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

Whakaoti mō I_1

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

Tautapa I_1

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

Pātaitai

Integration

5 raruraru e ōrite ana ki:

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/2016227/evaluating-the-improper-integral-int-0-infty-int-0-infty-e-2x7y-matho

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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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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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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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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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