Have you ever wondered why
? That is, why
? Let
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I=∫∞−∞e−x22dxI⋅I=(∫∞−∞e−x22dx)(∫∞−∞e−y22dy)I2=∫∞−∞∫∞−∞e−12(x2+y2)dxdy |
The double integral is easier to resolve in polar coordinate. In general,
In this particular case,
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\begin{align*}
g(x,y) &= e^{-\frac{1}{2}(x^2+y^2)} \\
\mathbf{T}(r,\theta) &= (r\cos{\theta}, r\sin{\theta}) = (x, y) \\
g(\mathbf{T}(r,\theta)) &= e^{-\frac{1}{2}((r\cos{\theta})^2+(r\sin{\theta})^2)} = e^{-\frac{1}{2}r^2} \\
D \mathbf{T}(r,\theta) &= \begin{bmatrix}\frac{d}{dr}r\cos{\theta} & \frac{d}{d\theta}r\cos{\theta}\\\frac{d}{dr}r\sin{\theta} & \frac{d}{d\theta}r\sin{\theta}\end{bmatrix}
= \begin{bmatrix}\cos{\theta} & -r\sin{\theta}\\\sin{\theta} & r\cos{\theta}\end{bmatrix} \\
\big| \det D \mathbf{T}(r,\theta)\,\big| &= \begin{vmatrix}\cos{\theta} & -r\sin{\theta}\\\sin{\theta} & r\cos{\theta}\end{vmatrix} = r \\
\end{align*}
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Therefore
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\begin{aligned}
\mathbf{I}^2 &= \int_0^{2\pi} \int_0^{\infty} e^{-\frac{1}{2}r^2} r\,dr\,d\theta \\
\end{aligned}
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Let
u = \frac{1}{2}r^2, so
du = r\,dr
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\begin{aligned}
\mathbf{I}^2 &= \int_0^{2\pi} \int_0^{\infty} e^{-u} \,du\,d\theta = \int_0^{2\pi} -e^{-u} \bigg|_{u=0}^{\infty} \,d\theta = \int_0^{2\pi}d\theta = 2\pi \\
\therefore \color{blue}{\mathbf{I}} &\color{blue}{= \sqrt{2\pi}} \\
\end{aligned}
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\Box
References:
# posted by rot13(Unafba Pune) @ 6:30 PM
