ya it's cool.you try to find the square of the quantity,with one variable x and the other y.then you integrate them simultaneously,a double integral.It's easy to evaluate once you transform it to polar coordinates.Nice One.Answer is root pi

As our function is an even function therefore split the limit of integration from -infinity to +infinity as 2 times 0 to infinity and use Gamma function by making a suitable substitution. It can also be done by converting the problem into polar form.

Although \(\int e^{-x^2}\,dx\) can't be expressed in terms of elementary functions, we can evaluate \(\int_{-\infty}^{+\infty} e^{-x^2}\,dx\). Doing so yields \(\sqrt{\pi}\) (for justification see http://en.wikipedia.org/wiki/Gaussian_integral#Computation).

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TopNewestya it's cool.you try to find the square of the quantity,with one variable x and the other y.then you integrate them simultaneously,a double integral.It's easy to evaluate once you transform it to polar coordinates.Nice One.Answer is root pi

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lower limit is negative infinity..

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is it 0

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You can also look up this pdf

www.stankova.net/statistics.2012/doubleintegration.pdfLog in to reply

another not-so-cool way is to use gamma function.use the substitution x=root u

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squareroot of Pi ? ( Error Function)

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As our function is an even function therefore split the limit of integration from -infinity to +infinity as 2 times 0 to infinity and use Gamma function by making a suitable substitution. It can also be done by converting the problem into polar form.

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This function is not integrable using the methods we learn till Undergraduate college level. I don't know about what we learn in college...

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Although \(\int e^{-x^2}\,dx\) can't be expressed in terms of elementary functions, we can evaluate \(\int_{-\infty}^{+\infty} e^{-x^2}\,dx\). Doing so yields \(\sqrt{\pi}\) (for justification see http://en.wikipedia.org/wiki/Gaussian_integral#Computation).

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It can be solved by GAMMA FUNCTION=integrate(0-infinity)e^-x.x^(n-1)dx for all x>=1 x belongs to Z+!!!

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I think ans is 0 .. I think we can solve it using integration by part

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