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

B(a,b)=∫_0 ^1 x^(a−1) (1−x)^(b−1) dx Γ(s)= ∫_0 ^∞ t^(s−1) e^(−t) dt Why B(a,b)= ((Γ(a)Γ(b))/(Γ(a+b))) ?

$$ \\ $$$$\:\:\:\:{B}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx}\: \\ $$$$\:\:\:\:\:\:\:\Gamma\left({s}\right)=\:\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$ \\ $$$$\:\:{Why}\:\:\:\:{B}\left({a},{b}\right)=\:\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}\:? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by aleks041103 last updated on 22/Jul/22

If f(x)=x^(a−1) and g(x)=x^(b−1) also h(t)=(f∗g)(t)=∫_0 ^( t) f(x)g(t−x)dx is the convolution ⇒h(t)=∫_0 ^( t) x^(a−1) (t−x)^(b−1) dx x=ty⇒dx=t dy ⇒h(t)=∫_0 ^( 1) t^(a−1) y^(a−1) t^(b−1) (1−y)^(b−1) tdy= =t^(a+b−1) ∫_0 ^( 1) t^(a−1) (1−y)^(b−1) dy= =B(a,b)t^(a+b−1) for convolution: H(s)=F(s)G(s) where F,G,H are the laplace transforms of f,g,h respectively. Since L{x^p }(s)=((Γ(p+1))/s^(p+1) ), then H(s)=B(a,b)((Γ(a+b))/s^(a+b) ) F(s)=((Γ(a))/s^a ) and G(s)=((Γ(b))/s^b ) ⇒((Γ(a)Γ(b))/s^(a+b) )=B(a,b)((Γ(a+b))/s^(a+b) ) ⇒B(a,b)=((Γ(a)Γ(b))/(Γ(a+b)))

$${If}\:{f}\left({x}\right)={x}^{{a}−\mathrm{1}} \:{and}\:{g}\left({x}\right)={x}^{{b}−\mathrm{1}} \\ $$$${also}\:{h}\left({t}\right)=\left({f}\ast{g}\right)\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} {f}\left({x}\right){g}\left({t}−{x}\right){dx}\:{is}\:{the}\:{convolution} \\ $$$$\Rightarrow{h}\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} {x}^{{a}−\mathrm{1}} \left({t}−{x}\right)^{{b}−\mathrm{1}} {dx} \\ $$$${x}={ty}\Rightarrow{dx}={t}\:{dy} \\ $$$$\Rightarrow{h}\left({t}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{a}−\mathrm{1}} {y}^{{a}−\mathrm{1}} {t}^{{b}−\mathrm{1}} \left(\mathrm{1}−{y}\right)^{{b}−\mathrm{1}} {tdy}= \\ $$$$={t}^{{a}+{b}−\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{a}−\mathrm{1}} \left(\mathrm{1}−{y}\right)^{{b}−\mathrm{1}} {dy}= \\ $$$$={B}\left({a},{b}\right){t}^{{a}+{b}−\mathrm{1}} \\ $$$${for}\:{convolution}: \\ $$$${H}\left({s}\right)={F}\left({s}\right){G}\left({s}\right) \\ $$$${where}\:{F},{G},{H}\:{are}\:{the}\:{laplace}\:{transforms} \\ $$$${of}\:{f},{g},{h}\:{respectively}. \\ $$$${Since}\:\mathscr{L}\left\{{x}^{{p}} \right\}\left({s}\right)=\frac{\Gamma\left({p}+\mathrm{1}\right)}{{s}^{{p}+\mathrm{1}} },\:{then} \\ $$$${H}\left({s}\right)={B}\left({a},{b}\right)\frac{\Gamma\left({a}+{b}\right)}{{s}^{{a}+{b}} } \\ $$$${F}\left({s}\right)=\frac{\Gamma\left({a}\right)}{{s}^{{a}} }\:{and}\:{G}\left({s}\right)=\frac{\Gamma\left({b}\right)}{{s}^{{b}} } \\ $$$$\Rightarrow\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{{s}^{{a}+{b}} }={B}\left({a},{b}\right)\frac{\Gamma\left({a}+{b}\right)}{{s}^{{a}+{b}} } \\ $$$$\Rightarrow{B}\left({a},{b}\right)=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)} \\ $$

Commented by savitar last updated on 22/Jul/22

Very nice sir

$${Very}\:{nice}\:{sir} \\ $$

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