$$
$$\displaystyle={\int_{{0}}^{{1}}}{u}^{m}{t}^{m}{\left({t}{\left({1}-{u}\right)}\right)}^{n}{t}{d}{u}$$
$$\displaystyle={t}^{{{m}+{n}+{1}}}{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}$$
(b) and (c)
Notice that
$$\displaystyle{\int_{{0}}^{{1}}}{u}^{m}{\left({1}-{u}\right)}^{n}{d}{u}={B}{\left({m},{n}\right)},$$
where B is $$\beta$$ function. Recall that,
$$\displaystyle{B}{\left({m},{n}\right)}=\frac{{\Gamma{\left({m}+{1}\right)}\Gamma{\left({n}+{1}\right)}}}{{\Gamma{\left({m}+{n}+{2}\right)}}},$$
where $$\displaystyle\Gamma{i}{s}\gamma$$ function. When m and n are positive integers,
$$\displaystyle{B}{\left({m},{n}\right)}=\frac{{{m}!{n}!}}{{{\left({m}+{n}+{1}\right)}!}}$$
The finally result for (a), consider the subtitution $$\displaystyle\tau={u}{t}$$.
For (b) and (c), notice that this integral is a $$\beta$$ function