Réitigh I _ { 1 } = ∫ (from 0 to infty) of int_0^inftye^-x^2e^-y^2 wrt xdy

Réitigh do I_1.

I_{1}=\int _{0}^{\infty}\int _{0}^{\infty}e^{-x^{2}-y^{2}}\mathrm{d}x\mathrm{d}y

Sann I_1

I_{1}≔\int _{0}^{\infty}\int _{0}^{\infty}e^{-x^{2}-y^{2}}\mathrm{d}x\mathrm{d}y

Tráth na gCeist

Integration

5 fadhbanna cosúil le:

I _ { 1 } = \int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x ^ { 2 } } e ^ { - y ^ { 2 } } d x d y

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https://math.stackexchange.com/questions/830755/evaluate-int-0-infty-int-0-inftye-x2-2xy-y2-dx-dy

\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}} ...

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https://math.stackexchange.com/questions/2016227/evaluating-the-improper-integral-int-0-infty-int-0-infty-e-2x7y-matho

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https://math.stackexchange.com/questions/2622068/how-to-calculate-iint-mathbbr2-exp-left-x2-y2ixy-ix-iy-right-dx

This is a Gaussian double-integral I ~:=~\iint_{\mathbb{R}^2} \! dx~dy~ e^{-S(x,y)} ,\tag{1} where S(x,y)~:=~x^2+y^2-ixy+ix+iy. \tag{2} One method is to change coordinates x^{\pm}=\frac{x\pm y}{\sqrt{2}} ...

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https://math.stackexchange.com/q/2662355

Observe that we can factor the joint density as f(x,y)=e^{-\lambda x}e^{-\lambda y}; \quad (x,y\geq 0)\tag{1}. Let I=\int_0^\infty\int_0^\infty f(x,y)\,dx\,dy and note that I=\int_0^\infty\int_0^\infty e^{-\lambda x}e^{-\lambda y}\, dx \,dy =\int_0^\infty e^{-\lambda y}\left[\int_0^\infty e^{-\lambda x}\, dx \right]\,dy =\left[\int_0^\infty e^{-\lambda x}\, dx \right]\left[\int_0^\infty e^{-\lambda y}\, dy \right] ...

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https://math.stackexchange.com/questions/3081376/need-help-understanding-why-int-0-infty-int-0-infty-x-ye-xe-y-dx

Note that \tag{*} \int_0^\infty\int_0^\infty |x-y|e^{-x}e^{-y} \, dx\, dy \\ = \int_0^\infty\int_y^\infty (x-y)e^{-x}e^{-y} \, dx\, dy + \int_0^\infty\int_0^y (y-x)e^{-x}e^{-y} \, dx\, dy The ...

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https://math.stackexchange.com/questions/968120/integral-of-exponent-iint-limits-0-infty-et-1-xt-2-y-y-dy-dx

Let's integrate with respect to x first. \int\limits_0^\infty \int\limits_0^\infty e^{(t_1x+t_2y-y) }\ dy\ dx = \int\limits_0^\infty \left( \int\limits_0^\infty e^{(t_1x+t_2y-y) }\ dx\right)dy \\ = \int\limits_0^\infty \left(\lim_{N \to \infty} \frac{1}{t_1}\left[e^{(t_1x+t_2y-y) } \right]\Big|_0^N \right)dy ...

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