Chardonnay Felix

2021-01-30

a) If $f\left(t\right)={t}^{m}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}g\left(t\right)={t}^{n}$ , where m and n are positive integers. show that
$f\ast g={t}^{m+n+1}{\int}_{0}^{1}{u}^{m}{(1-u)}^{n}du$

b) Use the convolution theorem to show that

$\int}_{0}^{1}{u}^{m}{(1-u)}^{n}du=\frac{m!n!}{(m+n+1)!$

c) Extend the result of part b to the case where m and n are positive numbers but not necessarily integers.

b) Use the convolution theorem to show that

c) Extend the result of part b to the case where m and n are positive numbers but not necessarily integers.

liingliing8

Skilled2021-01-31Added 95 answers

(a)

By definition,

$f\cdot g={\int}_{0}^{t}f\left(\tau \right)g(t-\tau )d\tau ={\int}_{0}^{t}{\tau}^{m}{(t-\tau )}^{n}d\tau$

Since we must get$u}^{m}{(1-u)}^{n$ , we can try with the substitution $\tau =ut$ :

$\int}_{0}^{t}{\tau}^{m}{(t-\tau )}^{n}d\tau =\{(\tau =ut\text{}0\mapsto 0),(d\tau =tdu\text{}t\mapsto 1)\$

$={\int}_{0}^{1}{\left(ut\right)}^{m}{(t-ut)}^{n}tdu$

$={\int}_{0}^{1}{u}^{m}{t}^{m}{\left(t(1-u)\right)}^{n}tdu$

$={t}^{m+n+1}{\int}_{0}^{1}{u}^{m}{(1-u)}^{n}du$

(b) and (c)

Notice that

${\int}_{0}^{1}{u}^{m}{(1-u)}^{n}du=B(m,n),$

where B is$\beta $ function. Recall that,

$B(m,n)=\frac{\mathrm{\Gamma}(m+1)\mathrm{\Gamma}(n+1)}{\mathrm{\Gamma}(m+n+2)},$

where$\mathrm{\Gamma}is\gamma$ function. When m and n are positive integers,

$B(m,n)=\frac{m!n!}{(m+n+1)!}$

The finally result for (a), consider the subtitution$\tau =ut$ .

For (b) and (c), notice that this integral is a$\beta $ function

By definition,

Since we must get

(b) and (c)

Notice that

where B is

where

The finally result for (a), consider the subtitution

For (b) and (c), notice that this integral is a