1) calculate f(x)=∫_(x+1) ^(x^2 +1) e^(−xt) arctan(t)dt 2) find lim


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Question Number 74345 by mathmax by abdo last updated on 22/Nov/19

$1) c a l c u l a t e f (x) = \int_{x + 1}^{x^{2} + 1} e^{- x t} a r c t a n (t) d t$ $2) f i n d {l i m}_{x \to 0} f (x)$
Commented by mathmax by abdo last updated on 24/Nov/19
$1)f(x)=∫_(x+1) ^(x^2 +1) e^(−xt) arctan(t)dt ⇒f(x)=_(xt=u) (1/x)∫_(x^2 +x) ^(x^3 +x) e^(−u) arctan((u/x))du ⇒xf(x)=∫_(x^2 +x) ^(x^3 +x) e^(−u) arctan((u/x))dx =_(by parts) [−e^(−u) arctan((u/x))]_(x^2 +x) ^(x^3 +x) +∫_(x^2 +x) ^(x^3 +x) e^(−u) ×(1/(x(1+(u^2 /x^2 ))))du =e^(−(x^2 +x)) arctan(x+1)−e^(−(x^3 +x)) arctan(x^2 +1) +x ∫_(x^2 +x) ^(x^3 +x) (e^(−u) /(x^2 +u^2 ))du ⇒ f(x)=(1/x){ e^(−(x^2 +x)) arctan(x+1)−e^(−(x^3 +x)) arctan(x^2 +1)} +∫_(x^2 +x) ^(x^3 +x) (e^(−u) /(x^2 +u^2 ))du ...be continued...$
$1) f (x) = \int_{x + 1}^{x^{2} + 1} e^{- x t} a r c t a n (t) d t \Rightarrow f (x) =_{x t = u} \frac{1}{x} \int_{x^{2} + x}^{x^{3} + x} e^{- u} a r c t a n (\frac{u}{x}) d u$ $\Rightarrow x f (x) = \int_{x^{2} + x}^{x^{3} + x} e^{- u} a r c t a n (\frac{u}{x}) d x =_{b y p a r t s} {[- e^{- u} a r c t a n (\frac{u}{x})]}_{x^{2} + x}^{x^{3} + x}$ $+ \int_{x^{2} + x}^{x^{3} + x} e^{- u} \times \frac{1}{x (1 + \frac{u^{2}}{x^{2}})} d u$ $= e^{- (x^{2} + x)} a r c t a n (x + 1) - e^{- (x^{3} + x)} a r c t a n (x^{2} + 1)$ $+ x \int_{x^{2} + x}^{x^{3} + x} \frac{e^{- u}}{x^{2} + u^{2}} d u \Rightarrow$ $f (x) = \frac{1}{x} {e^{- (x^{2} + x)} a r c t a n (x + 1) - e^{- (x^{3} + x)} a r c t a n (x^{2} + 1)}$ $+ \int_{x^{2} + x}^{x^{3} + x} \frac{e^{- u}}{x^{2} + u^{2}} d u . . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 24/Nov/19

$2) \exists c_{x} \in] x + 1, x^{2} + 1 [/ f (x) = {a r c t a n c}_{x} \int_{x + 1}^{x^{2} + 1} e^{- x t} d t$ $=_{x t = u} a r c t a n (c_{x}) \int_{x^{2} + x}^{x^{3} + x} e^{- u} \frac{d u}{x} = \frac{1}{x} a r c t a n (c_{x}) {[- e^{- u}]}_{x^{2} + x}^{x^{3} + x}$ $= a r c t a n (c_{x}) \times \frac{e^{- (x^{2} + x)} - e^{- (x^{3} + x)}}{x}$ $w e h a v e {l i m}_{x \to 0} c_{x} = \frac{π}{4} a l s o w e h a v e e^{- (x^{2} + x)} \sim 1 - (x^{2} + x)$ $a n d e^{- (x^{3} + x)} \sim 1 - (x^{3} + x) \Rightarrow$ $e^{- (x^{2} + x)} - e^{- (x^{3} + x)} = 1 - x^{2} - x - 1 + x^{3} + x = x^{3} - x^{2} \Rightarrow$ $\frac{e^{- (x^{2} + x)} - e^{- (x^{3} + x)}}{x} \sim x^{2} - x \to 0 (x \to 0) \Rightarrow {l i m}_{x \to 0} f (x) = 0$
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