Find out the value of J=∫_0 ^∞ ∫_0 ^1 (2e^(−2xy) −e^(−xy) )dxdy


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Question Number 73751 by ~blr237~ last updated on 15/Nov/19

$F i n d o u t t h e v a l u e o f$ $J = \int_{0}^{\infty} \int_{0}^{1} (2 e^{- 2 x y} - e^{- x y}) d x d y$
Commented by mathmax by abdo last updated on 15/Nov/19

$J = \int_{0}^{\infty} A (y) d y w i t h A (y) = \int_{0}^{1} (2 e^{- 2 x y} - e^{- x y}) d x \Rightarrow$ $A (y) = {[- \frac{1}{y} e^{- 2 x y}]}_{x = 0}^{x = 1} - {[- \frac{1}{y} e^{- x y}]}_{x = 0}^{x = 1}$ $= \frac{1}{y} (1 - e^{- 2 y}) + \frac{1}{y} (e^{- y} - 1) = \frac{1}{y} (1 - e^{- 2 y} + e^{- y} - 1) = \frac{e^{- x} - e^{- 2 y}}{y} \Rightarrow$ $J = \int_{0}^{\infty} \frac{e^{- y} - e^{- 2 y}}{y} d y l e t f (u) = \int_{0}^{\infty} \frac{e^{- y} - e^{- 2 y}}{y} e^{- y u} d y w i t h u ⩾ 0$ $w e h a v e f^{^{'}} (u) = - \int_{0}^{\infty} (e^{- y} - e^{- 2 y}) e^{- y u} d y$ $= \int_{0}^{\infty} (e^{- (u + 2) y} - e^{- (u + 1) y}) d y = {[\frac{- 1}{u + 2} e^{- (u + 2) y} + \frac{1}{u + 1} e^{- (u + 1) y}]}_{0}^{\infty}$ $= \frac{1}{u + 2} - \frac{1}{u + 1} \Rightarrow f (u) = l n ∣ \frac{u + 2}{u + 1} ∣ + c$ $d u e t o t h e c o n t i n u i t y \exists m > 0 / ∣ f (u) ∣⩽ m \int_{0}^{\infty} e^{- y u} d y = \frac{m}{u} \to 0 (u \to + \infty)$ $\Rightarrow c = 0 \Rightarrow f (u) = l n ∣ \frac{u + 2}{u + 1} ∣$ $J = f (0) = l n ∣ \frac{2}{1} ∣ = l n (2)$
	Answered by mind is power last updated on 15/Nov/19
	$j=∫_0 ^(+∞) [(e^(−2xy) /(−y))+(e^(−xy) /y)]_0 ^1 dy =∫_0 ^(+∞) {(e^(−2y) /(−y))+(e^(−y) /y)+(1/y)−(1/y)}dy =∫_0 ^(+∞) ((e^(−y) −e^(−2y) )/y)dy ∫_0 ^(+∞) ((f(ax)−f(bx))/x)dx=−f(0)ln((a/b)) if f continus withe limx→∞ f(x)=0 ∫_0 ^(+∞) ∫_b ^a f′(tx)dt.dx=∫_0 ^(+∞) [((f(tx))/x)]_b ^a dx=∫_0 ^(+∞) ((f(ax)−f(bx))/x).dx we use fubini ∫∫f(x,y)dxdy=∫∫f(x,y)dydx⇒ ∫((f(ax)−f(bx))/x)dx=∫_b ^a ∫_0 ^(+∞) f′(tx)dxdt=∫_b ^a [((f(tx))/t)]_0 ^(+∞) dt=∫_b ^a ((−f(0))/t)dt=−f(0)ln(∣(a/b)∣) in our exemple a=−1,b=−2 f(y)=e^(−y) ,f(0)=1,lim_(y→∞) f(y)=0 ⇒∫_0 ^(+∞) ((e^(−y) −e^(−2y) )/y)dy=−1ln(∣((−1)/(−2))∣)=ln(2)$
	$j = \int_{0}^{+ \infty} {[\frac{e^{- 2 x y}}{- y} + \frac{e^{- x y}}{y}]}_{0}^{1} d y$ $= \int_{0}^{+ \infty} {\frac{e^{- 2 y}}{- y} + \frac{e^{- y}}{y} + \frac{1}{y} - \frac{1}{y}} d y$ $= \int_{0}^{+ \infty} \frac{e^{- y} - e^{- 2 y}}{y} d y$ $\int_{0}^{+ \infty} \frac{f (a x) - f (b x)}{x} d x = - f (0) l n (\frac{a}{b})$ $i f f c o n t i n u s w i t h e l i m x \to \infty f (x) = 0$ $\int_{0}^{+ \infty} \int_{b}^{a} f^{'} (t x) d t . d x = \int_{0}^{+ \infty} {[\frac{f (t x)}{x}]}_{b}^{a} d x = \int_{0}^{+ \infty} \frac{f (a x) - f (b x)}{x} . d x$ $w e u s e f u b i n i \int \int f (x, y) d x d y = \int \int f (x, y) d y d x \Rightarrow$ $\int \frac{f (a x) - f (b x)}{x} d x = \int_{b}^{a} \int_{0}^{+ \infty} f^{'} (t x) d x d t = \int_{b}^{a} {[\frac{f (t x)}{t}]}_{0}^{+ \infty} d t = \int_{b}^{a} \frac{- f (0)}{t} d t = - f (0) l n (∣ \frac{a}{b} ∣)$ $i n o u r e x e m p l e a = - 1, b = - 2$ $f (y) = e^{- y}, f (0) = 1, \lim_{y \to \infty} f (y) = 0$ $\Rightarrow \int_{0}^{+ \infty} \frac{e^{- y} - e^{- 2 y}}{y} d y = - 1 l n (∣ \frac{- 1}{- 2} ∣) = l n (2)$
	Commented by ~blr237~ last updated on 15/Nov/19

	$T h a n k s s i r$
	Commented by mind is power last updated on 15/Nov/19

	$y^{'} r e w e l c o m$
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