let f(a) =∫_0 ^1 ((ln^2 (x))/((1−ax)^2 )) dx with ∣a∣<1 1) find a explicit form of f(a) 2) determine A(θ) =∫_0 ^1 ((ln^2 (x))/((1−(cosθ)x)^2 ))dx with 0<θ<(π/2)


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Question Number 60595 by maxmathsup by imad last updated on 22/May/19

$l e t f (a) = \int_{0}^{1} \frac{{l n}^{2} (x)}{{(1 - a x)}^{2}} d x w i t h ∣ a ∣< 1$ $1) f i n d a e x p l i c i t f o r m o f f (a)$ $2) d e t e r m i n e A (θ) = \int_{0}^{1} \frac{{l n}^{2} (x)}{{(1 - (c o s θ) x)}^{2}} d x w i t h 0 < θ < \frac{π}{2}$
Commented bymaxmathsup by imad last updated on 23/May/19
$1) we have for ∣x∣<1 Σ_(n=0) ^∞ x^n =(1/(1−x)) and Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒ (1/((1−ax)^2 )) =Σ_(n=1) ^∞ n(ax)^(n−1) =Σ_(n=1) ^∞ n a^(n−1) x^(n−1) ⇒ f(a) =∫_0 ^1 (Σ_(n=1) ^∞ na^(n−1) x^(n−1) )ln^2 (x)dx =Σ_(n=1) ^∞ na^(n−1) ∫_0 ^1 x^(n−1) ln^2 (x)dx =Σ_(n=1) ^∞ na^(n−1) w_n with w_n =∫_0 ^1 x^(n−1) ln^2 (x)dx by parts u^′ =x^(n−1) and v=ln^2 x w_n =[(1/n)x^n ln^2 x]_0 ^1 −∫_0 ^1 (1/n)x^n ((2lnx)/x)dx =−(2/n) ∫_0 ^1 x^(n−1) ln(x) =_(byparts) −(2/n){ [(1/n)x^n lnx]_0 ^1 −∫_0 ^1 (1/n)x^n (dx/x)}=−(2/n){−(1/n) ∫_0 ^1 x^(n−1) dx} =(2/n^3 ) ⇒f(a) =Σ_(n=1) ^∞ ((2na^(n−1) )/n^3 ) =2 Σ_(n=1) ^∞ (a^(n−1) /n^2 ) ⇒af(a) =2Σ_(n=1) ^∞ (a^n /n^2 ) let try to find s(x) =Σ_(n=1) ^∞ (x^n /n^2 ) if ∣x∣<1 ....$
$1) w e h a v e f o r ∣ x ∣< 1 \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} a n d \sum_{n = 1}^{\infty} {n x}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow$ $\frac{1}{{(1 - a x)}^{2}} = \sum_{n = 1}^{\infty} n {(a x)}^{n - 1} = \sum_{n = 1}^{\infty} n a^{n - 1} x^{n - 1} \Rightarrow$ $f (a) = \int_{0}^{1} (\sum_{n = 1}^{\infty} {n a}^{n - 1} x^{n - 1}) {l n}^{2} (x) d x = \sum_{n = 1}^{\infty} {n a}^{n - 1} \int_{0}^{1} x^{n - 1} {l n}^{2} (x) d x$ $= \sum_{n = 1}^{\infty} {n a}^{n - 1} w_{n} w i t h w_{n} = \int_{0}^{1} x^{n - 1} {l n}^{2} (x) d x b y p a r t s u^{^{'}} = x^{n - 1} a n d v = {l n}^{2} x$ $w_{n} = {[\frac{1}{n} x^{n} {l n}^{2} x]}_{0}^{1} - \int_{0}^{1} \frac{1}{n} x^{n} \frac{2 l n x}{x} d x = - \frac{2}{n} \int_{0}^{1} x^{n - 1} l n (x)$ $=_{b y p a r t s} - \frac{2}{n} {{[\frac{1}{n} x^{n} l n x]}_{0}^{1} - \int_{0}^{1} \frac{1}{n} x^{n} \frac{d x}{x}} = - \frac{2}{n} {- \frac{1}{n} \int_{0}^{1} x^{n - 1} d x}$ $= \frac{2}{n^{3}} \Rightarrow f (a) = \sum_{n = 1}^{\infty} \frac{2 {n a}^{n - 1}}{n^{3}} = 2 \sum_{n = 1}^{\infty} \frac{a^{n - 1}}{n^{2}} \Rightarrow a f (a) = 2 \sum_{n = 1}^{\infty} \frac{a^{n}}{n^{2}}$ $l e t t r y t o f i n d s (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n^{2}} i f ∣ x ∣< 1 . . . .$
Commented bymaxmathsup by imad last updated on 23/May/19

$w e h a v e S^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} \Rightarrow x S^{(1)} (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \Rightarrow$ ${(x S^{(1)} (x))}^{^{'}} = \sum_{n = 1}^{\infty} x^{n - 1} = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow S^{(1)} (x) + {x S}^{(2)} (x) = \frac{1}{1 - x} \Rightarrow$ $S i s s o l u t i o n o f (d e) {x y}^{^{″}} + y^{^{'}} = \frac{1}{1 - x} l e t y^{^{'}} = z \Rightarrow$ ${x z}^{^{'}} + z = \frac{1}{1 - x} (e)$ $(h e) \Rightarrow {x z}^{^{'}} + z = 0 \Rightarrow {x z}^{^{'}} = - z \Rightarrow \frac{z^{^{'}}}{z} = - \frac{1}{x} \Rightarrow l n ∣ z ∣ = - l n ∣ x ∣ + c \Rightarrow$ $z = \frac{k}{∣ x ∣} l e t d e t e r m i n e t h e s o l u t i o n o n] 0, + \infty [\Rightarrow z = \frac{k}{x} \Rightarrow m v c m e t h o d g i v e$ $z^{^{'}} = \frac{k^{^{'}}}{x} - \frac{k}{x^{2}} (e) \Rightarrow k^{^{'}} - \frac{k}{x} + \frac{k}{x} = \frac{1}{1 - x} \Rightarrow k^{^{'}} = \frac{1}{1 - x} \Rightarrow k (x) = - l n (1 - x) + c_{0}$ $\Rightarrow z (x) = - \frac{l n (1 - x)}{x} + \frac{c_{0}}{x}$ $y^{^{'}} = z \Rightarrow y^{^{'}} = - \frac{l n (1 - x)}{x} + \frac{c_{0}}{x} \Rightarrow y (x) = - \int_{0}^{x} \frac{l n (1 - t)}{t} d t + c_{0} l n (x) + λ \Rightarrow$ $S (x) = c_{0} l n (x) - \int_{0}^{x} \frac{l n (1 - t)}{t} d t (x > 0)$ $S (e) = c_{0} - \int_{0}^{e} \frac{l n (1 - t)}{t} d t = \sum_{n = 1}^{\infty} \frac{e^{n}}{n^{2}} \Rightarrow c_{0} = \sum_{n = 1}^{\infty} \frac{e^{n}}{n^{2}} + \int_{0}^{e} \frac{l n (1 - t)}{t} d t \Rightarrow$ $S (x) = (\sum_{n = 1}^{\infty} \frac{e^{n}}{n^{2}} + \int_{0}^{e} \frac{l n (1 - t)}{t} d t) l n (x) - \int_{0}^{x} \frac{l n (1 - t)}{t} d t . . . . b e c o n t i n u e d . . .$
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