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


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Question Number 35678 by abdo imad last updated on 21/May/18

$l e t f (t) = \int_{0}^{\infty} \frac{e^{- {t x}^{2}} a r c t a n (x^{2})}{x^{2}} d x w i t h t > 0$ $1) s t u d y t h e e x i s t e n c t e o f f (t)$ $2) c a l c u l a t e f^{^{'}} (t)$ $3) f i n d a s i m p l e f o r m o f f (t) .$
Commented by prof Abdo imad last updated on 23/May/18
$1) f(t) = ∫_0 ^1 ((e^(−tx^2 ) arctan(x^2 ))/x^2 )dx +∫_1 ^(+∞) ((e^(−tx^2 ) arctan(x^2 ))/x^2 )dx = I +J but we have ((e^(−tx^2 ) arctan(x^2 ))/x^2 )∼ e^(−tx^2 ) ⇒ lim_(x→0) ((e^(−tx^2 ) arctan(x^2 ))/x^2 ) = 1 so I converges also we have lim_(x→+∞) x^2 ((e^(−tx^2 ) arctan(x^2 ))/x^2 ) =0 the comvergence of J is assured so f(t)?exists for t>0 2) f^′ (t) = ∫_0 ^∞ (∂/∂t){ ((e^(−tx^2 ) arctan(x^2 ))/x^2 )}dx =−∫_0 ^∞ e^(−tx^2 ) arctan(x^2 )dx .$
$1) f (t) = \int_{0}^{1} \frac{e^{- {t x}^{2}} a r c t a n (x^{2})}{x^{2}} d x + \int_{1}^{+ \infty} \frac{e^{- {t x}^{2}} a r c t a n (x^{2})}{x^{2}} d x$ $= I + J b u t w e h a v e \frac{e^{- {t x}^{2}} a r c t a n (x^{2})}{x^{2}} \sim e^{- {t x}^{2}} \Rightarrow$ ${l i m}_{x \to 0} \frac{e^{- {t x}^{2}} a r c t a n (x^{2})}{x^{2}} = 1 s o I c o n v e r g e s$ $a l s o w e h a v e {l i m}_{x \to + \infty} x^{2} \frac{e^{- {t x}^{2}} a r c t a n (x^{2})}{x^{2}} = 0$ $t h e c o m v e r g e n c e o f J i s a s s u r e d s o f (t) ? e x i s t s$ $f o r t > 0$ $2) f^{^{'}} (t) = \int_{0}^{\infty} \frac{\partial}{\partial t} {\frac{e^{- {t x}^{2}} a r c t a n (x^{2})}{x^{2}}} d x$ $= - \int_{0}^{\infty} e^{- {t x}^{2}} a r c t a n (x^{2}) d x .$
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