....advanced mathematics.... please demonstrate that:: Φ =∫_0 ^( 1) xlog(1−x).log(1+x)= (1/4) − log(2) ... m.n.july 1970 #


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Question Number 111719 by mnjuly1970 last updated on 04/Sep/20

$. . . . a d v a n c e d m a t h e m a t i c s . . . .$ $p l e a s e d e m o n s t r a t e t h a t ::$ $Φ = \int_{0}^{1} x l o g (1 - x) . l o g (1 + x) = \frac{1}{4} - l o g (2) . . .$ $You can't use 'macro parameter character #' in math mode$
	Answered by mathmax by abdo last updated on 05/Sep/20

	$let take a try let A = \int_{0}^{1} xln (1 - x) \ln (1 + x) dx we have$ $\frac{d}{dx} \ln (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow \ln (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{n + 1}}{n + 1} + c (c = 0)$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} \Rightarrow A = \int_{0}^{1} xln (1 - x) \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n} dx$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{n + 1} \ln (1 - x) dx let U_{n} = \int_{0}^{1} x^{n + 1} \ln (1 - x) dx$ $by parts U_{n} = {[\frac{x^{n + 2} - 1}{n + 2} \ln (1 - x)]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 2} - 1}{n + 2} (\frac{- 1}{1 - x}) dx$ $= 0 + \frac{1}{n + 2} \int_{0}^{1} \frac{x^{n + 2} - 1}{1 - x} dx = - \frac{1}{n + 2} \int_{0}^{1} \frac{x^{n + 2} - 1}{x - 1} dx$ $= - \frac{1}{n + 2} \int_{0}^{1} \frac{(x - 1) (1 + x + x^{2} + . . . + x^{n + 1}}{x - 1} dx$ $= - \frac{1}{n + 2} \int (1 + x + x^{2} + . . . + x^{n + 1}) dx = - \frac{1}{n + 2} {[x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + . . . . + \frac{x^{n + 2}}{n + 2}]}_{0}^{1}$ $= - \frac{1}{n + 2} (1 + \frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{n + 2}) = - \frac{H_{n + 2}}{n + 2} (H_{n} = \sum_{k = 1}^{n} \frac{1}{k}) \Rightarrow$ $A = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} . \frac{H_{n + 2}}{n + 2} = \sum_{n = 1}^{\infty} {(- 1)}^{n} (\frac{1}{n} - \frac{1}{n + 2}) H_{n + 2}$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} H_{n + 2}}{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} H_{n + 2}}{n + 2}$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} H_{n + 2}}{n} =_{n + 2 = p} \sum_{p = 3}^{\infty} \frac{{(- 1)}^{p - 2} H_{p}}{p - 2} = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n} H_{n}}{n - 2}$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} H_{n + 2}}{n + 2} = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n} H_{n}}{n} \Rightarrow$ $A = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n}}{n - 2} H_{n} - \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n}}{n} H_{n}$ $rest to find vslues of those series . . . . be continued . . .$
	Answered by maths mind last updated on 07/Sep/20
	$log(1−x)log(1+x)=(1/4)[(log(1−x)+log(1+x))^2 −(log(1−x)−log(1+x))^2 ] =((log^2 (1−x^2 ))/4)−((log^2 (((1−x)/(1+x))))/4) Φ=(1/4)∫_0 ^1 xlog^2 (1−x^2 )dx−(1/4)∫_0 ^1 xlog^2 (((1−x)/(1+x)))dx Φ=(1/4)(I−J) I=∫_0 ^1 xlog^2 (1−x^2 )dx let u=x^2 ⇒xdx=(du/2) I=(1/2)∫_0 ^1 log^2 (1−u)du β(x,y)=∫_0 ^1 t^(x−1) (1−t)^(y−1) ⇒∂^2 yβ(1,1)=2I J=∫_0 ^1 xlog^2 (((1−x)/(1+x)))dx u=((1−x)/(1+x))⇒x=((1−u)/(1+u)) dx=−(2/((1+u)^2 ))du =2∫_0 ^1 (((1−u))/((1+u)^3 )).log^2 (u)du =2∫_0 ^1 (−(1/((1+u)^2 ))+(2/((1+u)^3 )))log^2 (u)du =lim_(y→0) 2[_y ^1 ((1/(1+u))−(1/((1+u)^2 ))).]log^2 (u)_(=0) −4∫_0 ^1 [((log(u))/(u(1+u)))−((log(u))/((1+u)^2 u))]du =−4∫_0 ^1 ((log(u))/((1+u)^2 ))du=4lim_(x→0) ([((log(u))/(1+u))]_x ^1 −∫_x ^1 (du/(u(1+u)))) =4lim_(x→0) [−((log(x))/(1+x))+log(x)+log(2)−log(1+x)] =4log(2) ∂_y β(x,y)=β(x,y){Ψ(y)−Ψ(x+y)} ∂^2 yβ(x,y)=β(x,y)[{Ψ(y)−Ψ(x+y)}^2 +(Ψ′(y)−Ψ′(x+y))] ={Ψ(1)−Ψ(2)}^2 +(Ψ′(1)−Ψ′(2))=(−γ−(−γ−1))+(ζ(2)−(ζ(2)−1))^2 =2 I=(1/2)∂^2 yβ(1,1)=1 Φ=(1/4)(I−J)=(1/4)(1−4log(2))=(1/4)−log(2)$
	$l o g (1 - x) l o g (1 + x) = \frac{1}{4} [{(l o g (1 - x) + \log (1 + x))}^{2} - {(\log (1 - x) - \log (1 + x))}^{2}]$ $= \frac{\log^{2} (1 - x^{2})}{4} - \frac{\log^{2} (\frac{1 - x}{1 + x})}{4}$ $Φ = \frac{1}{4} \int_{0}^{1} {x l o g}^{2} (1 - x^{2}) d x - \frac{1}{4} \int_{0}^{1} {x l o g}^{2} (\frac{1 - x}{1 + x}) d x$ $Φ = \frac{1}{4} (I - J)$ $I = \int_{0}^{1} {x l o g}^{2} (1 - x^{2}) d x l e t u = x^{2} \Rightarrow x d x = \frac{d u}{2}$ $I = \frac{1}{2} \int_{0}^{1} {l o g}^{2} (1 - u) d u$ $β (x, y) = \int_{0}^{1} t^{x - 1} {(1 - t)}^{y - 1} \Rightarrow \partial^{2} y β (1, 1) = 2 I$ $J = \int_{0}^{1} {x l o g}^{2} (\frac{1 - x}{1 + x}) d x$ $u = \frac{1 - x}{1 + x} \Rightarrow x = \frac{1 - u}{1 + u}$ $d x = - \frac{2}{{(1 + u)}^{2}} d u$ $= 2 \int_{0}^{1} \frac{(1 - u)}{{(1 + u)}^{3}} . {l o g}^{2} (u) d u$ $= 2 \int_{0}^{1} (- \frac{1}{{(1 + u)}^{2}} + \frac{2}{{(1 + u)}^{3}}) {l o g}^{2} (u) d u$ $Missing \left or extra \right$ $- 4 \int_{0}^{1} [\frac{l o g (u)}{u (1 + u)} - \frac{l o g (u)}{{(1 + u)}^{2} u}] d u$ $= - 4 \int_{0}^{1} \frac{l o g (u)}{{(1 + u)}^{2}} d u = 4 \lim_{x \to 0} ({[\frac{l o g (u)}{1 + u}]}_{x}^{1} - \int_{x}^{1} \frac{d u}{u (1 + u)})$ $= 4 \lim_{x \to 0} [- \frac{l o g (x)}{1 + x} + l o g (x) + l o g (2) - l o g (1 + x)]$ $= 4 l o g (2)$ $\partial_{y} β (x, y) = β (x, y) {Ψ (y) - Ψ (x + y)}$ $\partial^{2} y β (x, y) = β (x, y) [{Ψ (y) - Ψ (x + y)}^{2} + (Ψ^{'} (y) - Ψ^{'} (x + y))]$ $= {Ψ (1) - Ψ (2)}^{2} + (Ψ^{'} (1) - Ψ^{'} (2)) = (- γ - (- γ - 1)) + {(ζ (2) - (ζ (2) - 1))}^{2} = 2$ $I = \frac{1}{2} \partial^{2} y β (1, 1) = 1$ $Φ = \frac{1}{4} (I - J) = \frac{1}{4} (1 - 4 l o g (2)) = \frac{1}{4} - l o g (2)$
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