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 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.

Question
Integrals
asked 2021-01-30
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.

Answers (1)

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
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