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[tex]\int_{-\infty}^{\infty}e^{-\pi*t^2}dt = 1[/tex]
Integral
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Tror det er løst flere ganger tidligere her på forumet:
[tex]I = \int_{-\infty}^{\infty}e^{-\pi x^2}dx[/tex]
[tex]I^2 = \int_{-\infty}^{\infty}e^{-\pi x^2}dx\int_{-\infty}^{\infty}e^{-\pi y^2}dy[/tex]
[tex]I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\pi( x^2+y^2)}dxdy[/tex]
[tex]x^2+y^2=r^2,\qquad dxdy = rdrd\theta[/tex]
[tex]I^2 = 2\pi\int_0^{\infty}re^{-\pi r^2}dr[/tex]
[tex]u = \pi r^2,\qquad du = 2\pi rdr \rightarrow dr = du/(2\pi r)[/tex]
[tex]I^2 = \int_0^{\infty}e^{-u}du = [1-0][/tex]
[tex]I = \sqrt{1} = 1[/tex]
[tex]I = \int_{-\infty}^{\infty}e^{-\pi x^2}dx[/tex]
[tex]I^2 = \int_{-\infty}^{\infty}e^{-\pi x^2}dx\int_{-\infty}^{\infty}e^{-\pi y^2}dy[/tex]
[tex]I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\pi( x^2+y^2)}dxdy[/tex]
[tex]x^2+y^2=r^2,\qquad dxdy = rdrd\theta[/tex]
[tex]I^2 = 2\pi\int_0^{\infty}re^{-\pi r^2}dr[/tex]
[tex]u = \pi r^2,\qquad du = 2\pi rdr \rightarrow dr = du/(2\pi r)[/tex]
[tex]I^2 = \int_0^{\infty}e^{-u}du = [1-0][/tex]
[tex]I = \sqrt{1} = 1[/tex]