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

Evaluate: ∫_(−∞) ^∞ e^(−x^2 ) dx Please show and explain working

$$\mathrm{Evaluate}:\:\underset{−\infty} {\overset{\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\mathrm{Please}\:\mathrm{show}\:\mathrm{and}\:\mathrm{explain}\:\mathrm{working} \\ $$

Commented by Filup last updated on 22/Nov/15

My current working: (unsure if correct) I=∫_(−∞) ^∞ e^(−x^2 ) dx=2∫_0 ^∞ e^(−x^2 ) dx ∴I=∫_0 ^∞ e^(−x^2 ) dx+∫_0 ^∞ e^(−y^2 ) dy I=∫_0 ^∞ ∫_0 ^∞ (e^(−x^2 ) +e^(−y^2 ) )dx dy

$$\mathrm{My}\:\mathrm{current}\:\mathrm{working}:\:\:\left(\mathrm{unsure}\:\mathrm{if}\:\mathrm{correct}\right) \\ $$$${I}=\underset{−\infty} {\overset{\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dx}=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\therefore{I}=\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dx}+\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{y}^{\mathrm{2}} } {dy} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left({e}^{−{x}^{\mathrm{2}} } +{e}^{−{y}^{\mathrm{2}} } \right){dx}\:{dy} \\ $$

Commented by 123456 last updated on 22/Nov/15

∫_0 ^(+∞) e^(−x^2 ) dx+∫_0 ^(+∞) e^(−y^2 ) dy≠∫_0 ^(+∞) ∫_0 ^(+∞) e^(−x^2 ) +e^(−y^2 ) dxdy since Left=(√π) Right=∫_0 ^(+∞) ∫_0 ^(+∞) e^(−x^2 ) dxdy+∫_0 ^(+∞) ∫_0 ^(+∞) e^(−y^2 ) dxdy=diverge

$$\underset{\mathrm{0}} {\overset{+\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dx}+\underset{\mathrm{0}} {\overset{+\infty} {\int}}{e}^{−{y}^{\mathrm{2}} } {dy}\neq\underset{\mathrm{0}} {\overset{+\infty} {\int}}\underset{\mathrm{0}} {\overset{+\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } +{e}^{−{y}^{\mathrm{2}} } {dxdy} \\ $$$$\mathrm{since} \\ $$$$\mathrm{Left}=\sqrt{\pi} \\ $$$$\mathrm{Right}=\underset{\mathrm{0}} {\overset{+\infty} {\int}}\underset{\mathrm{0}} {\overset{+\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dxdy}+\underset{\mathrm{0}} {\overset{+\infty} {\int}}\underset{\mathrm{0}} {\overset{+\infty} {\int}}{e}^{−{y}^{\mathrm{2}} } {dxdy}=\mathrm{diverge} \\ $$

Answered by 123456 last updated on 22/Nov/15

I=∫_(−∞) ^(+∞) e^(−x^2 ) dx>0 since e^(−x^2 ) ≥0∀x∈R I^2 =(∫_(−∞) ^(+∞) e^(−x^2 ) dx)^2 =∫_(−∞) ^(+∞) e^(−x^2 ) dx∫_(−∞) ^(+∞) e^(−y^2 ) dy squaring it I^2 =∫_(−∞) ^(+∞) ∫_(−∞) ^(+∞) e^(−x^2 −y^2 ) dxdy fubini theorem x=rcos θ y=rsin θ dxdy=rdrdθ jacobian determinant (((∂x/∂r),(∂x/∂θ)),((∂y/∂r),(∂y/∂θ)))= determinant (((cos θ),(−rsin θ)),((sin θ),(rcos θ)))=r 0≤x<+∞,0≤y<+∞ r=(√(x^2 +y^2 )) θ=arctan_2 (x,y) 0≤r<+∞,0≤θ<2π (new limits) I^2 =∫_0 ^(+∞) ∫_0 ^(2π) re^(−r^2 ) dθdr I^2 =∫_0 ^(+∞) re^(−r^2 ) dr∫_0 ^(2π) dθ=(1/2)∙2π=π ∫_0 ^(+∞) re^(−r^2 ) dr is solved at coment I=(√π) (>0)

$$\mathrm{I}=\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dx}>\mathrm{0}\:\:\:\:\:\:\mathrm{since}\:{e}^{−{x}^{\mathrm{2}} } \geqslant\mathrm{0}\forall{x}\in\mathbb{R} \\ $$$$\mathrm{I}^{\mathrm{2}} =\left(\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dx}\right)^{\mathrm{2}} =\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } {dx}\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{y}^{\mathrm{2}} } {dy}\:\:\:\mathrm{squaring}\:\mathrm{it} \\ $$$$\mathrm{I}^{\mathrm{2}} =\underset{−\infty} {\overset{+\infty} {\int}}\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:\:\:\:\:\mathrm{fubini}\:\mathrm{theorem} \\ $$$${x}={r}\mathrm{cos}\:\theta \\ $$$${y}={r}\mathrm{sin}\:\theta \\ $$$${dxdy}={rdrd}\theta\:\:\:\:\:\mathrm{jacobian}\:\begin{vmatrix}{\frac{\partial{x}}{\partial{r}}}&{\frac{\partial{x}}{\partial\theta}}\\{\frac{\partial{y}}{\partial{r}}}&{\frac{\partial{y}}{\partial\theta}}\end{vmatrix}=\begin{vmatrix}{\mathrm{cos}\:\theta}&{−{r}\mathrm{sin}\:\theta}\\{\mathrm{sin}\:\theta}&{{r}\mathrm{cos}\:\theta}\end{vmatrix}={r} \\ $$$$\mathrm{0}\leqslant{x}<+\infty,\mathrm{0}\leqslant{y}<+\infty \\ $$$${r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\theta=\mathrm{arctan}_{\mathrm{2}} \left({x},{y}\right) \\ $$$$\mathrm{0}\leqslant{r}<+\infty,\mathrm{0}\leqslant\theta<\mathrm{2}\pi\:\:\:\:\left(\mathrm{new}\:\mathrm{limits}\right) \\ $$$$\mathrm{I}^{\mathrm{2}} =\underset{\mathrm{0}} {\overset{+\infty} {\int}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}{re}^{−{r}^{\mathrm{2}} } {d}\theta{dr} \\ $$$$\mathrm{I}^{\mathrm{2}} =\underset{\mathrm{0}} {\overset{+\infty} {\int}}{re}^{−{r}^{\mathrm{2}} } {dr}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{2}\pi=\pi \\ $$$$\underset{\mathrm{0}} {\overset{+\infty} {\int}}{re}^{−{r}^{\mathrm{2}} } {dr}\:\mathrm{is}\:\mathrm{solved}\:\mathrm{at}\:\mathrm{coment} \\ $$$$\mathrm{I}=\sqrt{\pi}\:\:\left(>\mathrm{0}\right) \\ $$

Commented by Filup last updated on 22/Nov/15

Wow! ThankYou very much!

$${Wow}!\:\mathscr{T}{hank}\mathscr{Y}{ou}\:{very}\:{much}! \\ $$

Commented by Filup last updated on 22/Nov/15

How did you get dxdy=rdrdθ?

$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:{dxdy}={rdrd}\theta? \\ $$

Commented by 123456 last updated on 22/Nov/15

jacobian of transformation J(r,θ)=((∂(r,θ))/(∂(x,y)))= determinant (((∂x/∂r),(∂x/∂θ)),((∂y/∂r),(∂y/∂θ))) dxdy=∣J(r,θ)∣drdθ

$$\mathrm{jacobian}\:\mathrm{of}\:\mathrm{transformation} \\ $$$$\mathrm{J}\left({r},\theta\right)=\frac{\partial\left({r},\theta\right)}{\partial\left({x},{y}\right)}=\begin{vmatrix}{\frac{\partial{x}}{\partial{r}}}&{\frac{\partial{x}}{\partial\theta}}\\{\frac{\partial{y}}{\partial{r}}}&{\frac{\partial{y}}{\partial\theta}}\end{vmatrix} \\ $$$${dxdy}=\mid\mathrm{J}\left({r},\theta\right)\mid{drd}\theta \\ $$

Commented by Filup last updated on 22/Nov/15

I have no idea what that is. It seems I have some research to conduct

$$\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{idea}\:\mathrm{what}\:\mathrm{that}\:\mathrm{is}.\:\mathrm{It}\:\mathrm{seems} \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{some}\:\mathrm{research}\:\mathrm{to}\:\mathrm{conduct} \\ $$

Commented by 123456 last updated on 22/Nov/15

its like a subtituition method for multivariable integral ∫_0 ^(+∞) e^(−r^2 ) rdr u=−r^2 ,du=−2rdr r=0⇒u=0 r→∞⇒u→−∞ −(1/2)∫_0 ^(−∞) e^u du=(1/2)∫_(−∞) ^0 e^u du=1/2 the jacobian generalize above idea to multivariable integrais like ∫_0 ^1 ∫_0 ^1 (x+y)(x−y)dxdy u=x+y v=x−y ∫∫_D uv ∣J(u,v)∣dudv

$$\mathrm{its}\:\mathrm{like}\:\mathrm{a}\:\mathrm{subtituition}\:\mathrm{method}\:\mathrm{for}\:\mathrm{multivariable}\:\mathrm{integral} \\ $$$$\underset{\mathrm{0}} {\overset{+\infty} {\int}}{e}^{−{r}^{\mathrm{2}} } {rdr} \\ $$$${u}=−{r}^{\mathrm{2}} ,{du}=−\mathrm{2}{rdr} \\ $$$${r}=\mathrm{0}\Rightarrow{u}=\mathrm{0} \\ $$$${r}\rightarrow\infty\Rightarrow{u}\rightarrow−\infty \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{−\infty} {\int}}{e}^{{u}} {du}=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\mathrm{0}} {\int}}{e}^{{u}} {du}=\mathrm{1}/\mathrm{2} \\ $$$$\mathrm{the}\:\mathrm{jacobian}\:\mathrm{generalize}\:\mathrm{above}\:\mathrm{idea}\:\mathrm{to} \\ $$$$\mathrm{multivariable}\:\mathrm{integrais}\:\mathrm{like} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({x}+{y}\right)\left({x}−{y}\right){dxdy} \\ $$$${u}={x}+{y} \\ $$$${v}={x}−{y} \\ $$$$\int\underset{\mathrm{D}} {\int}{uv}\:\mid\mathrm{J}\left({u},{v}\right)\mid{dudv} \\ $$

Commented by Filup last updated on 22/Nov/15

I just went and did a quick run−through. J(r,θ)=((∂(r,θ))/(∂(x,y)))= determinant (((∂x/∂r),(∂x/∂θ)),((∂y/∂r),(∂y/∂θ))) D=det(J(r,θ))=(∂x/∂r) (∂x/∂r)−(∂y/∂r) (∂y/∂θ)=r ∴∫∫e^(−x^2 −y^2 ) dxdy = ∫∫De^(−x^2 −y^2 ) drdθ =∫∫re^(−r^2 ) drdθ ?

$$\mathrm{I}\:\mathrm{just}\:\mathrm{went}\:\mathrm{and}\:\mathrm{did}\:\mathrm{a}\:\mathrm{quick}\:\mathrm{run}−\mathrm{through}. \\ $$$$ \\ $$$${J}\left({r},\theta\right)=\frac{\partial\left({r},\theta\right)}{\partial\left({x},{y}\right)}=\begin{vmatrix}{\frac{\partial{x}}{\partial{r}}}&{\frac{\partial{x}}{\partial\theta}}\\{\frac{\partial{y}}{\partial{r}}}&{\frac{\partial{y}}{\partial\theta}}\end{vmatrix} \\ $$$${D}={det}\left({J}\left({r},\theta\right)\right)=\frac{\partial{x}}{\partial{r}}\:\frac{\partial{x}}{\partial{r}}−\frac{\partial{y}}{\partial{r}}\:\frac{\partial{y}}{\partial\theta}={r} \\ $$$$ \\ $$$$\therefore\int\int{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:=\:\int\int{De}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {drd}\theta \\ $$$$=\int\int{re}^{−{r}^{\mathrm{2}} } {drd}\theta \\ $$$$? \\ $$

Commented by Filup last updated on 22/Nov/15

How do you determine the new limits?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{determine}\:\mathrm{the}\:\mathrm{new}\:\mathrm{limits}? \\ $$

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

$$−{r}^{\mathrm{2}} \\ $$$$−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−{r}^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \theta\right)=−{r}^{\mathrm{2}} \\ $$

Commented by 123456 last updated on 22/Nov/15

its a bit complex (mr.prakash could explain it better to you later) 0≤x<+∞∧0≤y<+∞ this limit give all the real plane R^2 in the new cordinates (polar) the same region is givd by 0≤r<+∞∧0≤θ<2π the radius varry fom 0 to +∞ and you rotate its to get the full R^2 in general drawing your region could help you to change the variables.

$$\mathrm{its}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{complex}\:\left(\mathrm{mr}.\mathrm{prakash}\:\mathrm{could}\:\mathrm{explain}\:\mathrm{it}\:\mathrm{better}\:\mathrm{to}\:\mathrm{you}\:\mathrm{later}\right) \\ $$$$\mathrm{0}\leqslant{x}<+\infty\wedge\mathrm{0}\leqslant{y}<+\infty \\ $$$$\mathrm{this}\:\mathrm{limit}\:\mathrm{give}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{plane}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{new}\:\mathrm{cordinates}\:\left(\mathrm{polar}\right)\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{region}\:\mathrm{is}\:\mathrm{givd}\:\mathrm{by} \\ $$$$\mathrm{0}\leqslant{r}<+\infty\wedge\mathrm{0}\leqslant\theta<\mathrm{2}\pi \\ $$$$\mathrm{the}\:\mathrm{radius}\:\mathrm{varry}\:\mathrm{fom}\:\mathrm{0}\:\mathrm{to}\:+\infty\:\mathrm{and}\:\mathrm{you} \\ $$$$\mathrm{rotate}\:\mathrm{its}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{full}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{in}\:\mathrm{general}\:\mathrm{drawing}\:\mathrm{your}\:\mathrm{region}\:\mathrm{could} \\ $$$$\mathrm{help}\:\mathrm{you}\:\mathrm{to}\:\mathrm{change}\:\mathrm{the}\:\mathrm{variables}. \\ $$

Commented by Filup last updated on 22/Nov/15

Whoops! 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! The limits are probably the most confusing right now. I′ll look into it!

$${Whoops}!\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{pointing}\:\mathrm{out}\:\mathrm{that} \\ $$$$\mathrm{typo}.\:\mathrm{Ok},\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}!\: \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{done}\:\mathrm{minimal}\:\mathrm{work}\:\mathrm{on}\:\mathrm{matricies} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{worked}\:\mathrm{it}\:\mathrm{all}\:\mathrm{out}\:\mathrm{just}\:\mathrm{now}! \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{helping}\:\mathrm{me}! \\ $$$$ \\ $$$${The}\:\mathrm{limits}\:\mathrm{are}\:\mathrm{probably}\:\mathrm{the}\:\mathrm{most}\:\mathrm{confusing} \\ $$$$\mathrm{right}\:\mathrm{now}.\:\mathrm{I}'\mathrm{ll}\:\mathrm{look}\:\mathrm{into}\:\mathrm{it}! \\ $$$$ \\ $$

Commented by Filup last updated on 22/Nov/15

I figured out what i was misunderstanding about the limits ∫_0 ^( ∞) re^(−r^2 ) dr→ −(1/2)∫_0 ^∞ −2re^(−r^2 ) dr u=−r^2 , du=−2rdr r=∞, u=−∞ r=0, u=0 =−(1/2)∫_0 ^(−∞) e^u du =(1/2)∫_(−∞) ^0 e^u du Simple problem now understood!

$$\mathrm{I}\:\mathrm{figured}\:\mathrm{out}\:\mathrm{what}\:\mathrm{i}\:\mathrm{was}\:\mathrm{misunderstanding} \\ $$$$\mathrm{about}\:\mathrm{the}\:\mathrm{limits} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {re}^{−{r}^{\mathrm{2}} } {dr}\rightarrow\:−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}−\mathrm{2}{re}^{−{r}^{\mathrm{2}} } {dr} \\ $$$${u}=−{r}^{\mathrm{2}} ,\:\:{du}=−\mathrm{2}{rdr} \\ $$$${r}=\infty,\:{u}=−\infty \\ $$$${r}=\mathrm{0},\:{u}=\mathrm{0} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{−\infty} {\int}}{e}^{{u}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\mathrm{0}} {\int}}{e}^{{u}} {du} \\ $$$$ \\ $$$$\mathrm{Simple}\:\mathrm{problem}\:\mathrm{now}\:\mathrm{understood}! \\ $$

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