let f(x)=∫_0 ^∞ ((1−cos(xt^2 ))/t^2 ) e^(−xt^2 ) dt with x>0 1) find a simple form of f(x) 2) calculate ∫_0 ^∞ ((1−cos(2t^2 ))/t^2 ) e^(−3t^2 ) dt .


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Question Number 38454 by maxmathsup by imad last updated on 25/Jun/18

$l e t f (x) = \int_{0}^{\infty} \frac{1 - c o s ({x t}^{2})}{t^{2}} e^{- {x t}^{2}} d t w i t h x > 0$ $1) f i n d a s i m p l e f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{1 - c o s (2 t^{2})}{t^{2}} e^{- 3 t^{2}} d t .$
Commented by math khazana by abdo last updated on 26/Jun/18

$t h e Q i s f (x) = \int_{0}^{\infty} \frac{1 - c o s ({a t}^{2})}{t^{2}} e^{- {x t}^{2}} d t$
Commented by abdo.msup.com last updated on 27/Jun/18
$wehave f(x)=∫_0 ^∞ ((1−cos(at^2 ))/t^2 ) e^(−xt^2 ) dt⇒ f^′ (x)= −∫_0 ^∞ (1−cos(at^2 )e^(−xt^2 ) dt =∫_0 ^∞ cos(at^2 )e^(−xt^2 ) dt −∫_0 ^∞ e^(−xt^2 ) dt but ∫_0 ^∞ e^(−xt^2 ) dt =_((√x)t=u) ∫_0 ^∞ e^(−u^2 ) (du/(√x)) =(1/(√x))∫_0 ^∞ e^(−u^2 ) du =(π/(2(√x))) and ∫_0 ^∞ cos(at^2 )e^(−xt^2 ) dt=(1/2) ∫_(−∞) ^(+∞) cos(at^2 )e^(−xt^2 ) dt =(1/2) Re( ∫_(−∞) ^∞ e^(iat^2 −xt^2 ) dt) but ∫_(−∞) ^(+∞) e^((−x+ia)t^2 ) dt =_((√(−x+ia))t=u) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√(−x+ia))) =((√π)/(√(−x+ia))) but −x+ia=(√(x^2 +a^2 )){((−x)/(√(x^2 +a^2 ))) +((ia)/(√(x^2 +a^2 )))} =r e^(iθ) ⇒r=(√(x^2 +a^2 )) and tanθ=−(a/x) ⇒ θ=−arctan((a/x))⇒−x+ia=r e^(−iarctan((a/x))) ⇒(√(−x+ia))=(x^2 +a^2 )^(1/4) e^(−(i/2)arctan((a/x))) ⇒ ∫_(−∞) ^(+∞) e^((−x+ia)t^2 ) dt=(√π) (x^2 +a^2 )^(−(1/4)) e^((i/2)arctan((a/x))) f^′ (x)=((√π)/2)(x^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/x)))⇒ f(x)=((√π)/2) ∫_. ^x (t^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/t)))dt +c$
$w e h a v e f (x) = \int_{0}^{\infty} \frac{1 - c o s ({a t}^{2})}{t^{2}} e^{- {x t}^{2}} d t \Rightarrow$ $f^{^{'}} (x) = - \int_{0}^{\infty} (1 - c o s ({a t}^{2}) e^{- {x t}^{2}} d t$ $= \int_{0}^{\infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t - \int_{0}^{\infty} e^{- {x t}^{2}} d t b u t$ $\int_{0}^{\infty} e^{- {x t}^{2}} d t =_{\sqrt{x} t = u} \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{x}}$ $= \frac{1}{\sqrt{x}} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{π}{2 \sqrt{x}} a n d$ $\int_{0}^{\infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t = \frac{1}{2} \int_{- \infty}^{+ \infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t$ $= \frac{1}{2} R e (\int_{- \infty}^{\infty} e^{{i a t}^{2} - {x t}^{2}} d t) b u t$ $\int_{- \infty}^{+ \infty} e^{(- x + i a) t^{2}} d t =_{\sqrt{- x + i a} t = u} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{- x + i a}}$ $= \frac{\sqrt{π}}{\sqrt{- x + i a}} b u t$ $- x + i a = \sqrt{x^{2} + a^{2}} {\frac{- x}{\sqrt{x^{2} + a^{2}}} + \frac{i a}{\sqrt{x^{2} + a^{2}}}}$ $= r e^{i θ} \Rightarrow r = \sqrt{x^{2} + a^{2}} a n d t a n θ = - \frac{a}{x} \Rightarrow$ $θ = - a r c t a n (\frac{a}{x}) \Rightarrow - x + i a = r e^{- i a r c t a n (\frac{a}{x})}$ $\Rightarrow \sqrt{- x + i a} = {(x^{2} + a^{2})}^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (\frac{a}{x})} \Rightarrow$ $\int_{- \infty}^{+ \infty} e^{(- x + i a) t^{2}} d t = \sqrt{π} {(x^{2} + a^{2})}^{- \frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{a}{x})}$ $f^{^{'}} (x) = \frac{\sqrt{π}}{2} {(x^{2} + a^{2})}^{- \frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{a}{x})) \Rightarrow$ $f (x) = \frac{\sqrt{π}}{2} \int_{.}^{x} {(t^{2} + a^{2})}^{- \frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{a}{t})) d t$ $+ c$
Commented by abdo.msup.com last updated on 27/Jun/18
$error at the final lines f^′ (x)=((√π)/2)(x^2 +a^2 )^(−(1/4)) cos{(1/2)arctan((a/x))} −(π/(2(√x))) ⇒ f(x)=((√π)/2) ∫_. ^x (t^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/t)))dt −π(√x) +c$
$e r r o r a t t h e f i n a l l i n e s$ $f^{^{'}} (x) = \frac{\sqrt{π}}{2} {(x^{2} + a^{2})}^{- \frac{1}{4}} c o s {\frac{1}{2} a r c t a n (\frac{a}{x})}$ $- \frac{π}{2 \sqrt{x}} \Rightarrow$ $f (x) = \frac{\sqrt{π}}{2} \int_{.}^{x} {(t^{2} + a^{2})}^{- \frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{a}{t})) d t$ $- π \sqrt{x} + c$
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