let F(x) =∫_x ^(x^2 +1) e^(−2t) sin(xt)dt determine F^′ (x) and calculate lim


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Question Number 68019 by mathmax by abdo last updated on 03/Sep/19

$l e t F (x) = \int_{x}^{x^{2} + 1} e^{- 2 t} s i n (x t) d t$ $d e t e r m i n e F^{^{'}} (x) a n d c a l c u l a t e {l i m}_{x \to 0} F (x) .$
Commented by kaivan.ahmadi last updated on 05/Sep/19

$t h a n k y o u s i r a b d o$
Commented by mathmax by abdo last updated on 05/Sep/19

$s i r a h m a d i y o u r a n s w e r i s n o t c o r r e c t b e c a u s e t h e f u n c t i o n$ $u n d e r i n t e g r a l c o n t a i n 2 v a r i b l e s x a n d t . . .!$
Commented by mathmax by abdo last updated on 05/Sep/19
$F(x) =∫_x ^(x^2 +1) e^(−2t) sin(xt)dt we have u(x)=x and v(x)=x^2 +1and g(x,t) =e^(−2t) sin(xt) we use the formulae F^, (x) =∫_(u(x)) ^(v(x)) (∂g/∂x)(x,t)dt +v^′ g(x,v)−u^′ g(x,u) =∫_x ^(x^2 +1) t e^(−2t) cos(xt)dt +2x g(x,x^2 +1)−g(x,x^2 ) =∫_x ^(x^2 +1) t e^(−2t) cos(xt)dt +2xe^(−2(x^2 +1)) sin(x(x^2 +1))−e^(−2x^2 ) sin(x^3 ) we have ∫_x ^(x^2 +1 ) t e^(−2t) cos(xt)dt =Re(∫_x ^(x^2 +1) t e^(−2t+ixt) dt) ∫_x ^(x^2 +1) t e^((−2+ix)t) dt =[(t/(−2+ix)) e^((−2+ix)t) ]_(t=x) ^(t=x^2 +1) −∫_x ^(x^2 +1) (1/(−2+ix))e^((−2+ix)t) dt =(1/(−2+ix)){(x^2 +1)e^((−2+ix)(x^2 +1)) −xe^((−2+ix)x) } −(1/(−2+ix)) ∫_x ^(x^2 +1) e^((−2+ix)t) dt ....rest to finich the calculus.$
$F (x) = \int_{x}^{x^{2} + 1} e^{- 2 t} s i n (x t) d t w e h a v e u (x) = x a n d v (x) = x^{2} + 1 a n d$ $g (x, t) = e^{- 2 t} s i n (x t)$ $w e u s e t h e f o r m u l a e F^{,} (x) = \int_{u (x)}^{v (x)} \frac{\partial g}{\partial x} (x, t) d t + v^{^{'}} g (x, v) - u^{^{'}} g (x, u)$ $= \int_{x}^{x^{2} + 1} t e^{- 2 t} c o s (x t) d t + 2 x g (x, x^{2} + 1) - g (x, x^{2})$ $= \int_{x}^{x^{2} + 1} t e^{- 2 t} c o s (x t) d t + 2 {x e}^{- 2 (x^{2} + 1)} s i n (x (x^{2} + 1)) - e^{- 2 x^{2}} s i n (x^{3})$ $w e h a v e \int_{x}^{x^{2} + 1} t e^{- 2 t} c o s (x t) d t = R e (\int_{x}^{x^{2} + 1} t e^{- 2 t + i x t} d t)$ $\int_{x}^{x^{2} + 1} t e^{(- 2 + i x) t} d t = {[\frac{t}{- 2 + i x} e^{(- 2 + i x) t}]}_{t = x}^{t = x^{2} + 1} - \int_{x}^{x^{2} + 1} \frac{1}{- 2 + i x} e^{(- 2 + i x) t} d t$ $= \frac{1}{- 2 + i x} {(x^{2} + 1) e^{(- 2 + i x) (x^{2} + 1)} - {x e}^{(- 2 + i x) x}}$ $- \frac{1}{- 2 + i x} \int_{x}^{x^{2} + 1} e^{(- 2 + i x) t} d t . . . . r e s t t o f i n i c h t h e c a l c u l u s .$
Commented by mathmax by abdo last updated on 06/Sep/19

$y o u a r e w e l c o m e s i r .$
Commented by mathmax by abdo last updated on 08/Sep/19
$we have F(x)=∫_x ^(x^2 +1) e^(−2t) sin(xt)dt ⇒F(x)=Im(∫_x ^(x^2 +1) e^(−2t+ixt) dt) ∫_x ^(x^2 +1) e^((−2+ix)t) dt =[(1/(−2+ix)) e^((−2+ix)t) ]_x ^(x^2 +1) =((−1)/(2−ix)){ e^((−2+ix)(x^2 +1)) −e^((−2+ix)x) } =((−1)/(2−ix)){ e^(−2(x^2 +1)) e^(i(x^3 +x)) −e^(−2x) e^(ix^2 ) } =((−1)/(2−ix)){ e^(−2x^2 −2) (cos(x^3 +x)+isin(x^3 +x)−e^(−2x) (cos(x^2 )+isin(x^2 )} =−((2+ix)/(4+x^2 )){ e^(−2x^2 −2) cos(x^3 +x)−e^(−2x) cos(x^2 ) +i( e^(−2x^2 −2) sin(x^3 +x)−e^(−2x) sin(x^2 )} =−(1/(4+x^2 ))(2+ix){......} =−(1/(4+x^2 )){ 2( e^(−2x^2 −2) cos(x^3 +x)−e^(−2x) cos(x^2 )) +2i(e^(−2x^2 −2) sin(x^3 +x)−e^(−2x) sin(x^2 )) +ix( e^(−2x^2 −2) cos(x^3 +x) −e^(−2x) cos(x^2 )−x(e^(−2x^2 −2) sin(x^3 +x)−e^(−2x) sin(x^2 )} ⇒ F(x) =−(1/(4+x^2 )){ 2(e^(−2x^2 −2) sin(x^3 +x)−e^(−2x) sin(x^2 )) +x(e^(−2x^2 −2) cos(x^3 +x)−e^(−2x) cos(x^2 )} ⇒lim_(x→0) F(x) =0$
$w e h a v e F (x) = \int_{x}^{x^{2} + 1} e^{- 2 t} s i n (x t) d t \Rightarrow F (x) = I m (\int_{x}^{x^{2} + 1} e^{- 2 t + i x t} d t)$ $\int_{x}^{x^{2} + 1} e^{(- 2 + i x) t} d t = {[\frac{1}{- 2 + i x} e^{(- 2 + i x) t}]}_{x}^{x^{2} + 1}$ $= \frac{- 1}{2 - i x} {e^{(- 2 + i x) (x^{2} + 1)} - e^{(- 2 + i x) x}}$ $= \frac{- 1}{2 - i x} {e^{- 2 (x^{2} + 1)} e^{i (x^{3} + x)} - e^{- 2 x} e^{{i x}^{2}}}$ $= \frac{- 1}{2 - i x} {e^{- 2 x^{2} - 2} (c o s (x^{3} + x) + i s i n (x^{3} + x) - e^{- 2 x} (c o s (x^{2}) + i s i n (x^{2})}$ $= - \frac{2 + i x}{4 + x^{2}} {e^{- 2 x^{2} - 2} c o s (x^{3} + x) - e^{- 2 x} c o s (x^{2})$ $+ i (e^{- 2 x^{2} - 2} s i n (x^{3} + x) - e^{- 2 x} s i n (x^{2})}$ $= - \frac{1}{4 + x^{2}} (2 + i x) {. . . . . .}$ $= - \frac{1}{4 + x^{2}} {2 (e^{- 2 x^{2} - 2} c o s (x^{3} + x) - e^{- 2 x} c o s (x^{2}))$ $+ 2 i (e^{- 2 x^{2} - 2} s i n (x^{3} + x) - e^{- 2 x} s i n (x^{2})) + i x (e^{- 2 x^{2} - 2} c o s (x^{3} + x)$ $- e^{- 2 x} c o s (x^{2}) - x (e^{- 2 x^{2} - 2} s i n (x^{3} + x) - e^{- 2 x} s i n (x^{2})} \Rightarrow$ $F (x) = - \frac{1}{4 + x^{2}} {2 (e^{- 2 x^{2} - 2} s i n (x^{3} + x) - e^{- 2 x} s i n (x^{2}))$ $+ x (e^{- 2 x^{2} - 2} c o s (x^{3} + x) - e^{- 2 x} c o s (x^{2})} \Rightarrow {l i m}_{x \to 0} F (x) = 0$
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