old and unanswered... Mr Mathdave??? ∫x^2 ln(1−x)ln(1+x)dx=?


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Question Number 114044 by Her_Majesty last updated on 16/Sep/20

$o l d a n d u n a n s w e r e d . . . M r M a t h d a v e ? ? ?$ $\int x^{2} l n (1 - x) l n (1 + x) d x = ?$
	Answered by mathdave last updated on 17/Sep/20

	$s o k u t i o n$ $p u t x = (2 y - 1) (w a t i d i d h e r e i s l o g i c a l)$ $I = \int_{0}^{1} {(2 y - 1)}^{2} \ln (2 - 2 y) \ln (2 y) d y$ $I = \ln^{2} (2) \int_{0}^{1} {(2 y - 1)}^{2} d y + \ln 2 \int_{0}^{1} {(2 y - 1)}^{2} \ln y d y + \ln 2 \int_{0}^{1} {(2 y - 1)}^{2} \ln (1 - y) d y + \int_{0}^{1} {(2 y - 1)}^{2} \ln y \ln (1 - y) d y$ $l e t I_{1} = \ln^{2} (2) \int_{0}^{1} {(2 y - 1)}^{2} d y = \frac{\ln^{2} (2)}{3} . . . . . (1)$ $I_{2} = \ln 2 \int_{0}^{1} {(2 y - 1)}^{2} \ln y d y + \ln 2 \int_{0}^{1} {(2 y - 1)}^{2} \ln (1 - y) d y$ $I_{2} = 2 \ln 2 \int_{0}^{1} {(2 y - 1)}^{2} \ln y d y$ $\frac{\partial}{\partial a} ∣_{a = 1} I_{2} = 2 \ln 2 \int_{0}^{1} (2 y - 1) y^{a - 1} d y$ $\frac{\partial}{\partial a} ∣_{a = 1} I_{2} = \frac{2 \ln 2 ∙ 4}{(2 + a)}, I_{2} = - \frac{8 \ln 2}{9} . . . . . . . (2)$ $I_{3} = \int_{0}^{1} {(2 y - 1)}^{2} \ln (1 - y) \ln y d y$ $b u t {(2 y - 1)}^{2} = 4 y^{2} - 4 y + 1$ $I_{3} = 4 \int_{0}^{1} y^{2} \ln (1 - y) \ln y d y - 4 \int_{0}^{1} y \ln (1 - y) \ln y d y + \int_{0}^{1} \ln (1 - y) \ln y d y$ $I_{3} = \frac{\partial}{\partial a} ∣_{a = 1} - 4 \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} y^{n} . y^{2} . y^{a - 1} d y + \frac{\partial}{\partial a} ∣_{a = 1} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} y^{n} . y . y^{a - 1} d y + \frac{\partial}{\partial a} ∣_{a = 1} \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} y^{n} . y^{a - 1} d y$ $I_{3} = \frac{\partial}{\partial a} ∣_{a = 1} [- 4 \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + 2 + a)} + 4 \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + 1 + a)} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + a)}]$ $I_{3} = 4 \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 3)}^{2}} - 4 \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 2)}^{2}} + \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 1)}^{2}}$ $I_{3} = 4 [\underset{n = 1}{\sum^{\infty}} (\frac{1}{9 n} - \frac{1}{9 (n + 3)} - \frac{1}{3 {(n + 3)}^{2}})] - 4 [\underset{n = 1}{\sum^{\infty}} (\frac{1}{4 n} - \frac{1}{4 (n + 2)} - \frac{1}{2 {(n + 2)}^{2}})] + \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{{(n + 1)}^{2}} - \frac{1}{(n + 1)})$ $I_{3} = \frac{71}{27} - \frac{4 π^{2}}{18} + \frac{π^{2}}{3} - 4 + 2 - \frac{π^{2}}{6} = (\frac{17}{27} - \frac{π^{2}}{18}) . . . . . (3)$ $b u t I = I_{1} + I_{2} + I_{3}$ $I = \frac{\ln^{2} (2)}{3} - \frac{8 \ln 2}{9} - \frac{π^{2}}{18} + \frac{17}{27}$ $∵ \int_{0}^{1} x^{2} \ln (1 - x) \ln (1 + x) d x = \frac{\ln^{2} (2)}{3} - \frac{8 \ln 2}{9} - \frac{π^{2}}{18} + \frac{17}{27}$ $m a t h d a v e (i t r y t o c o m p r e s s e d d w o r k a n d s k i p s o m e s t e p s)$ $a n y c o m p l a i n a b o u t t h e w o r k i n g i s$ $a l l o w$
	Commented by mnjuly1970 last updated on 17/Sep/20

	$n i c e v e r y n i c e m r d a v e$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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