Find K=∫_0 ^1 ((ln(1−t+t^2 ))/t) dt


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Question Number 68100 by ~ À ® @ 237 ~ last updated on 04/Sep/19

$F i n d K = \int_{0}^{1} \frac{l n (1 - t + t^{2})}{t} d t$
Commented by mind is power last updated on 05/Sep/19

$y, r e w e l c o m$
Commented by mind is power last updated on 05/Sep/19
$f(s)=∫_0 ^1 ((ln(1−(t−t^2 )s))/t) s∈[0,1] f′(s)=∫_0 ^1 ((−(t−t^2 ))/(t(1−(t−t^2 )s)))dt =∫_0 ^1 ((t−1)/(1−(t−t^2 )s))dt ∫_0 ^1 ((t−1)/(t^2 s−ts+1))dt =∫_0 ^1 ((t−(1/2))/(t^2 s−ts+1))−(1/2)∫_0 ^1 (1/(t^2 s−ts+1))dt =(1/(2s))[ln(t^2 s−ts+1)]_0 ^1 −(1/(2s))∫_0 ^1 (dt/(t^2 −t+(1/s))) −(1/(2s))∫_0 ^1 (dt/((t−(1/2))^2 +(1/s)−(1/4))) −(2/(4−s))∫_0 ^1 (dt/({((t−(1/2)).(√((4s)/(4−s))))^2 +1}.)) =−(1/(√(s(4−s))))[arctan((t−(1/2))(√((4s)/(4−s))))]_0 ^1 =−(2/(√(s(4−s))))[arctg((√(s/(4−s))))] f(s)=∫−(2/(√(s(4−s))))arctan((√(s/(4−s))))ds let u=(√(s/(4−s)))⇒s=((4u^2 )/(u^2 +1)) ds=((8u)/((1+u^2 )^2 ))du s(4−s)=((16u^2 )/((u^2 +1)^2 )) ∫((−2)/(√(s(4−s))))[arctan((√((4s)/(4−s))))]ds =∫((−2)/(√((16u^2 )/((u^2 +1)^2 )))).arctan(u).((8u)/((1+u^2 )^2 )) =∫((−2(u^2 +1))/(4u))arctan(u).((8u)/((1+u^2 )))du =∫((−4)/(1+u^2 ))arctan(u)du =−2(arctan(u))^2 +c =−2(arctan((√(s/(4−s)))))^2 +c=f(s) f(0)=0=>c=0 f(s)=−2(arctan((√(s/(4−s)))))^2 k=f(1)=−2(arctan((1/(√3))))^2 =−2((π/6))^2 =((−π^2 )/(18))$
$f (s) = \int_{0}^{1} \frac{l n (1 - (t - t^{2}) s)}{t}$ $s \in [0, 1]$ $f^{'} (s) = \int_{0}^{1} \frac{- (t - t^{2})}{t (1 - (t - t^{2}) s)} d t$ $= \int_{0}^{1} \frac{t - 1}{1 - (t - t^{2}) s} d t$ $\int_{0}^{1} \frac{t - 1}{t^{2} s - t s + 1} d t$ $= \int_{0}^{1} \frac{t - \frac{1}{2}}{t^{2} s - t s + 1} - \frac{1}{2} \int_{0}^{1} \frac{1}{t^{2} s - t s + 1} d t$ $= \frac{1}{2 s} {[l n (t^{2} s - t s + 1)]}_{0}^{1} - \frac{1}{2 s} \int_{0}^{1} \frac{d t}{t^{2} - t + \frac{1}{s}}$ $- \frac{1}{2 s} \int_{0}^{1} \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{1}{s} - \frac{1}{4}}$ $- \frac{2}{4 - s} \int_{0}^{1} \frac{d t}{{{((t - \frac{1}{2}) . \sqrt{\frac{4 s}{4 - s}})}^{2} + 1} .}$ $= - \frac{1}{\sqrt{s (4 - s)}} {[a r c t a n ((t - \frac{1}{2}) \sqrt{\frac{4 s}{4 - s}})]}_{0}^{1}$ $= - \frac{2}{\sqrt{s (4 - s)}} [a r c t g (\sqrt{\frac{s}{4 - s}})]$ $f (s) = \int - \frac{2}{\sqrt{s (4 - s)}} a r c t a n (\sqrt{\frac{s}{4 - s}}) d s$ $l e t u = \sqrt{\frac{s}{4 - s}} \Rightarrow s = \frac{4 u^{2}}{u^{2} + 1}$ $d s = \frac{8 u}{{(1 + u^{2})}^{2}} d u$ $s (4 - s) = \frac{16 u^{2}}{{(u^{2} + 1)}^{2}}$ $\int \frac{- 2}{\sqrt{s (4 - s)}} [a r c t a n (\sqrt{\frac{4 s}{4 - s}})] d s$ $= \int \frac{- 2}{\sqrt{\frac{16 u^{2}}{{(u^{2} + 1)}^{2}}}} . a r c t a n (u) . \frac{8 u}{{(1 + u^{2})}^{2}}$ $= \int \frac{- 2 (u^{2} + 1)}{4 u} a r c t a n (u) . \frac{8 u}{(1 + u^{2})} d u$ $= \int \frac{- 4}{1 + u^{2}} a r c t a n (u) d u$ $= - 2 {(a r c t a n (u))}^{2} + c$ $= - 2 {(a r c t a n (\sqrt{\frac{s}{4 - s}}))}^{2} + c = f (s)$ $f (0) = 0 => c = 0$ $f (s) = - 2 {(a r c t a n (\sqrt{\frac{s}{4 - s}}))}^{2}$ $k = f (1) = - 2 {(a r c t a n (\frac{1}{\sqrt{3}}))}^{2} = - 2 {(\frac{π}{6})}^{2} = \frac{- π^{2}}{18}$
Commented by ~ À ® @ 237 ~ last updated on 05/Sep/19

$T h a n k s y o u s i r$ $l e t c o n s i d e r g d e f i n e d b y g (a) = \int_{0}^{1} \frac{l n (1 + a t + t^{2})}{t} d t, w e h a v e D_{g} = [- 2; 2]$ $w h e n s t a t i n g t = \frac{1}{u} w e g e t g (a) = \int_{1}^{\infty} [\frac{- 2 l n t}{t} + \frac{l n (1 + a t + t^{2})}{t}] d t$ $N o w g^{'} (a) = \int_{1}^{\infty} \frac{1}{t} . \frac{t}{1 + a t + t^{2}} d t$ $u s i n g t^{2} + a t + 1 = [(t + \frac{a}{2}) + \frac{4 - a^{2}}{4}] = (\frac{4 - a^{2}}{4}) [{(\frac{2 t + a}{\sqrt{4 - a^{2}}})}^{2} + 1] c a u s e a \in D_{g}$ $S o g^{'} (a) = \frac{4}{4 - a^{2}} \int_{1}^{\infty} \frac{1}{1 + {(\frac{2 t + a}{\sqrt{4 - a^{2}}})}^{2}} d t = \frac{2}{\sqrt{4 - a^{2}}} {[a r c t a n (\frac{2 t + a}{\sqrt{4 - a^{2}}})]}_{1}^{\infty}$ $w e g e t g^{'} (a) = \frac{π - 2 a r c t a n (\frac{2 + a}{\sqrt{4 - a^{2}}})}{\sqrt{4 - a^{2}}}$ $N o w g (a) = \int \frac{π}{\sqrt{4 - a^{2}}} d a - 2 \int \frac{a r c t a n (\sqrt{\frac{2 + a}{2 - a}})}{\sqrt{4 - a^{2}}} d a$ $f o r t h e s e c o n d w h o l e l e t s t a t e u = \sqrt{\frac{2 + a}{2 - a}} \Rightarrow a = \frac{2 (u^{2} - 1)}{1 + u^{2}} a n d \sqrt{4 - a^{2}} = \frac{4 u}{1 + u^{2}}, d a = \frac{8 u d u}{{(1 + u^{2})}^{2}}$ $g (a) = \int π \frac{\frac{1}{2}}{\sqrt{1 - {(\frac{a}{2})}^{2}}} d a - 2 \int \frac{2 a r c t a n u}{1 + u^{2}} d u$ $t h e n g (a) = π a r c s i n (\frac{a}{2}) - 2 {(a r c t a n \sqrt{\frac{2 + a}{2 - a}})}^{2} + c$ $f i r s t l y g (- 2) = \int_{0}^{1} \frac{l n (1 - 2 t + t^{2})}{t} d t = 2 \int_{0}^{1} \frac{l n (1 - t)}{t} d t$ $g (- 2) = - 2 \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{t^{n - 1}}{n} = - 2 \underset{n = 1_{}}{\sum^{\infty}} \frac{1}{n^{2}} = - \frac{π^{2}}{3}$ $s e c o n d l y g (- 2) = - \frac{π^{2}}{2} + c$ $t h e n c = \frac{π^{2}}{6}$ $F i n a l l y w e w e r e s e a r c h i n g a b o u t$ $g (- 1) = π a r c s i n (\frac{- 1}{2}) - 2 {(a r c t a n (\frac{1}{\sqrt{3}}))}^{2} + \frac{π^{2}}{6}$ $g (- 1) = \frac{- π^{2}}{6} - \frac{2 π^{2}}{36} + \frac{π^{2}}{6} = - \frac{π^{2}}{18}$
Commented by mathmax by abdo last updated on 06/Sep/19
$let f(x) =∫_0 ^1 ((ln(1+xt +t^2 ))/t)dt we must have t^2 +xt +1>0 for all t⇒ x^2 −4<0 ⇒−2<x<2 we have f^′ (x)=∫_0 ^1 (t/(t(1+xt+t^2 )))dt =∫_0 ^1 (dt/(t^2 +xt +1)) =_(t=(1/u)) −∫_1 ^(+∞) ((−du)/(u^2 {(1/u^2 ) +(x/u)+1})) =∫_1 ^(+∞) (du/(1+xu+u^2 )) =∫_1 ^(+∞) (du/(u^2 +2(x/2)u +(x^2 /4)+1−(x^2 /4))) =∫_1 ^(+∞) (du/((u+(x/2))^2 +((4−x^2 )/4))) changement u+(x/2) =((√(4−x^2 ))/2)z givez=((2u+x)/(√(4−x^2 ))) f^′ (x) =(4/(4−x^2 )) ∫_((2+x)/(√(4−x^2 ))) ^(+∞) (1/(1+z^2 ))×((√(4−x^2 ))/2)dz =(2/(√(4−x^2 )))∫_((2+x)/(√(4−x^2 ))) ^(+∞) (dz/(1+z^2 )) =(2/(√(4−x^2 )))[arctanz]_((2+x)/(√(4−x^2 ))) ^(+∞) =(2/(√(4−x^2 ))){(π/2) −arctan(((2+x)/(√(4−x^2 ))))}=(π/(√(4−x^2 )))−(2/(√(4−x^2 ))) arctan(((2+x)/(√(4−x^2 )))) ⇒f(x) =∫ ((πdx)/(√(4−x^2 ))) −2 ∫ (1/(√(4−x^2 ))) arctan(((2+x)/(√(4−x^2 )))) +c changement x =2cosθ give ∫ (1/(√(4−x^2 ))) arctan(((2+x)/(√(4−x^2 )))) =∫ (1/(2sinθ)) arctan(((2+2cosθ)/(2sinθ)))(−2sinθ)cosθ =−∫ arctan(((2cos^2 ((θ/2)))/(2cos((θ/2))sin((θ/2))))) =−∫ arctan((1/(tanθ))) =−((π/2)−θ) =θ−(π/2) =arcos((x/2))−(π/2) ∫ ((πdx)/(√(4−x^2 ))) =_(x =2cosθ) ∫ ((−2πsinθ dθ)/(2sinθ)) =−πθ =−πarccos((x/2)) ⇒ f(x)=−π arcos((x/2))−2 arcos((x/2))+π +c =π−(π+2) arcos((x/2)) +c f(0) =π−(π+2)(π/2) +c =π−(π^2 /2)−π +c =c−(π^2 /2) ⇒c =(π^2 /2)+f(0) =(π^2 /2) +∫_0 ^1 ((ln(1+t^2 ))/t)dt we have ln^, (1+u) =(1/(1+u)) =Σ_(n=0) ^∞ (−1)^n u^n ⇒ ln(1+u) =Σ_(n=0) ^∞ (((−1)^n u^n )/(n+1)) +c(c=0)=Σ_(n=1) ^∞ (−1)^(n−1) (u^n /n) ⇒ ln(1+t^2 ) =Σ_(n=1) ^∞ (((−1)^(n−1) t^(2n) )/n) ⇒((ln(1+t^2 ))/t) =Σ_(n=1) ^∞ (((−1)^(n−1) t^(2n−1) )/n) ⇒∫_0 ^1 ((ln(1+t^2 ))/t)dt =Σ_(n=1) ^∞ (((−1)^n )/n) ∫_0 ^1 t^(2n−1) dt =Σ_(n=1) ^∞ (((−1)^n )/(2n^2 )) =(1/2)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(1/2)(2^(1−2) −1)ξ(2)=−(1/4)(π^2 /6) =−(π^2 /(24)) ⇒ c =(π^2 /2)−(π^2 /(24)) be continued...$
$l e t f (x) = \int_{0}^{1} \frac{l n (1 + x t + t^{2})}{t} d t w e m u s t h a v e t^{2} + x t + 1 > 0 f o r a l l t \Rightarrow$ $x^{2} - 4 < 0 \Rightarrow - 2 < x < 2 w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{t}{t (1 + x t + t^{2})} d t$ $= \int_{0}^{1} \frac{d t}{t^{2} + x t + 1} =_{t = \frac{1}{u}} - \int_{1}^{+ \infty} \frac{- d u}{u^{2} {\frac{1}{u^{2}} + \frac{x}{u} + 1}}$ $= \int_{1}^{+ \infty} \frac{d u}{1 + x u + u^{2}} = \int_{1}^{+ \infty} \frac{d u}{u^{2} + 2 \frac{x}{2} u + \frac{x^{2}}{4} + 1 - \frac{x^{2}}{4}}$ $= \int_{1}^{+ \infty} \frac{d u}{{(u + \frac{x}{2})}^{2} + \frac{4 - x^{2}}{4}} c h a n g e m e n t u + \frac{x}{2} = \frac{\sqrt{4 - x^{2}}}{2} z g i v e z = \frac{2 u + x}{\sqrt{4 - x^{2}}}$ $f^{^{'}} (x) = \frac{4}{4 - x^{2}} \int_{\frac{2 + x}{\sqrt{4 - x^{2}}}}^{+ \infty} \frac{1}{1 + z^{2}} \times \frac{\sqrt{4 - x^{2}}}{2} d z$ $= \frac{2}{\sqrt{4 - x^{2}}} \int_{\frac{2 + x}{\sqrt{4 - x^{2}}}}^{+ \infty} \frac{d z}{1 + z^{2}} = \frac{2}{\sqrt{4 - x^{2}}} {[a r c t a n z]}_{\frac{2 + x}{\sqrt{4 - x^{2}}}}^{+ \infty}$ $= \frac{2}{\sqrt{4 - x^{2}}} {\frac{π}{2} - a r c t a n (\frac{2 + x}{\sqrt{4 - x^{2}}})} = \frac{π}{\sqrt{4 - x^{2}}} - \frac{2}{\sqrt{4 - x^{2}}} a r c t a n (\frac{2 + x}{\sqrt{4 - x^{2}}})$ $\Rightarrow f (x) = \int \frac{π d x}{\sqrt{4 - x^{2}}} - 2 \int \frac{1}{\sqrt{4 - x^{2}}} a r c t a n (\frac{2 + x}{\sqrt{4 - x^{2}}}) + c$ $c h a n g e m e n t x = 2 c o s θ g i v e$ $\int \frac{1}{\sqrt{4 - x^{2}}} a r c t a n (\frac{2 + x}{\sqrt{4 - x^{2}}}) = \int \frac{1}{2 s i n θ} a r c t a n (\frac{2 + 2 c o s θ}{2 s i n θ}) (- 2 s i n θ) c o s θ$ $= - \int a r c t a n (\frac{2 {c o s}^{2} (\frac{θ}{2})}{2 c o s (\frac{θ}{2}) s i n (\frac{θ}{2})}) = - \int a r c t a n (\frac{1}{t a n θ})$ $= - (\frac{π}{2} - θ) = θ - \frac{π}{2} = a r c o s (\frac{x}{2}) - \frac{π}{2}$ $\int \frac{π d x}{\sqrt{4 - x^{2}}} =_{x = 2 c o s θ} \int \frac{- 2 π s i n θ d θ}{2 s i n θ} = - π θ = - π a r c c o s (\frac{x}{2}) \Rightarrow$ $f (x) = - π a r c o s (\frac{x}{2}) - 2 a r c o s (\frac{x}{2}) + π + c$ $= π - (π + 2) a r c o s (\frac{x}{2}) + c$ $f (0) = π - (π + 2) \frac{π}{2} + c = π - \frac{π^{2}}{2} - π + c = c - \frac{π^{2}}{2} \Rightarrow c = \frac{π^{2}}{2} + f (0)$ $= \frac{π^{2}}{2} + \int_{0}^{1} \frac{l n (1 + t^{2})}{t} d t$ $w e h a v e {l n}^{,} (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} \Rightarrow$ $l n (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} u^{n}}{n + 1} + c (c = 0) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{u^{n}}{n} \Rightarrow$ $l n (1 + t^{2}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} t^{2 n}}{n} \Rightarrow \frac{l n (1 + t^{2})}{t} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} t^{2 n - 1}}{n}$ $\Rightarrow \int_{0}^{1} \frac{l n (1 + t^{2})}{t} d t = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \int_{0}^{1} t^{2 n - 1} d t = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n^{2}}$ $= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \frac{1}{2} (2^{1 - 2} - 1) ξ (2) = - \frac{1}{4} \frac{π^{2}}{6} = - \frac{π^{2}}{24} \Rightarrow$ $c = \frac{π^{2}}{2} - \frac{π^{2}}{24} b e c o n t i n u e d . . .$
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