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For: $\int_{-1}^{1}{\frac{1}{\sqrt{2\pi}} e^{\frac{-x^{2}}{2}}}dx $

A method I saw (but did not get the right result) was to set:

$I = \int_{-1}^{1}{\frac{1}{\sqrt{2\pi}} e^{\frac{-x^{2}}{2}}}dx $

and

$I = \int_{-1}^{1}{\frac{1}{\sqrt{2\pi}} e^{\frac{-y^{2}}{2}}}dx $

then by calculating $I^2 $ I could make use of polar coordinates since $x^2+y^2=r^2$ and $dxdy=rdxd\theta$

Any other process I may use instead? I just need to be pointed in the right direction

Marco Castro
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1 Answers1

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The polar coordinates are useful only to calculate the integral: $$\int_{-\infty}^{\infty}{\frac{1}{\sqrt{2\pi}} e^{\frac{-x^{2}}{2}}}dx $$ You cannot calculate the integral $I$ exactly, because the primitive of the gaussian can't be expressed in terms of elementary functions. You can read here:The easiest way to evaluate Gaussian integral

Luca.b
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