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Question Number 45235 by maxmathsup by imad last updated on 10/Oct/18

l e t ∣ a ∣< 1 a n d f (a) = \int_{0}^{1} l n (x) l n (1 + a x) d x

1) f i n d a e x p l i c i t f o r m o f f (a)

2) c a l c u l a t e g (a) = \int_{0}^{1} \frac{x l n (x)}{1 + a x} d x

3) c a l c u l a t e \int_{0}^{1} l n (x) l n (2 + x) d x

4) c a l c u l a t e \int_{0}^{1} \frac{x l n (x)}{2 + x} d x

5) c a l c u l a t e u_{n} = \int_{0}^{1} \frac{x l n (x)}{n + x} d x w i t h n i n t e g r a n d n > 1

f i n d n a t u r e o f t h e s e r i e Σ u_{n}

Commented by maxmathsup by imad last updated on 13/Oct/18

1) w e h a v e {l n}^{^{'}} (1 + u) = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} w i t h ∣ u ∣< 1 \Rightarrow

l n (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} u^{n + 1}}{n + 1} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{u^{n}}{n} \Rightarrow

l n (1 + a x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{a^{n} x^{n}}{n} \Rightarrow f (a) = \int_{0}^{1} l n (x) {\sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{a^{n} x^{n}}{n}} d x

= \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{a^{n}}{n} \int_{0}^{1} x^{n} l n (x) d x b y p a r t s

A_{n} = \int_{0}^{1} x^{n} l n (x) d x = {[\frac{1}{n + 1} x^{n + 1} l n (x)]}_{0}^{1} - \int_{0}^{1} \frac{1}{n + 1} x^{n} d x

= - \frac{1}{{(n + 1)}^{2}} \Rightarrow f (a) = \sum_{n = 1}^{\infty} {(- 1)}^{n} a^{n} \frac{1}{n {(n + 1)}^{2}} l e t d e c o m p o s e

F (x) = \frac{1}{x {(x + 1)}^{2}} \Rightarrow F (x) = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{{(x + 1)}^{2}}

c = {l i m}_{x \to - 1} {(x + 1)}^{2} F (x) = - 1

a = {l i m}_{x \to 0} x F (x) = 1 \Rightarrow F (x) = \frac{1}{x} + \frac{b}{x + 1} - \frac{1}{{(x + 1)}^{2}}

F (2) = \frac{1}{18} = \frac{1}{2} + \frac{b}{3} - \frac{1}{9} \Rightarrow 1 = 9 + 6 b - 2 \Rightarrow 1 = 7 + 6 b \Rightarrow 1 - 7 = 6 b \Rightarrow b = - 1

\Rightarrow F (x) = \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{{(x + 1)}^{2}} \Rightarrow

f (a) = \sum_{n = 1}^{\infty} {(- a)}^{n} {\frac{1}{n} - \frac{1}{n + 1} - \frac{1}{{(n + 1)}^{2}}}

= \sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n} - \sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n + 1} - \sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{{(n + 1)}^{2}} b u t

\sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n} = - l n (1 + a)

\sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n + 1} = \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n} = - \frac{1}{a} {\sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n} - 1}

= - \frac{1}{a} {- l n (1 + a) - 1} = \frac{1}{a} + \frac{l n (1 + a)}{a} \Rightarrow

f (a) = - l n (1 + a) - \frac{l n (1 + a)}{a} - \frac{1}{a} - \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n^{2}}

f (a) = - (1 + \frac{1}{a}) l n (1 + a) - \frac{1}{a} - \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n^{2}}

Commented by maxmathsup by imad last updated on 13/Oct/18

2) w e h a v e f (a) = \int_{0}^{1} l n (x) l n (1 + a x) d x \Rightarrow

f^{^{'}} (a) = \int_{0}^{1} \frac{x l n (x)}{1 + a x} d x = g (a) b u t f (a) = - (1 + \frac{1}{a}) l n (1 + a) - \frac{1}{a} - \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n^{2}}

\Rightarrow f^{^{'}} (a) = \frac{1}{a^{2}} l n (1 + a) - (\frac{a + 1}{a}) \frac{1}{1 + a} + \frac{1}{a^{2}} - w^{^{'}} (a) w i t h w (a) = \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n^{2}}

= \frac{l n (1 + a)}{a^{2}} - \frac{1}{a} + \frac{1}{a^{2}} - w^{^{'}} (a) b u t

w (a) = \sum_{n = 2}^{\infty} {(- 1)}^{n - 1} \frac{a^{n - 1}}{n^{2}} d u e t o u n i f o r m c o n v e g e n c e w e g e t

w^{^{'}} (a) = \sum_{n = 2}^{\infty} (n - 1) {(- 1)}^{n - 1} \frac{a^{n - 2}}{n^{2}} = \sum_{n = 2}^{\infty} {(- 1)}^{n - 1} \frac{a^{n - 2}}{n}

- \sum_{n = 2}^{\infty} {(- 1)}^{n - 1} \frac{a^{n - 2}}{n^{2}} \dots . b e c o n t i n u e d \dots