let a>0 find the value of f(a) = ∫_0 ^(+∞) e^(−(t^2 +(a/t^2 ))) dt


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Question Number 37635 by math khazana by abdo last updated on 16/Jun/18

$l e t a > 0 f i n d t h e v a l u e o f$ $f (a) = \int_{0}^{+ \infty} e^{- (t^{2} + \frac{a}{t^{2}})} d t$
Commented byprof Abdo imad last updated on 17/Jun/18
$we have 2f(a)= ∫_(−∞) ^(+∞) e^(−{ (t −((√a)/t))^2 +2(√a)}) dt =e^(−2(√a)) ∫_(−∞) ^(+∞) e^(−(t−((√a)/t))^2 ) dt changement t −((√a)/t)=x give t^2 −(√a) =xt ⇒ t^2 −xt −(√a) =0 Δ=x^2 +4(√(a ))⇒t_1 = ((x +(√(x^2 +4(√a))))/2) t_2 =((x−(√(x^2 +4(√a))))/2) let take t= ((x +(√(x^(2 ) +4(√a))))/2) ⇒ dt = (1/2){1 + (x/(√(x^2 +4(√a))))} ⇒ 2f(a) = (e^(−2(√a)) /2) ∫_(−∞) ^(+∞) e^(−x^2 ) { 1+ (x/(√(x^2 +4(√a))))}dx = (e^(−2(√a)) /2) ∫_(−∞) ^(+∞) e^(−x^2 ) dx + (e^(−2(√a)) /2) ∫_(−∞) ^(+∞) ((x e^(−x^2 ) )/(√(x^2 +4(√a))))dx =(((√π) e^(−2(√a)) )/2) +0 ⇒ f(a) = (e^(−2(√a)) /4) (√(π )).$
$w e h a v e 2 f (a) = \int_{- \infty}^{+ \infty} e^{- {{(t - \frac{\sqrt{a}}{t})}^{2} + 2 \sqrt{a}}} d t$ $= e^{- 2 \sqrt{a}} \int_{- \infty}^{+ \infty} e^{- {(t - \frac{\sqrt{a}}{t})}^{2}} d t c h a n g e m e n t$ $t - \frac{\sqrt{a}}{t} = x g i v e t^{2} - \sqrt{a} = x t \Rightarrow$ $t^{2} - x t - \sqrt{a} = 0$ $Δ = x^{2} + 4 \sqrt{a} \Rightarrow t_{1} = \frac{x + \sqrt{x^{2} + 4 \sqrt{a}}}{2}$ $t_{2} = \frac{x - \sqrt{x^{2} + 4 \sqrt{a}}}{2} l e t t a k e t = \frac{x + \sqrt{x^{2} + 4 \sqrt{a}}}{2} \Rightarrow$ $d t = \frac{1}{2} {1 + \frac{x}{\sqrt{x^{2} + 4 \sqrt{a}}}} \Rightarrow$ $2 f (a) = \frac{e^{- 2 \sqrt{a}}}{2} \int_{- \infty}^{+ \infty} e^{- x^{2}} {1 + \frac{x}{\sqrt{x^{2} + 4 \sqrt{a}}}} d x$ $= \frac{e^{- 2 \sqrt{a}}}{2} \int_{- \infty}^{+ \infty} e^{- x^{2}} d x + \frac{e^{- 2 \sqrt{a}}}{2} \int_{- \infty}^{+ \infty} \frac{x e^{- x^{2}}}{\sqrt{x^{2} + 4 \sqrt{a}}} d x$ $= \frac{\sqrt{π} e^{- 2 \sqrt{a}}}{2} + 0 \Rightarrow$ $f (a) = \frac{e^{- 2 \sqrt{a}}}{4} \sqrt{π} .$
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