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Question Number 126631 by mnjuly1970 last updated on 23/Dec/20

$. . . c a l c u l u s . . .$ $p r o v e t h a t ::$ $:: Ω \overset{? ?}{=} \int_{0}^{1} x^{n - 1} {l n}^{2} (1 - x) d x$ $= \frac{2}{n} \underset{k = 1}{\sum^{n}} (\frac{H_{k}}{k}) ✓$
	Answered by mindispower last updated on 23/Dec/20

	$β (x, y) = \int_{0}^{1} t^{x - 1} {(1 - t)}^{y - 1} d t$ $\frac{\partial^{2} β}{\partial y^{2}} = \int_{0}^{1} t^{x - 1} {l n}^{2} (1 - t) y^{t - 1} d t$ $\frac{\partial^{2} β (x, y)}{\partial y^{2}} ∣_{(x, y) = (n, 1)} = \int_{0}^{1} t^{n - 1} {l n}^{2} (1 - t) d t$ $\frac{\partial β}{\partial y} = (Ψ (y) - Ψ (x + y)) β (x, y)$ $\frac{\partial^{2} β}{\partial y^{2}} = (Ψ^{1} (y) - Ψ^{1} (x + y)) β (x, y) + {(Ψ (y) - Ψ (x + y))}^{2} β (x, y)$ $Ω = (Ψ^{1} (1) - Ψ^{1} (1 + n)) β (1, n) + {(Ψ (1) - Ψ (1 + n))}^{2} β (1, n)$ $Ω = (\underset{k = 1}{\sum^{n}} \frac{1}{k^{2}}) \frac{1}{n} + {(\underset{k = 1}{\sum^{n}} \frac{1}{k})}^{2} . \frac{1}{n}$ $H_{n} = \sum_{k ⩽ n} \frac{1}{k}, H_{n}^{(2)} = \sum_{k ⩽ n} \frac{1}{k^{2}}$ $= \frac{H_{n}^{(2)} + H_{n}}{n}$
	Commented by mnjuly1970 last updated on 23/Dec/20

	$v e r y n i c e . . t h a n k i n g . . .$
	Answered by mnjuly1970 last updated on 23/Dec/20
	$solution: consider: I_n =∫_0 ^( 1) x^(n−1) ln(1−x)dx we can write: I_n =(1/n)∫(x^n −1) ^′ ln(1−x)dx =(1/n){[(x^n −1)ln(1−x)]_0 ^1 +∫_0 ^( 1) ((x^n −1)/(1−x))dx} =((−H_n )/n) .... (1) put:: Ω=∫_0 ^( 1) x^(n−1) ln^2 (1−x)dx =(1/n)∫_0 ^( 1) (x^n −1) ^′ ln^2 (1−x)dx =(1/n){[(x^n −1)ln^2 (1−x)]_0 ^1 +2∫_0 ^( 1) ((x^n −1)/(1−x))ln(1−x)dx =(2/n) ∫_0 ^( 1) (−Σ_(k=1) ^n x^(k−1) )ln(1−x)dx =((−2)/n^2 )Σ_(k=1) ^n ∫_0 ^( 1) x^(k−1) ln(1−x)dx=(2/n)Σ_(k=1) ^n (H_k /k) ✓$
	$s o l u t i o n :$ $c o n s i d e r :$ $I_{n} = \int_{0}^{1} x^{n - 1} l n (1 - x) d x$ $w e c a n w r i t e :$ $I_{n} = \frac{1}{n} \int (x^{n} - 1) \overset{^{'}}{} l n (1 - x) d x$ $= \frac{1}{n} {{[(x^{n} - 1) l n (1 - x)]}_{0}^{1} + \int_{0}^{1} \frac{x^{n} - 1}{1 - x} d x}$ $= \frac{- H_{n}}{n} . . . . (1)$ $p u t :: Ω = \int_{0}^{1} x^{n - 1} {l n}^{2} (1 - x) d x$ $= \frac{1}{n} \int_{0}^{1} (x^{n} - 1) \overset{^{'}}{} {l n}^{2} (1 - x) d x$ $= \frac{1}{n} {{[(x^{n} - 1) {l n}^{2} (1 - x)]}_{0}^{1} + 2 \int_{0}^{1} \frac{x^{n} - 1}{1 - x} l n (1 - x) d x$ $= \frac{2}{n} \int_{0}^{1} (- \underset{k = 1}{\sum^{n}} x^{k - 1}) l n (1 - x) d x$ $= \frac{- 2}{n^{2}} \underset{k = 1}{\sum^{n}} \int_{0}^{1} x^{k - 1} l n (1 - x) d x = \frac{2}{n} \underset{k = 1}{\sum^{n}} \frac{H_{k}}{k} ✓$
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