∫_0 ^1 ∫_0 ^1 (dy/(1+y(x^2 −x))) dx


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Question Number 63292 by aliesam last updated on 02/Jul/19

$\int_{0}^{1} \int_{0}^{1} \frac{d y}{1 + y (x^{2} - x)} d x$
Commented by mathmax by abdo last updated on 02/Jul/19

$l e t I = \int_{0}^{1} \int_{0}^{1} \frac{d y}{1 + (x^{2} - x) y} d x \Rightarrow I = \int_{0}^{1} (\int_{0}^{1} \frac{d y}{1 + (x^{2} - x) y}) d x$ $\int_{0}^{1} \frac{d y}{1 + (x^{2} - x) y} = \frac{1}{x^{2} - x} l n ∣ 1 + (x^{2} - x) y ∣_{y = 0}^{1} = \frac{1}{x^{2} - x} l n ∣ x^{2} - x + 1 ∣$ $= \frac{l n (x^{2} - x + 1)}{x^{2} - x} (b e c a u s e x^{2} - x + 1 > 0) \Rightarrow$ $I = \int_{0}^{1} \frac{l n (x^{2} - x + 1)}{x^{2} - x} d x l e t f (t) = \int_{0}^{1} \frac{l n ({t x}^{2} - x + 1)}{x^{2} - x} d x w i t h t > 0 w e h a v e$ $f^{^{'}} (t) = \int_{0}^{1} \frac{x^{2}}{(x^{2} - x) ({t x}^{2} - x + 1)} d x = \int_{0}^{1} \frac{x}{(x - 1) ({t x}^{2} - x + 1)} d x l e t d e c o m p o s e$
Commented by aliesam last updated on 02/Jul/19

$n i c e s o l u t i o n s i r$
Commented by mathmax by abdo last updated on 02/Jul/19
$let decompose F(x)=(x/((x−1)(tx^2 −x +1))) F(x)=(a/(x−1)) +((bx +c)/(tx^2 −x+1)) a =lim_(x→1) (x−1)F(x) = (1/t) lim_(x→+∞) xF(x) =0 = a+(b/t) ⇒at +b =0 ⇒b =−at =−1 ⇒ F(x) = (1/(t(x−1))) +((−x +c)/(tx^2 −x +1)) F(2) =(2/(4t−1)) =(1/t) +((−2+c)/(4t−1)) ⇒2 =((4t−1)/t) −2 +c ⇒c=4−((4t−1)/t) =(1/t) ⇒ F(t) =(1/(t(x−1))) +((−x+(1/t))/(tx^2 −x+1)) =(1/(t(x−1))) +((−tx +1)/(t(tx^2 −x+1))) ⇒ f^′ (t) =∫_0 ^1 (1/(t(x−1)))dx −(1/t) ∫_0 ^1 ((tx−1)/(tx^2 −x+1))dx =(1/t){∫_0 ^1 (dx/(x−1)) −(1/2)∫_0 ^1 ((2tx−1−1)/(tx^2 −x+1))dx} =(1/t) { [ln∣x−1∣ −(1/2)ln(tx^2 −x+1)]_(x=0) ^1 +(1/2) ∫_0 ^1 (dx/(tx^2 −x+1))} =(1/t){[ ln(((∣x−1∣)/(√(tx^2 −x+1))))]_(x=0) ^1 +(1/2) ∫_0 ^1 (dx/(tx^2 −x+1))}....perhaps the integral diverges... ...be continued....$
$l e t d e c o m p o s e F (x) = \frac{x}{(x - 1) ({t x}^{2} - x + 1)}$ $F (x) = \frac{a}{x - 1} + \frac{b x + c}{{t x}^{2} - x + 1}$ $a = {l i m}_{x \to 1} (x - 1) F (x) = \frac{1}{t}$ ${l i m}_{x \to + \infty} x F (x) = 0 = a + \frac{b}{t} \Rightarrow a t + b = 0 \Rightarrow b = - a t = - 1 \Rightarrow$ $F (x) = \frac{1}{t (x - 1)} + \frac{- x + c}{{t x}^{2} - x + 1}$ $F (2) = \frac{2}{4 t - 1} = \frac{1}{t} + \frac{- 2 + c}{4 t - 1} \Rightarrow 2 = \frac{4 t - 1}{t} - 2 + c \Rightarrow c = 4 - \frac{4 t - 1}{t} = \frac{1}{t} \Rightarrow$ $F (t) = \frac{1}{t (x - 1)} + \frac{- x + \frac{1}{t}}{{t x}^{2} - x + 1} = \frac{1}{t (x - 1)} + \frac{- t x + 1}{t ({t x}^{2} - x + 1)} \Rightarrow$ $f^{^{'}} (t) = \int_{0}^{1} \frac{1}{t (x - 1)} d x - \frac{1}{t} \int_{0}^{1} \frac{t x - 1}{{t x}^{2} - x + 1} d x$ $= \frac{1}{t} {\int_{0}^{1} \frac{d x}{x - 1} - \frac{1}{2} \int_{0}^{1} \frac{2 t x - 1 - 1}{{t x}^{2} - x + 1} d x}$ $= \frac{1}{t} {{[l n ∣ x - 1 ∣ - \frac{1}{2} l n ({t x}^{2} - x + 1)]}_{x = 0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{d x}{{t x}^{2} - x + 1}}$ $= \frac{1}{t} {{[l n (\frac{∣ x - 1 ∣}{\sqrt{{t x}^{2} - x + 1}})]}_{x = 0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{d x}{{t x}^{2} - x + 1}} . . . . p e r h a p s t h e i n t e g r a l d i v e r g e s . . .$ $. . . b e c o n t i n u e d . . . .$
Commented by mathmax by abdo last updated on 02/Jul/19
$let change the ordre of integral I =∫_0 ^1 (∫_0 ^1 (dx/(1+yx^2 −yx)))dy we have A(y)=∫_0 ^1 (dx/(yx^2 −yx +1)) →let solve yx^2 −yx +1 =0 Δ =y^2 −4y <0 due to 0≤y≤1 ⇒yx^2 −yx +1=y(x^2 −x +(1/y)) =y( x^2 −2x(1/2) +(1/4) +(1/y) −(1/4)) =y{ (x−(1/2))^2 +((4−y)/y)} ( y≠0) we use the changement x−(1/2) =(√((4−y)/y))t ⇒t =((2x−1)/2) (√(y/(4−y))) A(y) = ∫_(−(1/2)(√(y/(4−y)))) ^((1/2)(√(y/(4−y)))) (1/(y ((4−y)/y)(1+t^2 ))) (√((4−y)/y))dt =(1/(√(y(4−y)))) [arctan(t)]_(−(1/2)(√(y/(4−y)))) ^((1/2)(√(y/(4−y)))) =((2arctan((1/2)(√(y/(4−y)))))/(√(y(4−y)))) ⇒ I =2 ∫_0 ^1 ((arctan((1/2)(√(y/(4−y)))))/(√(y(4−y)))) dy ....be continued....$
$l e t c h a n g e t h e o r d r e o f i n t e g r a l$ $I = \int_{0}^{1} (\int_{0}^{1} \frac{d x}{1 + {y x}^{2} - y x}) d y w e h a v e$ $A (y) = \int_{0}^{1} \frac{d x}{{y x}^{2} - y x + 1} \to l e t s o l v e {y x}^{2} - y x + 1 = 0$ $Δ = y^{2} - 4 y < 0 d u e t o 0 ⩽ y ⩽ 1 \Rightarrow {y x}^{2} - y x + 1 = y (x^{2} - x + \frac{1}{y})$ $= y (x^{2} - 2 x \frac{1}{2} + \frac{1}{4} + \frac{1}{y} - \frac{1}{4}) = y {{(x - \frac{1}{2})}^{2} + \frac{4 - y}{y}} (y \neq 0)$ $w e u s e t h e c h a n g e m e n t x - \frac{1}{2} = \sqrt{\frac{4 - y}{y}} t \Rightarrow t = \frac{2 x - 1}{2} \sqrt{\frac{y}{4 - y}}$ $A (y) = \int_{- \frac{1}{2} \sqrt{\frac{y}{4 - y}}}^{\frac{1}{2} \sqrt{\frac{y}{4 - y}}} \frac{1}{y \frac{4 - y}{y} (1 + t^{2})} \sqrt{\frac{4 - y}{y}} d t$ $= \frac{1}{\sqrt{y (4 - y)}} {[a r c t a n (t)]}_{- \frac{1}{2} \sqrt{\frac{y}{4 - y}}}^{\frac{1}{2} \sqrt{\frac{y}{4 - y}}} = \frac{2 a r c t a n (\frac{1}{2} \sqrt{\frac{y}{4 - y}})}{\sqrt{y (4 - y)}} \Rightarrow$ $I = 2 \int_{0}^{1} \frac{a r c t a n (\frac{1}{2} \sqrt{\frac{y}{4 - y}})}{\sqrt{y (4 - y)}} d y . . . . b e c o n t i n u e d . . . .$
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