Question Number 224197 by klipto last updated on 24/Aug/25
$$\int_{\mathrm{0}} ^{\infty} \left(\boldsymbol{\mathrm{e}}^{−\boldsymbol{\varphi\mathrm{x}}^{\mathrm{2}} } +\boldsymbol{\mathrm{e}}^{−\boldsymbol{\delta\mathrm{x}}^{\mathrm{2}} } +\boldsymbol{\mathrm{e}}^{−\boldsymbol{\gamma\mathrm{x}}^{\mathrm{2}} } \right) \\ $$$$\boldsymbol{\gamma}−\boldsymbol{\mathrm{euler}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{mascheroni}}\:\boldsymbol{\mathrm{constant}} \\ $$$$\boldsymbol{\varphi}−\boldsymbol{\mathrm{golden}}\:\boldsymbol{\mathrm{ratio}} \\ $$$$\boldsymbol{\delta}−\boldsymbol{\mathrm{silver}}\:\boldsymbol{\mathrm{ratio}} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}}\spadesuit \\ $$
Answered by Ghisom_ last updated on 24/Aug/25
![∫_0 ^∞ e^(−px^2 ) dx dx − completes the syntax & tells us what to integrate p − number ∈R^+ ∫_0 ^∞ e^(−px^2 ) dx= [x=(t/( (√p)))] =(1/( (√p)))∫_0 ^∞ e^(−t^2 ) dt=((√π)/(2(√p))) ⇒ answer is ((√π)/2)((1/( (√ϕ)))+(1/( (√δ)))+(1/( (√γ)))) (c) 25.8.2025 Ghisom _♠^♦ ★_♥ ^♣](https://www.tinkutara.com/question/Q224199.png)
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{px}^{\mathrm{2}} } {dx} \\ $$$$\:\:\:\:\:{dx}\:−\:\mathrm{completes}\:\mathrm{the}\:\mathrm{syntax}\:\&\:\mathrm{tells}\:\mathrm{us} \\ $$$$\:\:\:\:\:\mathrm{what}\:\mathrm{to}\:\mathrm{integrate} \\ $$$$\:\:\:\:\:{p}\:−\:\mathrm{number}\:\in\mathbb{R}^{+} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{px}^{\mathrm{2}} } {dx}= \\ $$$$\:\:\:\:\:\left[{x}=\frac{{t}}{\:\sqrt{{p}}}\right] \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{p}}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{p}}} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\varphi}}+\frac{\mathrm{1}}{\:\sqrt{\delta}}+\frac{\mathrm{1}}{\:\sqrt{\gamma}}\right) \\ $$$$ \\ $$$$\left(\mathrm{c}\right)\:\mathrm{25}.\mathrm{8}.\mathrm{2025}\:\mathrm{Ghisom}\:_{\spadesuit} ^{\diamondsuit} \bigstar_{\heartsuit} ^{\clubsuit} \\ $$
Commented by fantastic last updated on 24/Aug/25
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Answered by fkwow344 last updated on 25/Aug/25
![1. ∫_0 ^( ∞) e^(−Kρ^2 ) dρ=(1/2)(√(π/K)) cus ∫_0 ^( ∞) e^(−Kx^2 ) dx∙∫_0 ^( ∞) e^(−Ky^2 ) dy=∫∫_([0,∞)×[0,∞)) e^(−K(x^2 +y^2 )) dx∧dy (Fubini′s Theorem), refer to the measure theory textbook for an accurate method of proof. Product space S^∗ =[0,∞)×[0,∞)∈R^2 ∫_0 ^( ∞) e^(−Kx^2 ) dx×∫_0 ^( ∞) e^(−Ky^2 ) dy=∫∫_S^∗ e^(−K(x^2 +y^2 )) dx∧dy Jacobian ∣J∣=∣∣((∂(x,y))/(∂(r,θ)))∣∣ dr∧dθ let′s consider radius R=r circle x^2 +y^2 =r^2 →^(parametrization) (((x(t))),((y(t))) )= (((r∙cos(t))),((r∙sin(t))) ) ∣J∣= determinant (((∂x/∂r),(∂x/∂θ)),((∂y/∂r),(∂y/∂θ)))dr∧dθ= determinant (((cos(t)),(−rsin(t))),((sin(t)),( rcos(t))))dr∧dθ ∣J∣=r∙dr∧dθ ∫∫_S^∗ r∙e^(−Kr^2 ) drdθ Since the integration section we want to integrate is from 0 to ∞ So, the integration section of r is from 0 to ∞. If it is difficult to understand imagine a circle with radius ′′r′′ and the value of ′′r′′ grows to infinity. ...... Now that the integral part of θ is an angle, imagine a 3-dimensional space. Obviously the integral part of x is from 0 to ∞ and the integral part of y is from 0 to ∞, Now lets imagine the xy-plane in a 3-dimesional space. Then of course θ has an inteval from 0 to (π/2) ∴ ∫∫_( S^∗ ) =∫_0 ^( (π/2)) ∫_0 ^( ∞) re^(−Kr^2 ) drdθ ∴ (∫_0 ^( ∞) e^(−Kρ^2 ) dρ)^2 =∫_0 ^( (π/2)) ∫_0 ^( ∞) re^(−Kr^2 ) drdθ ∫_0 ^( (π/2)) [−(1/(2K))e^(−Kr^2 ) ]_(r=0) ^(r=∞) dθ=∫_0 ^( (π/2)) (1/(2K)) dθ=(π/(4K)) ∴ ∫_0 ^( ∞) e^(−Kρ^2 ) dρ=(1/2)(√(π/K)) . ans.∫_0 ^( ∞) (e^(−ϕx^2 ) +e^(−δx^2 ) +e^(−γx^2 ) )dx=(1/2)((√(π/ϕ))+(√(π/δ))+(√(π/γ))) ϕ,δ,γ≠0](https://www.tinkutara.com/question/Q224202.png)
$$\mathrm{1}. \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:{e}^{−{K}\rho^{\mathrm{2}} } \mathrm{d}\rho=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{K}}}\:\:\mathrm{cus} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{Kx}^{\mathrm{2}} } \mathrm{d}{x}\centerdot\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{Ky}^{\mathrm{2}} } \mathrm{d}{y}=\int\int_{\left[\mathrm{0},\infty\right)×\left[\mathrm{0},\infty\right)} \:\:{e}^{−{K}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \:\mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$$$\left(\mathrm{Fubini}'\mathrm{s}\:\mathrm{Theorem}\right), \\ $$$$\mathrm{refer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{measure}\:\mathrm{theory}\:\mathrm{textbook}\:\mathrm{for}\:\mathrm{an} \\ $$$$\mathrm{accurate}\:\mathrm{method}\:\mathrm{of}\:\mathrm{proof}. \\ $$$$\mathrm{Product}\:\mathrm{space}\:\mathcal{S}^{\ast} =\left[\mathrm{0},\infty\right)×\left[\mathrm{0},\infty\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{Kx}^{\mathrm{2}} } \mathrm{d}{x}×\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{Ky}^{\mathrm{2}} } \mathrm{d}{y}=\int\int_{\mathcal{S}^{\ast} } \:{e}^{−{K}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \:\mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$$$\mathrm{Jacobian}\:\mid\boldsymbol{\mathrm{J}}\mid=\mid\mid\frac{\partial\left({x},{y}\right)}{\partial\left({r},\theta\right)}\mid\mid\:\mathrm{d}{r}\wedge\mathrm{d}\theta \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{consider}\:\mathrm{radius}\:{R}={r}\:\mathrm{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\overset{\mathrm{parametrization}} {\rightarrow}\:\begin{pmatrix}{{x}\left({t}\right)}\\{{y}\left({t}\right)}\end{pmatrix}=\begin{pmatrix}{{r}\centerdot\mathrm{cos}\left({t}\right)}\\{{r}\centerdot\mathrm{sin}\left({t}\right)}\end{pmatrix} \\ $$$$\mid\boldsymbol{\mathrm{J}}\mid=\begin{vmatrix}{\frac{\partial{x}}{\partial{r}}}&{\frac{\partial{x}}{\partial\theta}}\\{\frac{\partial{y}}{\partial{r}}}&{\frac{\partial{y}}{\partial\theta}}\end{vmatrix}\mathrm{d}{r}\wedge\mathrm{d}\theta=\begin{vmatrix}{\mathrm{cos}\left({t}\right)}&{−{r}\mathrm{sin}\left({t}\right)}\\{\mathrm{sin}\left({t}\right)}&{\:\:\:\:{r}\mathrm{cos}\left({t}\right)}\end{vmatrix}\mathrm{d}{r}\wedge\mathrm{d}\theta \\ $$$$\mid\boldsymbol{\mathrm{J}}\mid={r}\centerdot\mathrm{d}{r}\wedge\mathrm{d}\theta \\ $$$$\int\int_{\mathcal{S}^{\ast} } \:{r}\centerdot{e}^{−{Kr}^{\mathrm{2}} } \:\mathrm{d}{r}\mathrm{d}\theta\:\:\: \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{integration}\:\mathrm{section}\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{integrate}\:\mathrm{is} \\ $$$$\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\infty \\ $$$$\mathrm{So},\:\mathrm{the}\:\mathrm{integration}\:\mathrm{section}\:\mathrm{of}\:{r}\:\mathrm{is}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\infty. \\ $$$$\mathrm{If}\:\mathrm{it}\:\mathrm{is}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{imagine}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{with}\:\mathrm{radius}\:''{r}''\:\mathrm{and}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:''{r}''\:\mathrm{grows}\:\mathrm{to}\:\mathrm{infinity}. \\ $$$$…… \\ $$$$\mathrm{Now}\:\mathrm{that}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{part}\:\mathrm{of}\:\theta\:\mathrm{is}\:\mathrm{an}\:\mathrm{angle}, \\ $$$$\mathrm{imagine}\:\mathrm{a}\:\mathrm{3}-\mathrm{dimensional}\:\mathrm{space}. \\ $$$$\mathrm{Obviously}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{part}\:\mathrm{of}\: \\ $$$${x}\:\mathrm{is}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\infty\:\mathrm{and}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{part}\:\mathrm{of}\:{y}\:\mathrm{is}\:\mathrm{from}\: \\ $$$$\mathrm{0}\:\mathrm{to}\:\infty,\:\mathrm{Now}\:\mathrm{lets}\:\mathrm{imagine}\:\mathrm{the}\:\mathrm{xy}-\mathrm{plane}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{3}-\mathrm{dimesional}\:\mathrm{space}. \\ $$$$\mathrm{Then}\:\mathrm{of}\:\mathrm{course}\:\theta\:\mathrm{has}\:\mathrm{an}\:\mathrm{inteval}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\:\frac{\pi}{\mathrm{2}} \\ $$$$\therefore\:\int\int_{\:\mathcal{S}^{\ast} } =\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\infty} \:\:{re}^{−{Kr}^{\mathrm{2}} } \:\:\mathrm{d}{r}\mathrm{d}\theta \\ $$$$\therefore\:\left(\int_{\mathrm{0}} ^{\:\infty} \:\:{e}^{−{K}\rho^{\mathrm{2}} } \mathrm{d}\rho\right)^{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\infty} \:\:{re}^{−{Kr}^{\mathrm{2}} } \:\mathrm{d}{r}\mathrm{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:\left[−\frac{\mathrm{1}}{\mathrm{2}{K}}{e}^{−{Kr}^{\mathrm{2}} } \right]_{{r}=\mathrm{0}} ^{{r}=\infty} \:\mathrm{d}\theta=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}}{\mathrm{2}{K}}\:\mathrm{d}\theta=\frac{\pi}{\mathrm{4}{K}} \\ $$$$\therefore\:\:\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{K}\rho^{\mathrm{2}} } \:\mathrm{d}\rho=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{K}}}\:. \\ $$$$\mathrm{ans}.\int_{\mathrm{0}} ^{\:\infty} \:\:\left({e}^{−\varphi{x}^{\mathrm{2}} } +{e}^{−\delta{x}^{\mathrm{2}} } +{e}^{−\gamma{x}^{\mathrm{2}} } \right)\mathrm{d}{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{\pi}{\varphi}}+\sqrt{\frac{\pi}{\delta}}+\sqrt{\frac{\pi}{\gamma}}\right) \\ $$$$\varphi,\delta,\gamma\neq\mathrm{0} \\ $$