Find-out-the-value-of-J-0-0-1-2e-2xy-e-xy-dxdy-

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 = \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