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Question Number 228131 by Abod last updated on 29/Mar/26
Commented by Frix last updated on 29/Mar/26
∫_(−∞) ^(+∞) e^(x^2 /(2σ^2 )) dx=∞ ∀σ∈R but ∫_(−∞) ^(+∞) e^(−(x^2 /(2σ^2 ))) dx= [t=(x/( (√2)σ)) → dx=(√2)αdt] =(√2)σ∫_(−∞) ^(+∞) e^(−t^2 ) dt=σ(√(2π))
$$\underset{−\infty} {\overset{+\infty} {\int}}\mathrm{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dx}=\infty\:\forall\sigma\in\mathbb{R} \\ $$$$\mathrm{but} \\ $$$$\underset{−\infty} {\overset{+\infty} {\int}}\mathrm{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}}{\:\sqrt{\mathrm{2}}\sigma}\:\rightarrow\:{dx}=\sqrt{\mathrm{2}}\alpha{dt}\right] \\ $$$$=\sqrt{\mathrm{2}}\sigma\underset{−\infty} {\overset{+\infty} {\int}}\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}=\sigma\sqrt{\mathrm{2}\pi} \\ $$
Commented by Abod last updated on 30/Mar/26
yes but louk at this =∣σ∣(√(2π)) not σ(√(2π))
$${yes}\:{but}\:{louk}\:{at}\:{this} \\ $$$$=\mid\sigma\mid\sqrt{\mathrm{2}\pi} \\ $$$${not}\:\sigma\sqrt{\mathrm{2}\pi} \\ $$
Commented by Ghisom_ last updated on 30/Mar/26
no. the equation you gave is simply wrong. the lhs and rhs are not equal, no matter if ∣σ∣ or σ or whatever... lim_(a→+∞) ∫_(−a) ^(+a) e^(x^2 /(2σ^2 )) dx does not exist can we please return to answering exactly what was asked instead of trying to make sense of any nonsense.
$$\mathrm{no}.\:\mathrm{the}\:\mathrm{equation}\:\mathrm{you}\:\mathrm{gave}\:\mathrm{is}\:\mathrm{simply}\:\mathrm{wrong}. \\ $$$$\mathrm{the}\:\mathrm{lhs}\:\mathrm{and}\:\mathrm{rhs}\:\mathrm{are}\:\mathrm{not}\:\mathrm{equal},\:\mathrm{no}\:\mathrm{matter} \\ $$$$\mathrm{if}\:\mid\sigma\mid\:\mathrm{or}\:\sigma\:\mathrm{or}\:\mathrm{whatever}… \\ $$$$\underset{{a}\rightarrow+\infty} {\mathrm{lim}}\:\underset{−{a}} {\overset{+{a}} {\int}}\mathrm{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dx}\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist} \\ $$$$ \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{please}\:\mathrm{return}\:\mathrm{to}\:\mathrm{answering}\:\mathrm{exactly} \\ $$$$\mathrm{what}\:\mathrm{was}\:\mathrm{asked}\:\mathrm{instead}\:\mathrm{of}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{make} \\ $$$$\mathrm{sense}\:\mathrm{of}\:\mathrm{any}\:\mathrm{nonsense}. \\ $$
Commented by Abod last updated on 30/Mar/26
and that is the sulation lim_(a→∞) ∫_(−a) ^a e^(−(x^2 /(2σ^2 ))) dx = ∣σ∣(√(2π))
$${and}\:{that}\:{is}\:{the}\:{sulation} \\ $$$$\underset{{a}\rightarrow\infty} {\mathrm{lim}}\:\int_{−{a}} ^{{a}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dx}\:=\:\mid\sigma\mid\sqrt{\mathrm{2}\pi} \\ $$
Commented by Ghisom_ last updated on 31/Mar/26
yes to this different question
$$\mathrm{yes}\:\mathrm{to}\:\mathrm{this}\:{different}\:\mathrm{question} \\ $$
Answered by Lara2440 last updated on 30/Mar/26
for all σ>0 , σ∈R\{0} (∫_R -)^2 =∫_R -×∫_R -=∫∫_(R×R) - (∫_R e^(−(t^2 /(2σ^2 ))) dt)^2 =∫_R e^(−(x^2 /(2σ^2 ))) dx×∫_R e^(−(x^2 /(2σ^2 ))) dy=∫∫_(R×R) e^(−(1/(2σ^2 ))(x^2 +y^2 )) da ∣∣J∣∣=∣∣((∂(x′,y′))/(∂(r,θ)))∣∣drdθ (Jacobian) ∫∫_( J(R)) r∙e^(−(r^2 /(2σ^2 ))) drdθ=∫_0 ^( 2π) ∫_0 ^( ∞) r∙e^(−(r^2 /(2σ^2 ))) drdθ=2πσ^2 ∴I=σ(√(2π))
$$\mathrm{for}\:\mathrm{all}\:\sigma>\mathrm{0}\:,\:\sigma\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\left(\int_{\mathbb{R}} \:-\right)^{\mathrm{2}} =\int_{\mathbb{R}} -×\int_{\mathbb{R}} -=\int\int_{\mathbb{R}×\mathbb{R}} – \\ $$$$\left(\int_{\mathbb{R}} \:{e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} \mathrm{d}{t}\right)^{\mathrm{2}} =\int_{\mathbb{R}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} \mathrm{d}{x}×\int_{\mathbb{R}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} \mathrm{d}{y}=\int\int_{\mathbb{R}×\mathbb{R}} \:{e}^{−\frac{\mathrm{1}}{\mathrm{2}\sigma^{\mathrm{2}} }\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \mathrm{da} \\ $$$$\mid\mid{J}\mid\mid=\mid\mid\frac{\partial\left({x}',{y}'\right)}{\partial\left({r},\theta\right)}\mid\mid\mathrm{d}{r}\mathrm{d}\theta\:\left(\mathrm{Jacobian}\right) \\ $$$$\int\int_{\:\boldsymbol{\mathcal{J}}\left(\mathbb{R}\right)} \:{r}\centerdot{e}^{−\frac{{r}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} \mathrm{d}{r}\mathrm{d}\theta=\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:\infty} \:\:{r}\centerdot{e}^{−\frac{{r}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} \mathrm{d}{r}\mathrm{d}\theta=\mathrm{2}\pi\sigma^{\mathrm{2}} \\ $$$$\therefore{I}=\sigma\sqrt{\mathrm{2}\pi} \\ $$$$\: \\ $$
Commented by Frix last updated on 30/Mar/26
There′s no minus sign until your 5^(th) line. Where and how did you get it?
$$\mathrm{There}'\mathrm{s}\:\mathrm{no}\:\mathrm{minus}\:\mathrm{sign}\:\mathrm{until}\:\mathrm{your}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{line}. \\ $$$$\mathrm{Where}\:\mathrm{and}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{it}? \\ $$
Commented by Lara2440 last updated on 30/Mar/26
Oh. Frix Actually I was assuming the case with the negative exponent. For a function like f(z)=e^(z^2 /(2σ)) it only convergence if you integrate from −∞ to 0. If the limit from 0 to ∞ the integral clearly diverges and i′m so Sorry for my typos....
$$\mathrm{Oh}.\:\mathrm{Frix} \\ $$$$\mathrm{Actually}\:\mathrm{I}\:\mathrm{was}\:\mathrm{assuming}\:\mathrm{the}\:\mathrm{case}\:\mathrm{with}\:\mathrm{the}\:\mathrm{negative} \\ $$$$\mathrm{exponent}.\:\mathrm{For}\:\mathrm{a}\:\mathrm{function}\:\mathrm{like}\:\:\:{f}\left({z}\right)={e}^{\frac{{z}^{\mathrm{2}} }{\mathrm{2}\sigma}} \:\: \\ $$$$\mathrm{it}\:\mathrm{only}\:\:\mathrm{convergence}\:\mathrm{if}\:\mathrm{you}\:\mathrm{integrate}\:\mathrm{from}\:−\infty\:\mathrm{to}\:\mathrm{0}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\infty\:\mathrm{the}\:\mathrm{integral}\:\mathrm{clearly}\:\mathrm{diverges} \\ $$$$\mathrm{and}\:\mathrm{i}'\mathrm{m}\:\mathrm{so}\:\mathrm{Sorry}\:\mathrm{for}\:\mathrm{my}\:\mathrm{typos}…. \\ $$
Commented by Abod last updated on 30/Mar/26
Legend sulation but .... I=∣σ∣(√(2π))
$${Legend}\:{sulation}\:{but}\:…. \\ $$$${I}=\mid\sigma\mid\sqrt{\mathrm{2}\pi} \\ $$
Answered by mr W last updated on 31/Mar/26
I=∫_(−∞) ^∞ e^(−(x^2 /(2σ^2 ))) dx >0 I=∫_(−∞) ^∞ e^(−(y^2 /(2σ^2 ))) dy I^2 =∫_(−∞) ^∞ e^(−(x^2 /(2σ^2 ))) dx∫_(−∞) ^∞ e^(−(y^2 /(2σ^2 ))) dy =∫_(−∞) ^∞ ∫_(−∞) ^∞ e^(−(x^2 /(2σ^2 ))) e^(−(y^2 /(2σ^2 ))) dydx =∫_(−∞) ^∞ ∫_(−∞) ^∞ e^(−((x^2 +y^2 )/(2σ^2 ))) dydx =∫_0 ^(2π) ∫_0 ^∞ e^(−(r^2 /(2σ^2 ))) rdrdθ =2πσ^2 ∫_0 ^∞ e^(−(r^2 /(2σ^2 ))) d((r^2 /(2σ^2 ))) =2πσ^2 ∫_0 ^∞ e^(−t) dt =2πσ^2 [e^(−t) ]_∞ ^0 =2πσ^2 ⇒I=∣σ∣(√(2π))
$${I}=\int_{−\infty} ^{\infty} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dx}\:>\mathrm{0} \\ $$$${I}=\int_{−\infty} ^{\infty} {e}^{−\frac{{y}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dy} \\ $$$${I}^{\mathrm{2}} =\int_{−\infty} ^{\infty} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dx}\int_{−\infty} ^{\infty} {e}^{−\frac{{y}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dy} \\ $$$$\:\:=\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {e}^{−\frac{{y}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dydx} \\ $$$$\:\:=\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} {e}^{−\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dydx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\infty} {e}^{−\frac{{r}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {rdrd}\theta \\ $$$$\:\:=\mathrm{2}\pi\sigma^{\mathrm{2}} \int_{\mathrm{0}} ^{\infty} {e}^{−\frac{{r}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {d}\left(\frac{{r}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }\right) \\ $$$$\:\:=\mathrm{2}\pi\sigma^{\mathrm{2}} \int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {dt} \\ $$$$\:\:=\mathrm{2}\pi\sigma^{\mathrm{2}} \left[{e}^{−{t}} \right]_{\infty} ^{\mathrm{0}} \\ $$$$\:\:=\mathrm{2}\pi\sigma^{\mathrm{2}} \\ $$$$\Rightarrow{I}=\mid\sigma\mid\sqrt{\mathrm{2}\pi} \\ $$
Commented by Abod last updated on 30/Mar/26
all is correct :)
$$\left.{all}\:{is}\:{correct}\::\right) \\ $$
Commented by Ghisom_ last updated on 30/Mar/26
it′s not correct because it doesn′t answer your question but a different one including a “−” where there is none in your question.
$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{correct}\:\mathrm{because}\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{answer} \\ $$$$\mathrm{your}\:\mathrm{question}\:\mathrm{but}\:\mathrm{a}\:\mathrm{different}\:\mathrm{one}\:\mathrm{including} \\ $$$$\mathrm{a}\:“−''\:\mathrm{where}\:\mathrm{there}\:\mathrm{is}\:\mathrm{none}\:\mathrm{in}\:\mathrm{your}\:\mathrm{question}. \\ $$
Answered by Jyrgen last updated on 01/Apr/26
∫_(−∞) ^∞ e^(x^2 /(2σ^2 )) dx≠σ(√(2π))
$$\underset{−\infty} {\overset{\infty} {\int}}\mathrm{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}\sigma^{\mathrm{2}} }} {dx}\neq\sigma\sqrt{\mathrm{2}\pi} \\ $$

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