Show that:: Ω = ∫_0 ^( 1) ((Li_2 (x)log(x))/(1+x))dx = −(3/(16))ζ(4) Goodluck


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Question Number 126065 by Lordose last updated on 17/Dec/20

$Show that ::$ $Ω = \int_{0}^{1} \frac{{Li}_{2} (x) \log (x)}{1 + x} dx = - \frac{3}{16} ζ (4)$ $Goodluck$
	Answered by mnjuly1970 last updated on 17/Dec/20
	$Ω=∫_0 ^( 1) (Σ_(n=1) ^∞ (x^n /n^2 ))((log(x))/(1+x))dx =Σ_(n=1) ^∞ (1/n^2 )∫_0 ^( 1) ((x^n log(x))/(1+x))dx =Σ_(n=) ^∞ (1/n^2 )∫_0 ^( 1) {log(x)Σ_(m=0) ^∞ (−1)^m x^(n+m) }dx =Σ_(n=1) ^∞ (1/n^2 )Σ_(m=0) ^∞ (−1)^(m+1) (1/((n+m+1)^2 )) =Σ_(n=1) ^∞ Σ_(m=1) ^∞ (((−1)^m )/((n+m)^2 n^2 )) =_(property) ^(symmetrey) (1/2)Σ_(n=1) ^∞ Σ_(m=n) ^∞ (((−1)^(m−n) )/((mn)^2 )) =−(1/2)(Σ_(n=1) ^∞ (((−1)^n )/n^2 ))^2 =−(1/2)(((−π^2 )/(12)))^2 =−(1/2) ∗(π^4 /(144)) =−(1/2)∗((90)/(144))ζ(4)=−((15)/(48))ζ(4)=−(3/(16))ζ(4)✓$
	$Ω = \int_{0}^{1} (\underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n^{2}}) \frac{l o g (x)}{1 + x} d x$ $= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} \int_{0}^{1} \frac{x^{n} l o g (x)}{1 + x} d x$ $= \underset{n =}{\sum^{\infty}} \frac{1}{n^{2}} \int_{0}^{1} {l o g (x) \underset{m = 0}{\sum^{\infty}} {(- 1)}^{m} x^{n + m}} d x$ $= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} \underset{m = 0}{\sum^{\infty}} {(- 1)}^{m + 1} \frac{1}{{(n + m + 1)}^{2}}$ $= \underset{n = 1}{\sum^{\infty}} \underset{m = 1}{\sum^{\infty}} \frac{{(- 1)}^{m}}{{(n + m)}^{2} n^{2}}$ $\underset{p r o p e r t y}{\overset{s y m m e t r e y}{=}} \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \underset{m = n}{\sum^{\infty}} \frac{{(- 1)}^{m - n}}{{(m n)}^{2}}$ $= - \frac{1}{2} {(\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}})}^{2} = - \frac{1}{2} {(\frac{- π^{2}}{12})}^{2} = - \frac{1}{2} * \frac{π^{4}}{144}$ $= - \frac{1}{2} * \frac{90}{144} ζ (4) = - \frac{15}{48} ζ (4) = - \frac{3}{16} ζ (4) ✓$
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