∫_0 ^( 1) ((ln (x+(√(1−x^2 ))))/x) dx =?


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Question Number 140961 by iloveisrael last updated on 14/May/21

$\underset{0}{\int^{1}} \frac{\ln (x + \sqrt{1 - x^{2}})}{x} d x = ?$
	Answered by qaz last updated on 14/May/21
	$∫_0 ^1 ((ln(x+(√(1−x^2 ))))/x)dx =∫_0 ^1 dy∫_0 ^1 (dx/(yx+(√(1−x^2 ))))+(1/2)∫_0 ^1 ((ln(1−x)+ln(1+x))/x)dx =∫_0 ^1 dy∫_0 ^1 (dx/(yx+(√(1−x^2 ))))+(1/2)(−(π^2 /6)+(π^2 /(12))) =∫_0 ^1 dy∫_0 ^(π/2) ((cos z)/(ysin z+cos z))dz−(π^2 /(24)) =∫_0 ^1 dy∫_0 ^(π/2) ((d(tan z))/((ytan z+1)(tan^2 z+1)))−(π^2 /(24)) =∫_0 ^1 dy∫_0 ^∞ (du/((yu+1)(u^2 +1)))−(π^2 /(24)) =∫_0 ^1 (1/(1+y^2 ))∫_0 ^∞ ((y^2 /(yu+1))+((−yu+1)/(u^2 +1)))du−(π^2 /(24)) =∫_0 ^1 (1/(1+y^2 ))(yln((yu+1)/( (√(u^2 +1))))∣_0 ^∞ +(π/2))dy−(π^2 /(24)) =∫_0 ^1 (1/(1+y^2 ))(ylny+(π/2))dy−(π^2 /(24)) =−(π^2 /(48))+(π^2 /8)−(π^2 /(24)) =(π^2 /(16)) −−−−−−−−−−−−−−−−−− ∫_0 ^1 ((ylny)/(1+y^2 ))dy ={(1/2)ln(1+y^2 )lny−(1/2)∫((ln(1+y^2 ))/y)dy}_0 ^1 =−(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 y^(2n−1) dy =−(1/4)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )=−(π^2 /(48))$
	$\int_{0}^{1} \frac{l n (x + \sqrt{1 - x^{2}})}{x} d x$ $= \int_{0}^{1} d y \int_{0}^{1} \frac{d x}{y x + \sqrt{1 - x^{2}}} + \frac{1}{2} \int_{0}^{1} \frac{l n (1 - x) + l n (1 + x)}{x} d x$ $= \int_{0}^{1} d y \int_{0}^{1} \frac{d x}{y x + \sqrt{1 - x^{2}}} + \frac{1}{2} (- \frac{π^{2}}{6} + \frac{π^{2}}{12})$ $= \int_{0}^{1} d y \int_{0}^{π / 2} \frac{\cos z}{y \sin z + \cos z} d z - \frac{π^{2}}{24}$ $= \int_{0}^{1} d y \int_{0}^{π / 2} \frac{d (\tan z)}{(y \tan z + 1) (\tan^{2} z + 1)} - \frac{π^{2}}{24}$ $= \int_{0}^{1} d y \int_{0}^{\infty} \frac{d u}{(y u + 1) (u^{2} + 1)} - \frac{π^{2}}{24}$ $= \int_{0}^{1} \frac{1}{1 + y^{2}} \int_{0}^{\infty} (\frac{y^{2}}{y u + 1} + \frac{- y u + 1}{u^{2} + 1}) d u - \frac{π^{2}}{24}$ $= \int_{0}^{1} \frac{1}{1 + y^{2}} (y l n \frac{y u + 1}{\sqrt{u^{2} + 1}} ∣_{0}^{\infty} + \frac{π}{2}) d y - \frac{π^{2}}{24}$ $= \int_{0}^{1} \frac{1}{1 + y^{2}} (y l n y + \frac{π}{2}) d y - \frac{π^{2}}{24}$ $= - \frac{π^{2}}{48} + \frac{π^{2}}{8} - \frac{π^{2}}{24}$ $= \frac{π^{2}}{16}$ $- - - - - - - - - - - - - - - - - -$ $\int_{0}^{1} \frac{y l n y}{1 + y^{2}} d y$ $= {\frac{1}{2} l n (1 + y^{2}) l n y - \frac{1}{2} \int \frac{l n (1 + y^{2})}{y} d y}_{0}^{1}$ $= - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} y^{2 n - 1} d y$ $= - \frac{1}{4} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{2}} = - \frac{π^{2}}{48}$
	Answered by iloveisrael last updated on 15/May/21

	$w i t h i n t e g r a l p a r a m e t e r i z e d b y a > 0$ $I (a) = \int_{0}^{1} \frac{\ln (a x + \sqrt{1 - x^{2}})}{x} d x$ $I^{'} (a) = \int_{0}^{1} \frac{1}{a x + \sqrt{1 - x^{2}}} d x$ $I^{'} (a) = {[\frac{a \ln (\frac{[(a^{2} + 1) x^{2} - 1] [\sqrt{1 - x^{2}} + a x]}{\sqrt{1 - x^{2}} - a x}) + 2 \arcsin x}{2 (a^{2} + 1)}]}_{x = 0}^{x = 1}$ $= \frac{a \ln a}{a^{2} + 1} + \frac{π}{2} \frac{1}{a^{2} + 1}$ $I (1) - I (0) = \int_{0}^{1} \frac{a \ln a}{a^{2} + 1} d a + \frac{π}{2} \int_{0}^{1} \frac{1}{a^{2} + 1} d a$ $(1) \int_{0}^{1} \frac{a \ln a}{a^{2} + 1} d a = \underset{k = 1}{\sum^{\infty}} {(- 1)}^{k} \int_{0}^{1} a^{2 k + 1} \ln a d a$ $= - \frac{1}{4} \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k}}{{(k + 1)}^{2}} = - \frac{1}{4} η (2) = - \frac{π^{2}}{48}$ $(2) \frac{π}{2} \int_{0}^{1} \frac{d a}{1 + a^{2}} = \frac{π^{2}}{8}$ $I (0) = \frac{1}{2} \int_{0}^{1} \frac{\ln (1 - x^{2})}{x} d x$ $= - \frac{1}{2} \underset{k = 1}{\sum^{\infty}} \frac{1}{k} \int_{0}^{1} x^{2 k - 1} d x$ $= - \frac{1}{4} \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{2}} = - \frac{1}{4} ζ (2) = - \frac{π^{2}}{24}$ $I (1) = - \frac{π^{2}}{48} + \frac{π^{2}}{8} - \frac{π^{2}}{24} = \frac{π^{2}}{16}$
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