Evaluate-e-x-2-dx-Please-show-and-explain-working-

Question Number 2526 by Filup last updated on 22/Nov/15

Evaluate : \underset{- \infty}{\int^{\infty}} e^{- x^{2}} d x

Please show and explain working

Commented by Filup last updated on 22/Nov/15

My current working : (unsure if correct)

I = \underset{- \infty}{\int^{\infty}} e^{- x^{2}} d x = 2 \underset{0}{\int^{\infty}} e^{- x^{2}} d x

∴ I = \underset{0}{\int^{\infty}} e^{- x^{2}} d x + \underset{0}{\int^{\infty}} e^{- y^{2}} d y

I = \underset{0}{\int^{\infty}} \underset{0}{\int^{\infty}} (e^{- x^{2}} + e^{- y^{2}}) d x d y

Commented by 123456 last updated on 22/Nov/15

\underset{0}{\int^{+ \infty}} e^{- x^{2}} d x + \underset{0}{\int^{+ \infty}} e^{- y^{2}} d y \neq \underset{0}{\int^{+ \infty}} \underset{0}{\int^{+ \infty}} e^{- x^{2}} + e^{- y^{2}} d x d y

since

Left = \sqrt{π}

Right = \underset{0}{\int^{+ \infty}} \underset{0}{\int^{+ \infty}} e^{- x^{2}} d x d y + \underset{0}{\int^{+ \infty}} \underset{0}{\int^{+ \infty}} e^{- y^{2}} d x d y = diverge

Answered by 123456 last updated on 22/Nov/15

I = \underset{- \infty}{\int^{+ \infty}} e^{- x^{2}} d x > 0 since e^{- x^{2}} ⩾ 0 \forall x \in R

I^{2} = {(\underset{- \infty}{\int^{+ \infty}} e^{- x^{2}} d x)}^{2} = \underset{- \infty}{\int^{+ \infty}} e^{- x^{2}} d x \underset{- \infty}{\int^{+ \infty}} e^{- y^{2}} d y squaring it

I^{2} = \underset{- \infty}{\int^{+ \infty}} \underset{- \infty}{\int^{+ \infty}} e^{- x^{2} - y^{2}} d x d y fubini theorem

x = r \cos θ

y = r \sin θ

d x d y = r d r d θ jacobian | \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} \end{matrix} | = | \begin{matrix} \cos θ & - r \sin θ \\ \sin θ & r \cos θ \end{matrix} | = r

0 ⩽ x < + \infty, 0 ⩽ y < + \infty

r = \sqrt{x^{2} + y^{2}}

θ = \arctan_{2} (x, y)

0 ⩽ r < + \infty, 0 ⩽ θ < 2 π (new limits)

I^{2} = \underset{0}{\int^{+ \infty}} \underset{0}{\int^{2 π}} {r e}^{- r^{2}} d θ d r

I^{2} = \underset{0}{\int^{+ \infty}} {r e}^{- r^{2}} d r \underset{0}{\int^{2 π}} d θ = \frac{1}{2} \cdot 2 π = π

\underset{0}{\int^{+ \infty}} {r e}^{- r^{2}} d r is solved at coment

I = \sqrt{π} (> 0)

Commented by Filup last updated on 22/Nov/15

W o w! T h a n k Y o u v e r y m u c h!

Commented by Filup last updated on 22/Nov/15

How did you get d x d y = r d r d θ ?

Commented by 123456 last updated on 22/Nov/15

jacobian of transformation

J (r, θ) = \frac{\partial (r, θ)}{\partial (x, y)} = | \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} \end{matrix} |

d x d y =∣ J (r, θ) ∣ d r d θ

Commented by Filup last updated on 22/Nov/15

I have no idea what that is . It seems

I have some research to conduct

Commented by 123456 last updated on 22/Nov/15

its like a subtituition method for multivariable integral

\underset{0}{\int^{+ \infty}} e^{- r^{2}} r d r

u = - r^{2}, d u = - 2 r d r

r = 0 \Rightarrow u = 0

r \to \infty \Rightarrow u \to - \infty

- \frac{1}{2} \underset{0}{\int^{- \infty}} e^{u} d u = \frac{1}{2} \underset{- \infty}{\int^{0}} e^{u} d u = 1 / 2

the jacobian generalize above idea to

multivariable integrais like

\underset{0}{\int^{1}} \underset{0}{\int^{1}} (x + y) (x - y) d x d y

u = x + y

v = x - y

\int \int_{D} u v ∣ J (u, v) ∣ d u d v

Commented by Filup last updated on 22/Nov/15

I just went and did a quick run - through .

J (r, θ) = \frac{\partial (r, θ)}{\partial (x, y)} = | \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} \end{matrix} |

D = d e t (J (r, θ)) = \frac{\partial x}{\partial r} \frac{\partial x}{\partial r} - \frac{\partial y}{\partial r} \frac{\partial y}{\partial θ} = r

∴ \int \int e^{- x^{2} - y^{2}} d x d y = \int \int {D e}^{- x^{2} - y^{2}} d r d θ

= \int \int {r e}^{- r^{2}} d r d θ

?

Commented by Filup last updated on 22/Nov/15

How do you determine the new limits ?

Commented by 123456 last updated on 22/Nov/15

- r^{2}

- x^{2} - y^{2} = - r^{2} (\cos^{2} θ + \sin^{2} θ) = - r^{2}

Commented by 123456 last updated on 22/Nov/15

its a bit complex (mr . prakash could explain it better to you later)

0 ⩽ x < + \infty \land 0 ⩽ y < + \infty

this limit give all the real plane R^{2}

in the new cordinates (polar) the same

region is givd by

0 ⩽ r < + \infty \land 0 ⩽ θ < 2 π

the radius varry fom 0 to + \infty and you

rotate its to get the full R^{2}

in general drawing your region could

help you to change the variables .

Commented by Filup last updated on 22/Nov/15

W h o o p s! Thanks for pointing out that

typo . Ok, I understand now!

I have done minimal work on matricies

but I worked it all out just now!

Thank you for helping me!

T h e limits are probably the most confusing

right now . I^{'} ll look into it!

Commented by Filup last updated on 22/Nov/15

I figured out what i was misunderstanding

about the limits

\int_{0}^{\infty} {r e}^{- r^{2}} d r \to - \frac{1}{2} \underset{0}{\int^{\infty}} - 2 {r e}^{- r^{2}} d r

u = - r^{2}, d u = - 2 r d r

r = \infty, u = - \infty

r = 0, u = 0

= - \frac{1}{2} \underset{0}{\int^{- \infty}} e^{u} d u

= \frac{1}{2} \underset{- \infty}{\int^{0}} e^{u} d u

Simple problem now understood!

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