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Question Number 137668 by mnjuly1970 last updated on 05/Apr/21
.......advanced ... ... ... ... calculus.... prove that::::: ∗∗∗∗∗ 𝛀=Σ_(n=0) ^∞ ((Γ(n+1)Γ(x+1))/((n+1)Γ(n+x+2)))=ψ′(x+1) proof:: Ω=Σ_(n=0 ) ^∞ ((β (n+1,x+1))/(n+1))=Σ_(n=0) ^∞ {(1/((n+1)))∫_0 ^( 1) t^n .(1−t)^x dt} =∫_0 ^( 1) {(1−t)^x Σ_(n=0 ) ^∞ (t^n /(n+1))dt} =∫_0 ^( 1) {(1−t)^x Σ_(n=1) ^∞ (t^(n−1) /n)dt} =−∫_0 ^( 1) (((1−t)^x ln(1−t))/t)dt=−∫_0 ^( 1) ((t^x ln(t))/(1−t))dt = (∂/∂x)∫_0 ^( 1) ((1−t^x )/(1−t))dt=ψ′(1+x) .....✓✓ ....... 𝛀= ψ′(1+x) ......
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….{advanced}\:…\:…\:…\:…\:{calculus}…. \\ $$$$\:\:{prove}\:{that}:::::\:\ast\ast\ast\ast\ast \\ $$$$\:\:\:\:\boldsymbol{\Omega}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)\Gamma\left({x}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)\Gamma\left({n}+{x}+\mathrm{2}\right)}=\psi'\left({x}+\mathrm{1}\right) \\ $$$${proof}:: \\ $$$$\:\:\:\:\Omega=\underset{{n}=\mathrm{0}\:} {\overset{\infty} {\sum}}\frac{\beta\:\left({n}+\mathrm{1},{x}+\mathrm{1}\right)}{{n}+\mathrm{1}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{n}} .\left(\mathrm{1}−{t}\right)^{{x}} {dt}\right\} \\ $$$$\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\left(\mathrm{1}−{t}\right)^{{x}} \underset{{n}=\mathrm{0}\:} {\overset{\infty} {\sum}}\frac{{t}^{{n}} }{{n}+\mathrm{1}}{dt}\right\} \\ $$$$\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\left(\mathrm{1}−{t}\right)^{{x}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{t}^{{n}−\mathrm{1}} }{{n}}{dt}\right\} \\ $$$$\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left(\mathrm{1}−{t}\right)^{{x}} {ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{t}^{{x}} {ln}\left({t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$$=\:\frac{\partial}{\partial{x}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{t}^{{x}} }{\mathrm{1}−{t}}{dt}=\psi'\left(\mathrm{1}+{x}\right)\:…..\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….\:\:\boldsymbol{\Omega}=\:\psi'\left(\mathrm{1}+{x}\right)\:…… \\ $$
Commented by Dwaipayan Shikari last updated on 05/Apr/21
Σ_(n=0) ^∞ ((Γ(n+1)Γ(1))/((n+1)Γ(n+2)))=Σ_(n=0) ^∞ (1/n^2 )=(π^2 /6)=ψ′(1) Σ_(n=0) ^∞ ((Γ(n+1)Γ(2))/((n+1)Γ(n+3)))=Σ_(n=1) ^∞ (1/(n^2 (n+1)))=(π^2 /6)−1 =ψ′(2) ...
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)\Gamma\left(\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)\Gamma\left({n}+\mathrm{2}\right)}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\psi'\left(\mathrm{1}\right) \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)\Gamma\left(\mathrm{2}\right)}{\left({n}+\mathrm{1}\right)\Gamma\left({n}+\mathrm{3}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\:=\psi'\left(\mathrm{2}\right) \\ $$$$… \\ $$
Commented by mnjuly1970 last updated on 05/Apr/21
grateful.. thank you so much mr payan...
$$\:\:\:{grateful}.. \\ $$$$\:\:{thank}\:{you}\:{so}\:{much}\:{mr}\:{payan}… \\ $$

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