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Question Number 173976 by savitar last updated on 22/Jul/22

$$$$$$B(a,b)={\int }_0^1{x}^{a-1}{(1-x)}^{b-1}dx$$$$\mathrm{\Gamma }\left(s\right)={\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t}dt$$$$$$$$WhyB(a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}?$$$$$$$$$$$$$$$$$$

Answered by aleks041103 last updated on 22/Jul/22

$$Iff\left(x\right)=x^{a-1}andg\left(x\right)=x^{b-1}$$$$alsoh\left(t\right)=(f\ast g)\left(t\right)={\int }_0^{t}f\left(x\right)g(t-x)dxistheconvolution$$$$\Rightarrow h\left(t\right)={\int }_0^t{x}^{a-1}{(t-x)}^{b-1}dx$$$$x=ty\Rightarrow dx=tdy$$$$\Rightarrow h\left(t\right)={\int }_0^1{t}^{a-1}{y}^{a-1}{t}^{b-1}{(1-y)}^{b-1}tdy=$$$$=t^{a+b-1}{\int }_0^1{t}^{a-1}{(1-y)}^{b-1}dy=$$$$=B(a,b)t^{a+b-1}$$$$forconvolution:$$$$H\left(s\right)=F\left(s\right)G\left(s\right)$$$$whereF,G,Harethelaplacetransforms$$$$off,g,hrespectively.$$$$Since\mathcal{L}\left\{x^p\right\}\left(s\right)=\frac{\mathrm{\Gamma }(p+1)}{s^{p+1}},then$$$$H\left(s\right)=B(a,b)\frac{\mathrm{\Gamma }(a+b)}{s^{a+b}}$$$$F\left(s\right)=\frac{\mathrm{\Gamma }\left(a\right)}{s^a}andG\left(s\right)=\frac{\mathrm{\Gamma }\left(b\right)}{s^b}$$$$\Rightarrow \frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{s^{a+b}}=B(a,b)\frac{\mathrm{\Gamma }(a+b)}{s^{a+b}}$$$$\Rightarrow B(a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}$$

Commented by savitar last updated on 22/Jul/22

$$Verynicesir$$

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