∫_1 ^2 ((logu)/(((√(u−1)))((√(u−1))+1)))du


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Question Number 101793 by Dwaipayan Shikari last updated on 04/Jul/20

$\int_{1}^{2} \frac{l o g u}{(\sqrt{u - 1}) (\sqrt{u - 1} + 1)} d u$
	Answered by mathmax by abdo last updated on 05/Jul/20
	$I =∫_1 ^(2 ) ((ln(u))/((√(u−1))((√(u−1))+1)))du changement (√(u−1))=x give u−1 =x^2 ⇒ I =∫_0 ^1 ((ln(1+x^2 ))/(x(x+1))) (2x)dx =2 ∫_0 ^1 ((ln(1+x^2 ))/(1+x)) dx =2 ∫_0 ^1 ln(1+x^2 )Σ_(n=0) ^∞ (−1)^n x^n =2 Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^n ln(1+x^2 )dx =2 Σ_(n=0) ^∞ (−1)^n A_n with A_n =∫_0 ^1 x^n ln(1+x^2 )dx by parts we get A_n =[(x^(n+1) /(n+1))ln(1+x^2 )]_0 ^1 −∫_0 ^1 (x^(n+1) /(n+1))×((2x)/(1+x^2 ))dx =((ln(2))/(n+1))−(2/(n+1))∫_0 ^1 (x^(n+2) /(1+x^2 ))dx ∫_0 ^1 (x^(n+2) /(1+x^2 ))dx =∫_0 ^1 (((x^2 +1−1)x^n )/(1+x^2 ))dx =∫_0 ^1 x^n dx−∫_0 ^1 (x^n /(x^2 +1))dx =(1/(n+1))−∫_0 ^(1 ) (x^n /(x^2 +1))dx we have ∫_0 ^1 (x^n /(x^2 +1))dx =∫_0 ^1 x^n (Σ_(k=0) ^∞ (−1)^k x^(2k) ) =Σ_(k=0) ^∞ (−1)^k ∫_0 ^1 x^(n+2k) dx =Σ_(k=0) ^∞ (−1)^k ×(1/(n+2k+1)) ⇒ A_n =((ln2)/(n+1))−(2/(n+1)){(1/(n+1))−Σ_(k=0) ^∞ (((−1)^k )/(2k+n+1))} =((ln2)/(n+1))−(2/((n+1)^2 )) −(2/(n+1)) Σ_(k=0) ^∞ (((−1)^k )/(2k+n+1)) ⇒ I =2ln2 Σ_(n=0) ^∞ (((−1)^n )/(n+1))−4 Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) −4 Σ_(n=0) ^∞ (((−1)^n )/(n+1))(Σ_(k=0) ^∞ (((−1)^k )/(2k+n+1))) Σ_(n=0) ^∞ (((−1)^n )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =ln(2) Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−{2^(1−2) −1)ξ(2) =−(−(1/2))×(π^2 /6) =(π^2 /(12)) Σ_(n=0) ^∞ (((−1)^n )/(n+1))Σ_(k=0) ^∞ (((−1)^k )/(2k+n+1)) =Σ_(n=0) ^∞ Σ_(k=0) ^∞ (((−1)^(n+k) )/((n+1)(2k+n+1))) ...be continued...$
	$I = \int_{1}^{2} \frac{\ln (u)}{\sqrt{u - 1} (\sqrt{u - 1} + 1)} du changement \sqrt{u - 1} = x give u - 1 = x^{2} \Rightarrow$ $I = \int_{0}^{1} \frac{\ln (1 + x^{2})}{x (x + 1)} (2 x) dx = 2 \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x} dx$ $= 2 \int_{0}^{1} \ln (1 + x^{2}) \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} = 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{n} \ln (1 + x^{2}) dx$ $= 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n} with A_{n} = \int_{0}^{1} x^{n} \ln (1 + x^{2}) dx by parts we get$ $A_{n} = {[\frac{x^{n + 1}}{n + 1} \ln (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} \times \frac{2 x}{1 + x^{2}} dx$ $= \frac{\ln (2)}{n + 1} - \frac{2}{n + 1} \int_{0}^{1} \frac{x^{n + 2}}{1 + x^{2}} dx$ $\int_{0}^{1} \frac{x^{n + 2}}{1 + x^{2}} dx = \int_{0}^{1} \frac{(x^{2} + 1 - 1) x^{n}}{1 + x^{2}} dx = \int_{0}^{1} x^{n} dx - \int_{0}^{1} \frac{x^{n}}{x^{2} + 1} dx$ $= \frac{1}{n + 1} - \int_{0}^{1} \frac{x^{n}}{x^{2} + 1} dx we have$ $\int_{0}^{1} \frac{x^{n}}{x^{2} + 1} dx = \int_{0}^{1} x^{n} (\sum_{k = 0}^{\infty} {(- 1)}^{k} x^{2 k})$ $= \sum_{k = 0}^{\infty} {(- 1)}^{k} \int_{0}^{1} x^{n + 2 k} dx = \sum_{k = 0}^{\infty} {(- 1)}^{k} \times \frac{1}{n + 2 k + 1} \Rightarrow$ $A_{n} = \frac{\ln 2}{n + 1} - \frac{2}{n + 1} {\frac{1}{n + 1} - \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + n + 1}}$ $= \frac{\ln 2}{n + 1} - \frac{2}{{(n + 1)}^{2}} - \frac{2}{n + 1} \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + n + 1} \Rightarrow$ $I = 2 \ln 2 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} - 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} - 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} (\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + n + 1})$ $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = \ln (2)$ $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = - {2^{1 - 2} - 1) ξ (2)$ $= - (- \frac{1}{2}) \times \frac{π^{2}}{6} = \frac{π^{2}}{12}$ $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + n + 1} = \sum_{n = 0}^{\infty} \sum_{k = 0}^{\infty} \frac{{(- 1)}^{n + k}}{(n + 1) (2 k + n + 1)}$ $. . . be continued . . .$
	Commented by 1549442205 last updated on 06/Jul/20

	$Thank you sir . Please, can you show me$ $about the function ξ (n) that used in solution ?$
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