Question-110543

Question Number 110543 by mnjuly1970 last updated on 29/Aug/20

Answered by mathdave last updated on 29/Aug/20

s o l u t i o n

l e t y = x^{2} a n d d x = \frac{1}{2 \sqrt{y}} d y t h e n p u t t i n g i n t o

I = \frac{1}{2} \int_{0}^{1} \frac{\ln (1 - y) \ln (1 + y)}{y} d y

f r o m a l g e b r a i c u d i n g t h i s n o t a t i o n

a b = \frac{1}{4} {(a + b)}^{2} - \frac{1}{4} {(a - b)}^{2}

\ln (1 - y) \ln (1 + y) = \frac{1}{4} \ln^{2} (1 - y^{2}) - \frac{1}{4} \ln^{2} (\frac{1 - y}{1 + y})

I = \frac{1}{8} \int_{0}^{1} \frac{\ln^{2} (1 - y^{2})}{y} d y - \frac{1}{8} \int_{0}^{1} \frac{\ln^{2} (\frac{1 - y}{1 + y})}{y} d y

l e t I_{x} = \int_{0}^{1} \frac{\ln^{2} (1 - y^{2})}{y} d y (l e t x = 1 - y^{2}

a n d d y = - \frac{d x}{2 \sqrt{1 - x}}

I_{x} = \frac{1}{2} \int_{0}^{1} \frac{\ln^{2} x}{1 - x} d x

l e t

I_{x} (a) = \frac{1}{2} \int_{0}^{1} \frac{\ln^{2} x}{1 - x} d x = \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} x^{n - 1} \ln^{2} x d x

\frac{\partial^{2}}{\partial a^{2}} ∣_{a = 0} I_{x} (a) = \frac{1}{2} \frac{\partial^{2}}{\partial a^{2}} \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} x^{n - 1 + a} d x = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}}

I_{x}^{^{″}} (0) = ζ (3) \dots \dots (x)

l e t I_{x x} = \int_{0}^{1} \frac{\ln^{2} (\frac{1 - y}{1 + y})}{y} d y l e t x = \frac{1 - y}{1 + y} a n d d y = - \frac{2}{{(1 + x)}^{2}}

I_{x x} = 2 \int_{0}^{1} \frac{\ln^{2} x}{1 - x^{2}} d x = 2 \int_{0}^{1} \frac{\ln^{2} (x)}{(1 + x) (1 - x)} d x

I_{x x} = \int_{0}^{1} \frac{\ln^{2} x}{1 + x} d x + \int_{0}^{1} \frac{\ln^{2} x}{1 - x} d x t h e r e f o r e

I_{x x}^{^{″}} (0) = {(- 1)}^{n} 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} + 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}}

I_{x x}^{^{″}} (0) = 2 \times \frac{3}{4} ζ (3) + 2 ζ (3) = \frac{7}{2} ζ (3) \dots \dots . (2)

b u t

I = \frac{1}{8} [I_{x}^{^{″}} (0) - I_{x x}^{^{″}} (0)] = \frac{1}{8} [ζ (3) - \frac{7}{2} ζ (3)]

I = \frac{1}{8} [- \frac{5}{2} ζ (3)] = - \frac{5}{16} ζ (3)

∵ \int_{0}^{1} \frac{\ln (1 - x^{2}) \ln (1 + x^{2})}{x} d x = - \frac{5}{16} ζ (3)

w h e r e ζ (3) \Rightarrow {a p e r y}^{'} s c o n s t a n t

b y m a t h d a v e (29 / 08 / 2020)

Commented by mnjuly1970 last updated on 29/Aug/20

t h a n k y o u v e r y m u c h \dots s i r \dots

Commented by mathdave last updated on 29/Aug/20

y o u a r e w e l c o m e