Chardonnay Felix

## Answered question

2021-01-30

a) If $f\left(t\right)={t}^{m}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(t\right)={t}^{n}$, where m and n are positive integers. show that $f\ast g={t}^{m+n+1}{\int }_{0}^{1}{u}^{m}{\left(1-u\right)}^{n}du$
b) Use the convolution theorem to show that
${\int }_{0}^{1}{u}^{m}{\left(1-u\right)}^{n}du=\frac{m!n!}{\left(m+n+1\right)!}$
c) Extend the result of part b to the case where m and n are positive numbers but not necessarily integers.

### Answer & Explanation

liingliing8

Skilled2021-01-31Added 95 answers

(a)
By definition,
$f\cdot g={\int }_{0}^{t}f\left(\tau \right)g\left(t-\tau \right)d\tau ={\int }_{0}^{t}{\tau }^{m}{\left(t-\tau \right)}^{n}d\tau$
Since we must get ${u}^{m}{\left(1-u\right)}^{n}$, we can try with the substitution $\tau =ut$:

$={\int }_{0}^{1}{\left(ut\right)}^{m}{\left(t-ut\right)}^{n}tdu$
$={\int }_{0}^{1}{u}^{m}{t}^{m}{\left(t\left(1-u\right)\right)}^{n}tdu$
$={t}^{m+n+1}{\int }_{0}^{1}{u}^{m}{\left(1-u\right)}^{n}du$
(b) and (c)
Notice that
${\int }_{0}^{1}{u}^{m}{\left(1-u\right)}^{n}du=B\left(m,n\right),$
where B is $\beta$ function. Recall that,
$B\left(m,n\right)=\frac{\mathrm{\Gamma }\left(m+1\right)\mathrm{\Gamma }\left(n+1\right)}{\mathrm{\Gamma }\left(m+n+2\right)},$
where $\mathrm{\Gamma }is\gamma$ function. When m and n are positive integers,
$B\left(m,n\right)=\frac{m!n!}{\left(m+n+1\right)!}$
The finally result for (a), consider the subtitution $\tau =ut$.
For (b) and (c), notice that this integral is a $\beta$ function

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