......Advanced .....Calculus........ ..... ∫_( R) ^ (e^(−x^2 ) /((1+x^2 )^2 )) dx=?


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Question Number 141868 by mnjuly1970 last updated on 24/May/21

$. . . . . . A d v a n c e d . . . . . C a l c u l u s . . . . . . . .$ $. . . . . \int_{R}^{} \frac{e^{- x^{2}}}{{(1 + x^{2})}^{2}} d x = ?$
Commented by mindispower last updated on 24/May/21

$s a m a s 141859 Q u a t i o n s i r, a = 1$
	Answered by Dwaipayan Shikari last updated on 24/May/21

	$\int_{R} \frac{e^{- x^{2}}}{{(1 + x^{2})}^{2}} d x = 2 \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(1 + x^{2})}^{2}} d x$ $= 2 \int_{0}^{\infty} \int_{0}^{\infty} {u e}^{- x^{2} - {u x}^{2}} e^{- u} d x d u$ $= \sqrt{π} \int_{0}^{\infty} \frac{u}{\sqrt{u + 1}} e^{- u} d u$ $= 2 \sqrt{π} \int_{1}^{\infty} (t^{2} - 1) e^{- t^{2} + 1} d t$ $= 2 e \sqrt{π} (\int_{0}^{\infty} t^{2} e^{- t^{2}} - \int_{0}^{1} t^{2} e^{- t^{2}} d t - \int_{0}^{\infty} e^{- t^{2}} + \int_{0}^{1} e^{- t^{2}} d t)$ $= 2 e \sqrt{π} (- \frac{\sqrt{π}}{4}) - 2 e \sqrt{π} (\frac{\sqrt{π}}{4} e r f (1) - \frac{1}{2 e} - \frac{\sqrt{π}}{2} e r f (1))$ $= \sqrt{π} - \frac{π e}{2} + \frac{π e}{2} e r f (1) = \sqrt{π} - \frac{π e}{2} (1 - e r f (1)) = \sqrt{π} - \frac{π e}{2} e r f c (1)$ $1 - e r f (x) = e r f c (x)$ $\int e^{- t^{2}} d t = \frac{\sqrt{π}}{2} e r f c (t) + C$ $\int e^{- t^{2}} d t = {t e}^{- t^{2}} + 2 \int t^{2} e^{- t^{2}} d t$ $\Rightarrow \frac{\sqrt{π}}{4} e r f c (t) - \frac{t}{2} e^{- t^{2}} = \int t^{2} e^{- t^{2}} d t$
	Commented by mathmax by abdo last updated on 24/May/21
	$f(a)=∫_0 ^∞ (e^(−x^2 ) /(x^2 +a^2 ))dx ⇒f(a)=∫_0 ^∞ ∫_0 ^∞ (e^(−t(x^2 +a^2 )) dt)e^(−x^2 ) dx =∫_0 ^∞ ( ∫_0 ^∞ e^(−tx^2 −x^2 ) dx)e^(−ta^2 ) dt [but ∫_0 ^∞ e^(−tx^2 −x^2 ) dx =∫_0 ^∞ e^(−(t+1)x^2 ) dx =_((√(t+1))x=z) ∫_0 ^∞ e^(−z^2 ) (dz/( (√(t+1))))=((√π)/(2(√(t+1)))) ⇒ f(a)=((√π)/2)∫_0 ^∞ (e^(−ta^2 ) /( (√(t+1))))dt =_((√(t+1))=u) ((√π)/2)∫_1 ^∞ (e^(−a^2 (u^2 −1)) /u)(2u)du =(√π)e^a^2 ∫_1 ^∞ e^(−a^2 u^2 ) du ⇒ f^′ (a)=2a(√π)e^a^2 ∫_1 ^∞ e^(−a^2 u^2 ) du +(√π)e^a^2 ∫_1 ^∞ −2au^2 e^(−a^2 u^2 ) du =2a(√π)e^a^2 ∫_1 ^∞ e^(−a^2 u^2 ) du −2a(√π)e^a^2 ∫_1 ^∞ u^2 e^(−a^2 u^2 ) du [also ∫_1 ^∞ e^(−a^2 u^2 ) du =_(au=z) ∫_a ^∞ e^(−z^2 ) (dz/a) and ∫_1 ^∞ u^2 e^(−a^2 u^2 ) du =_(au=z) ∫_a ^∞ (z^2 /a^2 )e^(−z^2 ) (dz/a)=(1/a^3 )∫_a ^∞ z^2 e^(−z^2 ) dz ⇒ f^′ (a)=2(√π)e^a^2 ∫_a ^∞ e^(−z^2 ) dz−((2(√π))/a^2 ) e^a^2 ∫_a ^∞ z^2 e^(−z^2 ) dz f^′ (a)=−2a ∫_0 ^∞ (e^(−x^2 ) /((x^2 +a^2 )^2 ))dx ⇒ −2a ∫_0 ^∞ (e^(−x^2 ) /((x^2 +a^2 )^2 ))dx =2(√π)e^a^2 (∫_a ^∞ e^(−z^2 ) dz−(1/a^2 )∫_a ^∞ z^2 e^(−z^2 ) dz) ⇒ ∫_0 ^∞ (e^(−x^2 ) /((x^2 +a^2 )^2 ))dx =(√π)e^a^2 ((1/a^3 )∫_a ^∞ z^2 e^(−z^2 ) dz−∫_a ^∞ e^(−z^2 ) dz) a=(√3) ⇒∫_0 ^∞ (e^(−x^2 ) /((x^2 +3)^2 ))dx =(√π)e^3 ((1/(3(√3)))∫_(√3) ^∞ z^2 e^(−z^2 ) dz−∫_(√3) ^∞ e^(−z^2 ) dz) a=1 ⇒∫_0 ^∞ (e^(−x^2 ) /((x^2 +1)^2 ))dx=(√π)e(∫_1 ^∞ z^2 e^(−z^2 ) dz−∫_1 ^∞ e^(−z^2 ) dz) we have ∫_1 ^∞ z(ze^(−z^2 ) )dz =−(1/2)∫_1 ^∞ z(−2z e^(−z^2 ) )dz =−(1/2){ [ze^(−z^2 ) ]_1 ^∞ −∫_0 ^∞ ze^(−z^2 ) dz} =−(1/2){e^(−1) −[−(1/2)e^(−z^2 ) ]_1 ^∞ }=−(1/2)(e^(−1) −(1/2)e^(−1) ) =−(1/2)×(1/2)e^(−1) =−(1/(4e)) ....$
	$f (a) = \int_{0}^{\infty} \frac{e^{- x^{2}}}{x^{2} + a^{2}} dx \Rightarrow f (a) = \int_{0}^{\infty} \int_{0}^{\infty} (e^{- t (x^{2} + a^{2})} dt) e^{- x^{2}} dx$ $= \int_{0}^{\infty} (\int_{0}^{\infty} e^{- {tx}^{2} - x^{2}} dx) e^{- {ta}^{2}} dt [but$ $\int_{0}^{\infty} e^{- {tx}^{2} - x^{2}} dx = \int_{0}^{\infty} e^{- (t + 1) x^{2}} dx =_{\sqrt{t + 1} x = z} \int_{0}^{\infty} e^{- z^{2}} \frac{dz}{\sqrt{t + 1}} = \frac{\sqrt{π}}{2 \sqrt{t + 1}} \Rightarrow$ $f (a) = \frac{\sqrt{π}}{2} \int_{0}^{\infty} \frac{e^{- {ta}^{2}}}{\sqrt{t + 1}} dt =_{\sqrt{t + 1} = u} \frac{\sqrt{π}}{2} \int_{1}^{\infty} \frac{e^{- a^{2} (u^{2} - 1)}}{u} (2 u) du$ $= \sqrt{π} e^{a^{2}} \int_{1}^{\infty} e^{- a^{2} u^{2}} du \Rightarrow$ $f^{^{'}} (a) = 2 a \sqrt{π} e^{a^{2}} \int_{1}^{\infty} e^{- a^{2} u^{2}} du + \sqrt{π} e^{a^{2}} \int_{1}^{\infty} - {2 au}^{2} e^{- a^{2} u^{2}} du$ $= 2 a \sqrt{π} e^{a^{2}} \int_{1}^{\infty} e^{- a^{2} u^{2}} du - 2 a \sqrt{π} e^{a^{2}} \int_{1}^{\infty} u^{2} e^{- a^{2} u^{2}} du [also$ $\int_{1}^{\infty} e^{- a^{2} u^{2}} du =_{au = z} \int_{a}^{\infty} e^{- z^{2}} \frac{dz}{a} and$ $\int_{1}^{\infty} u^{2} e^{- a^{2} u^{2}} du =_{au = z} \int_{a}^{\infty} \frac{z^{2}}{a^{2}} e^{- z^{2}} \frac{dz}{a} = \frac{1}{a^{3}} \int_{a}^{\infty} z^{2} e^{- z^{2}} dz \Rightarrow$ $f^{^{'}} (a) = 2 \sqrt{π} e^{a^{2}} \int_{a}^{\infty} e^{- z^{2}} dz - \frac{2 \sqrt{π}}{a^{2}} e^{a^{2}} \int_{a}^{\infty} z^{2} e^{- z^{2}} dz$ $f^{^{'}} (a) = - 2 a \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + a^{2})}^{2}} dx \Rightarrow$ $- 2 a \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + a^{2})}^{2}} dx = 2 \sqrt{π} e^{a^{2}} (\int_{a}^{\infty} e^{- z^{2}} dz - \frac{1}{a^{2}} \int_{a}^{\infty} z^{2} e^{- z^{2}} dz) \Rightarrow$ $\int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + a^{2})}^{2}} dx = \sqrt{π} e^{a^{2}} (\frac{1}{a^{3}} \int_{a}^{\infty} z^{2} e^{- z^{2}} dz - \int_{a}^{\infty} e^{- z^{2}} dz)$ $a = \sqrt{3} \Rightarrow \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + 3)}^{2}} dx = \sqrt{π} e^{3} (\frac{1}{3 \sqrt{3}} \int_{\sqrt{3}}^{\infty} z^{2} e^{- z^{2}} dz - \int_{\sqrt{3}}^{\infty} e^{- z^{2}} dz)$ $a = 1 \Rightarrow \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + 1)}^{2}} dx = \sqrt{π} e (\int_{1}^{\infty} z^{2} e^{- z^{2}} dz - \int_{1}^{\infty} e^{- z^{2}} dz)$ $we have \int_{1}^{\infty} z ({ze}^{- z^{2}}) dz = - \frac{1}{2} \int_{1}^{\infty} z (- 2 z e^{- z^{2}}) dz$ $= - \frac{1}{2} {{[{ze}^{- z^{2}}]}_{1}^{\infty} - \int_{0}^{\infty} {ze}^{- z^{2}} dz}$ $= - \frac{1}{2} {e^{- 1} - {[- \frac{1}{2} e^{- z^{2}}]}_{1}^{\infty}} = - \frac{1}{2} (e^{- 1} - \frac{1}{2} e^{- 1})$ $= - \frac{1}{2} \times \frac{1}{2} e^{- 1} = - \frac{1}{4 e} . . . .$
	Commented by mathmax by abdo last updated on 24/May/21

	$sorry \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + a^{2})}^{2}} dx = \sqrt{π} e^{a^{2}} (\frac{1}{a^{3}} \int_{a}^{\infty} z^{2} e^{- z^{2}} dz - \frac{1}{a} \int_{a}^{\infty} e^{- z^{2}} dz)$
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