(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

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