nice integral ∫_0 ^1 (1/x)ln(Σ_(m=0) ^n x^m )dx=? −−−−−−−−−−−−−by MATH.AMIN


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Question Number 165255 by amin96 last updated on 28/Jan/22

$nice integral$ $\int_{0}^{1} \frac{1}{x} \ln (\underset{m = 0}{\sum^{n}} x^{m}) dx = ?$ $- - - - - - - - - - - - - by MATH . AMIN$
	Answered by mindispower last updated on 28/Jan/22
	$=∫_0 ^1 ((ln(((1−x^(n+1) )/(1−x))))/x)dx =∫_0 ^1 ((ln(1−x^(n+1) ))/x^(n+1) ).x^n dx−∫_0 ^1 ((ln(1−x))/x) =−(1/(n+1))(−∫_0 ^1 ((ln(1−x))/x)dx)−∫_0 ^1 ((ln(1−x))/x)dx =(n/(n+1)){−∫_0 ^1 ((ln(1−x))/x)dx}=(n/(n+1))Li_2 (1) =(π^2 /6).(n/(n+1))$
	$= \int_{0}^{1} \frac{l n (\frac{1 - x^{n + 1}}{1 - x})}{x} d x$ $= \int_{0}^{1} \frac{l n (1 - x^{n + 1})}{x^{n + 1}} . x^{n} d x - \int_{0}^{1} \frac{l n (1 - x)}{x}$ $= - \frac{1}{n + 1} (- \int_{0}^{1} \frac{l n (1 - x)}{x} d x) - \int_{0}^{1} \frac{l n (1 - x)}{x} d x$ $= \frac{n}{n + 1} {- \int_{0}^{1} \frac{l n (1 - x)}{x} d x} = \frac{n}{n + 1} {L i}_{2} (1)$ $= \frac{π^{2}}{6} . \frac{n}{n + 1}$
	Commented by amin96 last updated on 28/Jan/22

	$nice solution sir . greatefull$
	Commented by mindispower last updated on 28/Jan/22

	$W i t h e P l e a s u r H a v e a n i c e D a y$
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