let ∣a∣<1 and f(a)=∫_0 ^1 ln(x)ln(1+ax)dx 1) find a explicit form of f(a) 2) calculate g(a) =∫_0 ^1 ((xln(x))/(1+ax))dx 3) calculate ∫_0 ^1 ln(x)ln(2+x)dx 4) calculate ∫_0 ^1 ((xln(x))/(2+x))dx 5) calculate u_n =∫_0 ^1 ((xln(x))/(n+x))dx with n integr and n>1 find nature of the serie Σ u


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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 bymaxmathsup by imad last updated on 13/Oct/18
$1) we have ln^′ (1+u)=Σ_(n=0) ^∞ (−1)^n u^n with ∣u∣<1 ⇒ ln(1+u) =Σ_(n=0) ^∞ (((−1)^n u^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (−1)^(n−1) (u^n /n) ⇒ ln(1+ax) =Σ_(n=1) ^∞ (−1)^(n−1) ((a^n x^n )/n) ⇒f(a) =∫_0 ^1 ln(x){Σ_(n=1) ^∞ (−1)^(n−1) ((a^n x^n )/n)}dx =Σ_(n=1) ^∞ (−1)^(n−1) (a^n /n) ∫_0 ^1 x^n ln(x)dx by parts A_n =∫_0 ^1 x^n ln(x)dx =[(1/(n+1))x^(n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/(n+1)) x^n dx =−(1/((n+1)^2 )) ⇒f(a) = Σ_(n=1) ^∞ (−1)^n a^n (1/(n(n+1)^2 )) let decompose F(x) =(1/(x(x+1)^2 )) ⇒F(x) = (a/x) +(b/(x+1)) +(c/((x+1)^2 )) c =lim_(x→−1) (x+1)^2 F(x) =−1 a =lim_(x→0) xF(x) =1 ⇒F(x) =(1/x) +(b/(x+1)) −(1/((x+1)^2 )) F(2) =(1/(18)) =(1/2) +(b/3) −(1/9) ⇒1=9+6b−2 ⇒1=7+6b ⇒1−7=6b ⇒b=−1 ⇒F(x)=(1/x)−(1/(x+1)) −(1/((x+1)^2 )) ⇒ f(a) =Σ_(n=1) ^∞ (−a)^n {(1/n) −(1/(n+1)) −(1/((n+1)^2 ))} =Σ_(n=1) ^∞ (((−a)^n )/n) −Σ_(n=1) ^∞ (((−a)^n )/(n+1)) −Σ_(n=1) ^∞ (((−a)^n )/((n+1)^2 )) but Σ_(n=1) ^∞ (((−a)^n )/n) = −ln(1+a) Σ_(n=1) ^∞ (((−a)^n )/(n+1)) =Σ_(n=2) ^∞ (((−a)^(n−1) )/n) =−(1/a){Σ_(n=1) ^∞ (((−a)^n )/n) −1} =−(1/a){−ln(1+a)−1} =(1/a) +((ln(1+a))/a) ⇒ f(a) =−ln(1+a)−((ln(1+a))/a) −(1/a) −Σ_(n=2) ^∞ (((−a)^(n−1) )/n^2 ) f(a) =−(1+(1/a))ln(1+a) −(1/a) −Σ_(n=2) ^∞ (((−a)^(n−1) )/n^2 )$
$1) w e h a v e {l n}^{^{'}} (1 + u) = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} w i t h ∣ u ∣< 1 \Rightarrow$ $l n (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} u^{n + 1}}{n + 1} = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{u^{n}}{n} \Rightarrow$ $l n (1 + a x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{a^{n} x^{n}}{n} \Rightarrow f (a) = \int_{0}^{1} l n (x) {\sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{a^{n} x^{n}}{n}} d x$ $= \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{a^{n}}{n} \int_{0}^{1} x^{n} l n (x) d x b y p a r t s$ $A_{n} = \int_{0}^{1} x^{n} l n (x) d x = {[\frac{1}{n + 1} x^{n + 1} l n (x)]}_{0}^{1} - \int_{0}^{1} \frac{1}{n + 1} x^{n} d x$ $= - \frac{1}{{(n + 1)}^{2}} \Rightarrow f (a) = \sum_{n = 1}^{\infty} {(- 1)}^{n} a^{n} \frac{1}{n {(n + 1)}^{2}} l e t d e c o m p o s e$ $F (x) = \frac{1}{x {(x + 1)}^{2}} \Rightarrow F (x) = \frac{a}{x} + \frac{b}{x + 1} + \frac{c}{{(x + 1)}^{2}}$ $c = {l i m}_{x \to - 1} {(x + 1)}^{2} F (x) = - 1$ $a = {l i m}_{x \to 0} x F (x) = 1 \Rightarrow F (x) = \frac{1}{x} + \frac{b}{x + 1} - \frac{1}{{(x + 1)}^{2}}$ $F (2) = \frac{1}{18} = \frac{1}{2} + \frac{b}{3} - \frac{1}{9} \Rightarrow 1 = 9 + 6 b - 2 \Rightarrow 1 = 7 + 6 b \Rightarrow 1 - 7 = 6 b \Rightarrow b = - 1$ $\Rightarrow F (x) = \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{{(x + 1)}^{2}} \Rightarrow$ $f (a) = \sum_{n = 1}^{\infty} {(- a)}^{n} {\frac{1}{n} - \frac{1}{n + 1} - \frac{1}{{(n + 1)}^{2}}}$ $= \sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n} - \sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n + 1} - \sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{{(n + 1)}^{2}} b u t$ $\sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n} = - l n (1 + a)$ $\sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n + 1} = \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n} = - \frac{1}{a} {\sum_{n = 1}^{\infty} \frac{{(- a)}^{n}}{n} - 1}$ $= - \frac{1}{a} {- l n (1 + a) - 1} = \frac{1}{a} + \frac{l n (1 + a)}{a} \Rightarrow$ $f (a) = - l n (1 + a) - \frac{l n (1 + a)}{a} - \frac{1}{a} - \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n^{2}}$ $f (a) = - (1 + \frac{1}{a}) l n (1 + a) - \frac{1}{a} - \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n^{2}}$
Commented bymaxmathsup by imad last updated on 13/Oct/18

$2) w e h a v e f (a) = \int_{0}^{1} l n (x) l n (1 + a x) d x \Rightarrow$ $f^{^{'}} (a) = \int_{0}^{1} \frac{x l n (x)}{1 + a x} d x = g (a) b u t f (a) = - (1 + \frac{1}{a}) l n (1 + a) - \frac{1}{a} - \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n^{2}}$ $\Rightarrow f^{^{'}} (a) = \frac{1}{a^{2}} l n (1 + a) - (\frac{a + 1}{a}) \frac{1}{1 + a} + \frac{1}{a^{2}} - w^{^{'}} (a) w i t h w (a) = \sum_{n = 2}^{\infty} \frac{{(- a)}^{n - 1}}{n^{2}}$ $= \frac{l n (1 + a)}{a^{2}} - \frac{1}{a} + \frac{1}{a^{2}} - w^{^{'}} (a) b u t$ $w (a) = \sum_{n = 2}^{\infty} {(- 1)}^{n - 1} \frac{a^{n - 1}}{n^{2}} d u e t o u n i f o r m c o n v e g e n c e w e g e t$ $w^{^{'}} (a) = \sum_{n = 2}^{\infty} (n - 1) {(- 1)}^{n - 1} \frac{a^{n - 2}}{n^{2}} = \sum_{n = 2}^{\infty} {(- 1)}^{n - 1} \frac{a^{n - 2}}{n}$ $- \sum_{n = 2}^{\infty} {(- 1)}^{n - 1} \frac{a^{n - 2}}{n^{2}} . . . . b e c o n t i n u e d . . .$
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