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

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).

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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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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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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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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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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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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squareroot of Pi ? ( Error Function)

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

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You can also look up this pdf

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

lower limit is negative infinity..

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is it 0

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

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