a) If f(t)=tmandg(t)=tn, where m and n are positive integers. show that f∗g=tm+n+1∫01um(1−u)ndu b) Use the convolution theorem to show that ∫01um(1−u)ndu=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.

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a) If f(t)=tmandg(t)=tn, where m and n are positive integers. show that f∗g=tm+n+1∫01um(1−u)ndu  b) Use the convolution theorem to show that  ∫01um(1−u)ndu=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.

liingliing8 · Accepted Answer

(a)  By definition,  f⋅g=∫0tf(τ)g(t−τ)dτ=∫0tτm(t−τ)ndτ  Since we must get um(1−u)n, we can try with the substitution τ=ut:  ∫0tτm(t−τ)ndτ={(τ=ut 0↦0),(dτ=tdu t↦1)}  =∫01(ut)m(t−ut)ntdu  =∫01umtm(t(1−u))ntdu  =tm+n+1∫01um(1−u)ndu  (b) and (c)  Notice that  ∫01um(1−u)ndu=B(m,n),  where B is β function. Recall that,  B(m,n)=Γ(m+1)Γ(n+1)Γ(m+n+2),  where Γisγ function. When m and n are positive integers,  B(m,n)=m!n!(m+n+1)!  The finally result for (a), consider the subtitution τ=ut.  For (b) and (c), notice that this integral is a β function

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

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