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Question Number 206522 by MrGHK last updated on 17/Apr/24

	Answered by Berbere last updated on 17/Apr/24

	$u^{'} = {x l n}^{2} (x) \Rightarrow u = \frac{1}{2} (x^{2} {l n}^{2} (x) - x^{2} l n (x) + \frac{x^{2}}{2})$ $v = {L i}_{2} (\frac{1 + x}{x}); v^{'} = - \frac{1}{x^{2}} . - \frac{l n (- \frac{1}{x})}{\frac{1 + x}{x}} = - \frac{l n (- x)}{x (1 + x)}$ $I B P$ $= \frac{1}{2} {[(x^{2} {l n}^{2} (x) - \frac{x^{2}}{2} l n (x) + \frac{x^{2}}{2}) {L i}_{2} (\frac{1 + x}{x})]}_{0}^{1}$ $+ \frac{1}{2} \int_{0}^{1} \frac{x}{1 + x} l n (- x) ({x l n}^{2} (x) - x l n (x) + \frac{x}{2}) d x$ $‘ ‘ {L i}_{2} (z) + {L i}_{2} (1 - \frac{1}{z}) = - \frac{{(l n (z))}^{2}}{2}^{″}$ $\Rightarrow {L i}_{2} (\frac{1 + z}{z}) = {L i}_{2} (- z) - \frac{1}{2} {(l n (- z))}^{2} \Rightarrow \lim_{x \to 0} {x L i}_{2} (\frac{1 + x}{x}) = 0$ $A = \frac{1}{4} {L i}_{2} (2) + \frac{1}{2} \int_{0}^{1} \frac{x^{2} {l n}^{2} (x) l n (- x)}{1 + x} - \frac{1}{2} \int_{0}^{1} \frac{x^{2} l n (x) l n (- x)}{1 + x} d x$ $+ \frac{1}{4} \int_{0}^{1} \frac{x^{2}}{1 + x} l n (- x) d x$ $l n (- x) = l n (x) + i π$ $\int_{0}^{1} \frac{x^{2} {l n}^{n} (x) l n (- x)}{1 + x} d x$ $= \int_{0}^{1} \frac{x^{2} {l n}^{n + 1} (x)}{1 + x} d x + i π \int_{0}^{1} \frac{x^{2} {l n}^{n} (x) d x}{1 + x} =$ $\int_{0}^{1} \frac{x^{2} {l n}^{m} (x)}{1 + x} = \sum_{n ⩾ 0} {(- 1)}^{n} x^{n + 2} {l n}^{m} (x) d x = H (m)$ $= \sum_{n ⩾ 0} {(- 1)}^{n} \int_{0}^{1} x^{n + 2} {l n}^{m} (x) = \sum_{n ⩾ 0} {(- 1)}^{n} \int_{0}^{\infty} {(- t)}^{m} e^{- (n + 2) t} d t$ $= {(- 1)}^{m} \sum_{n ⩾ 0} \frac{{(- 1)}^{n} Γ (m + 1)}{{(n + 2)}^{m + 1}} = {(- 1)}^{m} m! (\sum_{n ⩾ 0} \frac{{(- 1)}^{n + 2}}{{(n + 2)}^{m + 1}})$ $= {(- 1)}^{m} m! (\sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n^{m + 1}} + 1)$ $\sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n^{m + 1}} = (1 - \frac{1}{2^{m}}) ζ (m + 1); \forall m ⩾ 1$ $H = {(- 1)}^{m} m! ((1 - \frac{1}{2^{m}}) ζ (m + 1) + 1)$ $A = \frac{{L i}_{2} (2)}{4} + \frac{1}{2} \int_{0}^{1} \frac{x^{2} {l n}^{3} (x)}{1 + x} + \frac{i π}{2} \int_{0}^{1} \frac{x^{2} {l n}^{2} (x)}{1 + x} d x - \frac{1}{2} \int_{0}^{1} \frac{x^{2} {l n}^{2} (x)}{1 + x} + \frac{1}{2} \int_{0}^{1} x^{2} \frac{i π l n (x)}{1 + x} d x$ $+ \frac{1}{4} \int_{0}^{1} \frac{x^{2} l n (x)}{1 + x} d x + \frac{i π}{4} \int_{0}^{1} \frac{x^{2}}{1 + x} d x$ $= \frac{{L i}_{2} (2)}{4} + \frac{1}{2} H (3) - \frac{H (2)}{2} + \frac{1}{4} H (1) + \frac{i π}{2} (H (2) + H (1))$ $+ i \frac{π}{2} \int_{0}^{1} \frac{x^{2}}{1 + x} d x e a s y T o c o n t i n u e$
	Commented by MrGHK last updated on 18/Apr/24

	$thanks sir Nice solution$
	Commented by Berbere last updated on 18/Apr/24

	$w i t h e P l e a s u r$
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