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Question Number 107639 by mathdave last updated on 11/Aug/20

	Answered by mathmax by abdo last updated on 12/Aug/20
	$I =∫_0 ^1 ((xln(x))/((1+x)^2 ))dx we have (1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n for ∣x∣<1 ⇒ −(1/((1+x)^2 )) =Σ_(n=1) ^∞ n(−1)^n x^(n−1) =Σ_(n=0) ^∞ (n+1)(−1)^(n+1) x^n ⇒ (1/((1+x^2 ))) =Σ_(n=0) ^∞ (n+1)(−1)^n x^n ⇒ I =∫_0 ^1 xln(x){Σ_(n=0) ^∞ (n+1)(−1)^n x^n }dx =Σ_(n=0) ^∞ (n+1)(−1)^n ∫_0 ^(1 ) x^(n+1) ln(x)dx but by parts u_n =∫_0 ^1 x^(n+1) ln(x)dx =[(1/(n+2))x^(n+2) ln(x)]_0 ^1 −∫_0 ^1 (x^(n+2) /(n+2))(dx/x) =−(1/(n+2)) ∫_0 ^1 x^(n+1) dx =−(1/((n+2)^2 )) ⇒I =−Σ_(n=0) ^∞ (n+1)(−1)^n ×(1/((n+2)^2 )) =−Σ_(n=0) ^∞ (−1)^n ×((n+1)/((n+2)^2 )) =_(n+2 =p) −Σ_(p=2) ^∞ (−1)^(p−2) ×((p−1)/p^2 ) =−Σ_(p=2) ^∞ (−1)^p {(1/p)−(1/p^2 )} =−Σ_(p=2) ^∞ (((−1)^p )/p) +Σ_(p=2) ^∞ (((−1)^p )/p^2 ) we have Σ_(p=2) ^∞ (((−1)^p )/p) =Σ_(p=1) ^∞ (((−1)^p )/p) +1 =−ln(2)+1 Σ_(p=2) ^∞ (((−1)^p )/p^2 ) =Σ_(p=1) ^∞ (((−1)^p )/p^2 ) +1 =δ(2)+1 =(2^(1−2) −1)ξ(2)+1 =−(π^2 /(12)) +1 ⇒ I =ln(2)−1−(π^2 /(12)) +1 ⇒I =ln(2)−(π^2 /(12))$
	$I = \int_{0}^{1} \frac{xln (x)}{{(1 + x)}^{2}} dx we have \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} for ∣ x ∣< 1 \Rightarrow$ $- \frac{1}{{(1 + x)}^{2}} = \sum_{n = 1}^{\infty} n {(- 1)}^{n} x^{n - 1} = \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n + 1} x^{n} \Rightarrow$ $\frac{1}{(1 + x^{2})} = \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} x^{n} \Rightarrow$ $I = \int_{0}^{1} xln (x) {\sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} x^{n}} dx$ $= \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} \int_{0}^{1} x^{n + 1} \ln (x) dx but by parts$ $u_{n} = \int_{0}^{1} x^{n + 1} \ln (x) dx = {[\frac{1}{n + 2} x^{n + 2} \ln (x)]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 2}}{n + 2} \frac{dx}{x}$ $= - \frac{1}{n + 2} \int_{0}^{1} x^{n + 1} dx = - \frac{1}{{(n + 2)}^{2}} \Rightarrow I = - \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} \times \frac{1}{{(n + 2)}^{2}}$ $= - \sum_{n = 0}^{\infty} {(- 1)}^{n} \times \frac{n + 1}{{(n + 2)}^{2}} =_{n + 2 = p} - \sum_{p = 2}^{\infty} {(- 1)}^{p - 2} \times \frac{p - 1}{p^{2}}$ $= - \sum_{p = 2}^{\infty} {(- 1)}^{p} {\frac{1}{p} - \frac{1}{p^{2}}}$ $= - \sum_{p = 2}^{\infty} \frac{{(- 1)}^{p}}{p} + \sum_{p = 2}^{\infty} \frac{{(- 1)}^{p}}{p^{2}} we have$ $\sum_{p = 2}^{\infty} \frac{{(- 1)}^{p}}{p} = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p}}{p} + 1 = - \ln (2) + 1$ $\sum_{p = 2}^{\infty} \frac{{(- 1)}^{p}}{p^{2}} = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p}}{p^{2}} + 1 = δ (2) + 1 = (2^{1 - 2} - 1) ξ (2) + 1$ $= - \frac{π^{2}}{12} + 1 \Rightarrow I = \ln (2) - 1 - \frac{π^{2}}{12} + 1 \Rightarrow I = \ln (2) - \frac{π^{2}}{12}$
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