calculate f(α)=∫_0 ^∞ ((e^(−2x) −e^(−x) )/x^2 ) e^(−αx^2 ) dx with α>0 1) find the value of ∫_0 ^∞ ((e^(−2x) −e^(−x) )/x^2 ) e^(−2x^2 ) dx


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Question Number 42708 by prof Abdo imad last updated on 01/Sep/18

$c a l c u l a t e f (α) = \int_{0}^{\infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- α x^{2}} d x w i t h α > 0$ $1) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- 2 x^{2}} d x$
Commented bymaxmathsup by imad last updated on 02/Sep/18
$1) f^′ (α) = −∫_0 ^(+∞) (e^(−2x) −e^(−x) )e^(−αx^2 ) dx =∫_0 ^∞ e^(−αx^2 −x) dx −∫_0 ^∞ e^(−αx^2 −2x) dx but ∫_0 ^∞ e^(−αx^2 −x) dx =∫_0 ^∞ e^(−{((√α)x)^2 +2 (((√α)x)/(2(√α))) + (1/(4α))−(1/(4α)))) dx =∫_0 ^∞ e^(−{((√α)x +(1/(2(√α))))^2 } +(1/(4α))) dx =_((√α)x +(1/(2(√α))) =t) e^(1/(4α)) ∫_0 ^∞ e^(−t^2 ) (dt/(√α)) =((√π)/2) (e^(1/(4α)) /(√α)) ⇒ f(α) = ∫_. ^α ((√π)/2) (e^(1/(4x)) /(√x))dx +c =_((√x)=u) ((√π)/2) ∫_. ^(√α) (e^(1/(4u^2 )) /u)(2u)du +c = (√π) ∫_. ^(√α) e^(1/(4u^2 )) du +c ⇒f(α) =(√π) ∫_1 ^(√α) e^(1/(4u^2 )) +c c =f(1) = ∫_0 ^∞ ((e^(−x^2 −2x) −e^(−x^2 −x) )/x^2 )dx 2) ∫_0 ^∞ ((e^(−2x) −e^(−x) )/x^2 ) e^(−2x^2 ) dx =f(2) = (√π) ∫_1 ^(√2) e^(1/(4u^2 )) du +f(1) ....$
$1) f^{^{'}} (α) = - \int_{0}^{+ \infty} (e^{- 2 x} - e^{- x}) e^{- α x^{2}} d x = \int_{0}^{\infty} e^{- α x^{2} - x} d x - \int_{0}^{\infty} e^{- α x^{2} - 2 x} d x$ $b u t \int_{0}^{\infty} e^{- α x^{2} - x} d x = \int_{0}^{\infty} e^{- {{(\sqrt{α} x)}^{2} + 2 \frac{\sqrt{α} x}{2 \sqrt{α}} + \frac{1}{4 α} - \frac{1}{4 α})} d x = \int_{0}^{\infty} e^{- {{(\sqrt{α} x + \frac{1}{2 \sqrt{α}})}^{2}} + \frac{1}{4 α}} d x$ $=_{\sqrt{α} x + \frac{1}{2 \sqrt{α}} = t} e^{\frac{1}{4 α}} \int_{0}^{\infty} e^{- t^{2}} \frac{d t}{\sqrt{α}} = \frac{\sqrt{π}}{2} \frac{e^{\frac{1}{4 α}}}{\sqrt{α}} \Rightarrow f (α) = \int_{.}^{α} \frac{\sqrt{π}}{2} \frac{e^{\frac{1}{4 x}}}{\sqrt{x}} d x + c$ $=_{\sqrt{x} = u} \frac{\sqrt{π}}{2} \int_{.}^{\sqrt{α}} \frac{e^{\frac{1}{4 u^{2}}}}{u} (2 u) d u + c = \sqrt{π} \int_{.}^{\sqrt{α}} e^{\frac{1}{4 u^{2}}} d u + c \Rightarrow f (α) = \sqrt{π} \int_{1}^{\sqrt{α}} e^{\frac{1}{4 u^{2}}} + c$ $c = f (1) = \int_{0}^{\infty} \frac{e^{- x^{2} - 2 x} - e^{- x^{2} - x}}{x^{2}} d x$ $2) \int_{0}^{\infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- 2 x^{2}} d x = f (2) = \sqrt{π} \int_{1}^{\sqrt{2}} e^{\frac{1}{4 u^{2}}} d u + f (1) . . . .$
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