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September 30, 2014

September 30, 2014

Posted by **mathstudent** on Monday, January 14, 2008 at 1:46pm.

Definite Integral of e^(-x^2) dx over [0,infinity] = sqrt(pi)/2

Solve for

Definite Integral of e^(-ax^2) dx over [-infinity,infinity]

I don't know how to approach the new "a" term. I can't use u-substitution, integration by parts, partial fractions, or trig substitution. How do I do this?

- calculus -
**Count Iblis**, Monday, January 14, 2008 at 6:37pmSubstitute x = t/sqrt(a). The integral then becomes:

a^(-1/2)Integral of e^(-t^2) dt over [-infinity,infinity] =

2 a^(-1/2)Integral of e^(-t^2) dt over [0,infinity] = sqrt(pi/a)

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