.......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) ......


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Question Number 137668 by mnjuly1970 last updated on 05/Apr/21
$.......advanced ... ... ... ... calculus.... prove that::::: ∗∗∗∗∗ 𝛀=Σ_(n=0) ^∞ ((Γ(n+1)Γ(x+1))/((n+1)Γ(n+x+2)))=ψ′(x+1) proof:: Ω=Σ_(n=0 ) ^∞ ((β (n+1,x+1))/(n+1))=Σ_(n=0) ^∞ {(1/((n+1)))∫_0 ^( 1) t^n .(1−t)^x dt} =∫_0 ^( 1) {(1−t)^x Σ_(n=0 ) ^∞ (t^n /(n+1))dt} =∫_0 ^( 1) {(1−t)^x Σ_(n=1) ^∞ (t^(n−1) /n)dt} =−∫_0 ^( 1) (((1−t)^x ln(1−t))/t)dt=−∫_0 ^( 1) ((t^x ln(t))/(1−t))dt = (∂/∂x)∫_0 ^( 1) ((1−t^x )/(1−t))dt=ψ′(1+x) .....✓✓ ....... 𝛀= ψ′(1+x) ......$
$. . . . . . . a d v a n c e d . . . . . . . . . . . . c a l c u l u s . . . .$ $p r o v e t h a t ::::: * * * * *$ $Ω = \underset{n = 0}{\sum^{\infty}} \frac{Γ (n + 1) Γ (x + 1)}{(n + 1) Γ (n + x + 2)} = ψ^{'} (x + 1)$ $p r o o f ::$ $Ω = \underset{n = 0}{\sum^{\infty}} \frac{β (n + 1, x + 1)}{n + 1} = \underset{n = 0}{\sum^{\infty}} {\frac{1}{(n + 1)} \int_{0}^{1} t^{n} . {(1 - t)}^{x} d t}$ $= \int_{0}^{1} {{(1 - t)}^{x} \underset{n = 0}{\sum^{\infty}} \frac{t^{n}}{n + 1} d t}$ $= \int_{0}^{1} {{(1 - t)}^{x} \underset{n = 1}{\sum^{\infty}} \frac{t^{n - 1}}{n} d t}$ $= - \int_{0}^{1} \frac{{(1 - t)}^{x} l n (1 - t)}{t} d t = - \int_{0}^{1} \frac{t^{x} l n (t)}{1 - t} d t$ $= \frac{\partial}{\partial x} \int_{0}^{1} \frac{1 - t^{x}}{1 - t} d t = ψ^{'} (1 + x) . . . . . ✓ ✓$ $. . . . . . . Ω = ψ^{'} (1 + x) . . . . . .$
Commented by Dwaipayan Shikari last updated on 05/Apr/21

$\underset{n = 0}{\sum^{\infty}} \frac{Γ (n + 1) Γ (1)}{(n + 1) Γ (n + 2)} = \underset{n = 0}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6} = ψ^{'} (1)$ $\underset{n = 0}{\sum^{\infty}} \frac{Γ (n + 1) Γ (2)}{(n + 1) Γ (n + 3)} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (n + 1)} = \frac{π^{2}}{6} - 1 = ψ^{'} (2)$ $. . .$
Commented by mnjuly1970 last updated on 05/Apr/21

$g r a t e f u l . .$ $t h a n k y o u s o m u c h m r p a y a n . . .$
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