let f(t) = ∫_0 ^∞ ((e^(−ax) −e^(−bx) )/x^2 ) e^(−tx^2 ) dx with t>0 1) calculate f^′ (t) 2)find a simple form of f(t) 3) find the value of ∫_0 ^∞ ((e^(−2x) −e^(−x) )/x^2 ) e^(−3x^2 ) dx


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Question Number 35821 by prof Abdo imad last updated on 24/May/18

$l e t f (t) = \int_{0}^{\infty} \frac{e^{- a x} - e^{- b x}}{x^{2}} e^{- {t x}^{2}} d x w i t h t > 0$ $1) c a l c u l a t e f^{^{'}} (t)$ $2) f i n d a s i m p l e f o r m o f f (t)$ $3) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- 3 x^{2}} d x$
Commented by prof Abdo imad last updated on 31/May/18

$f (t) = \int_{- \infty}^{+ \infty} \frac{e^{- a x} - e^{- b x}}{x^{2}} e^{- {t x}^{2}} w i t h t > 0$
Commented by prof Abdo imad last updated on 31/May/18

$3) f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- 3 x^{2}} d x$
Commented by abdo mathsup 649 cc last updated on 01/Jun/18
$we have f^′ (t) = −∫_(−∞) ^(+∞) (e^(−ax) −e^(−bx) ) e^(−tx^2 ) dx = ∫_(−∞) ^(+∞) e^(−bx −tx^2 ) dx −∫_(−∞) ^(+∞) e^(−ax −tx^2 ) dx but ∫_(−∞) ^(+∞) e^(−tx^2 −ax) dx =∫_(−∞) ^(+∞) e^(−{ ((√t)x)^2 +2 (a/(2(√t)))x +(a^2 /(4t)) −(a^2 /(4t))}) dx = ∫_(−∞) ^(+∞) e^(−{ ((√t) x +(a/(2(√t))))^2 } +(a^2 /(4t))) dx = e^(a^2 /(4t)) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√t)) (chang. (√t)x +(a/(2(√t))) =u) = (e^(a^2 /(4t)) /(√t)) (√π) ⇒ f(t) = (√π) ∫_. ^t (e^(a^2 /(4u)) /(√u))du +λ changement (√u)=xgive f(t) = (√π) ∫_. ^(√t) (e^(a^2 /(4x^2 )) /x) 2xdx +λ = 2(√π) ∫_. ^(√t) e^(a^2 /(4x^2 )) dx +λ$
$w e h a v e f^{^{'}} (t) = - \int_{- \infty}^{+ \infty} (e^{- a x} - e^{- b x}) e^{- {t x}^{2}} d x$ $= \int_{- \infty}^{+ \infty} e^{- b x - {t x}^{2}} d x - \int_{- \infty}^{+ \infty} e^{- a x - {t x}^{2}} d x b u t$ $\int_{- \infty}^{+ \infty} e^{- {t x}^{2} - a x} d x = \int_{- \infty}^{+ \infty} e^{- {{(\sqrt{t} x)}^{2} + 2 \frac{a}{2 \sqrt{t}} x + \frac{a^{2}}{4 t} - \frac{a^{2}}{4 t}}} d x$ $= \int_{- \infty}^{+ \infty} e^{- {{(\sqrt{t} x + \frac{a}{2 \sqrt{t}})}^{2}} + \frac{a^{2}}{4 t}} d x$ $= e^{\frac{a^{2}}{4 t}} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{t}} (c h a n g . \sqrt{t} x + \frac{a}{2 \sqrt{t}} = u)$ $= \frac{e^{\frac{a^{2}}{4 t}}}{\sqrt{t}} \sqrt{π} \Rightarrow f (t) = \sqrt{π} \int_{.}^{t} \frac{e^{\frac{a^{2}}{4 u}}}{\sqrt{u}} d u + λ$ $c h a n g e m e n t \sqrt{u} = x g i v e$ $f (t) = \sqrt{π} \int_{.}^{\sqrt{t}} \frac{e^{\frac{a^{2}}{4 x^{2}}}}{x} 2 x d x + λ$ $= 2 \sqrt{π} \int_{.}^{\sqrt{t}} e^{\frac{a^{2}}{4 x^{2}}} d x + λ$
Commented by abdo mathsup 649 cc last updated on 01/Jun/18
$error from line 6 ∫_(−∞) ^(+∞) e^(−tx^2 −ax) dx = ((√π)/(√t)) e^(a^2 /(4t)) also ∫_(−∞) ^(+∞) e^(−tx^2 −bx) dx =((√π)/(√t)) e^(b^2 /(4t)) ⇒ f^′ (t) = (√π){ (e^(b^2 /(4t)) /(√t)) −(e^(a^2 /(4t)) /(√t))} ⇒ f(t) = (√π) ∫_1 ^t ((e^(b^2 /(4u)) −e^(a^2 /(4u)) )/(√u)) du +c =_((√u) =x) (√π) ∫_1 ^(√t) ((e^(b^2 /(4x^2 )) − e^(a^2 /(4x^2 )) )/x) 2x dx +c = 2(√π) ∫_1 ^(√t) { e^(b^2 /(4x^2 )) − e^(a^2 /(4x^2 )) } dx +c c =f(1)$
$e r r o r f r o m l i n e 6$ $\int_{- \infty}^{+ \infty} e^{- {t x}^{2} - a x} d x = \frac{\sqrt{π}}{\sqrt{t}} e^{\frac{a^{2}}{4 t}} a l s o$ $\int_{- \infty}^{+ \infty} e^{- {t x}^{2} - b x} d x = \frac{\sqrt{π}}{\sqrt{t}} e^{\frac{b^{2}}{4 t}} \Rightarrow$ $f^{^{'}} (t) = \sqrt{π} {\frac{e^{\frac{b^{2}}{4 t}}}{\sqrt{t}} - \frac{e^{\frac{a^{2}}{4 t}}}{\sqrt{t}}} \Rightarrow$ $f (t) = \sqrt{π} \int_{1}^{t} \frac{e^{\frac{b^{2}}{4 u}} - e^{\frac{a^{2}}{4 u}}}{\sqrt{u}} d u + c$ $=_{\sqrt{u} = x} \sqrt{π} \int_{1}^{\sqrt{t}} \frac{e^{\frac{b^{2}}{4 x^{2}}} - e^{\frac{a^{2}}{4 x^{2}}}}{x} 2 x d x + c$ $= 2 \sqrt{π} \int_{1}^{\sqrt{t}} {e^{\frac{b^{2}}{4 x^{2}}} - e^{\frac{a^{2}}{4 x^{2}}}} d x + c$ $c = f (1)$
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