Advanced-Calculus-R-e-x-2-1-x-2-2-dx-

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

\dots \dots A d v a n c e d \dots . . C a l c u l u s \dots \dots . .

\dots . . \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{\boldsymbol π} - \frac{\boldsymbol π e}{2} \boldsymbol 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) = \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} \dots .

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)

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)