∫−∞∞e−12x2dx and ∫−∞∞x2e−12x2dx how i compute these integrals via Gauss Integral?

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∫−∞∞e−12x2dx and ∫−∞∞x2e−12x2dx  how i compute these integrals via Gauss Integral?

Jonathan Burroughs · Accepted Answer

Make the substitution x=2t in both, and use integration by parts in the second, by writing  2t2e−t2=−t×de−t2dt

Stuart Rountree · Accepted Answer

An interesting alternate way to evaluate the second integral ∫−∞∞x2e−12x2dx using the results form the first one is to utilize a technique called &#039;&#039;differentiation under the integral sign&#039;&#039;.From a substitution and our first integral&#039;s value in hand, we can conclude that ∀a&amp;gt;0∈R∫−∞∞e−ax2dx=πaNow we differentiate both sides with respect to a (under the integral sign on the left), and we obtain∫−∞∞−x2e−ax2dx=−π2aaFactoring out the −1 from both sides and plugging in a=12 gives us our answer:∫−∞∞x2e−12x2dx=π12=2π

karton · Accepted Answer

The one on the left is what I usually think of when I hear the term &#039;&#039;Gaussian integral&#039;&#039;. But some people, maybe especially physicists, omit the fraction 1/2. Including the 1/2 makes sense because then the variance of the probability distribution that you get when you normalize the function to be a probability density is 1. If you know that ∫e−w2dw=π then you can write x=2w and dx=2dw. As w goes from −∞ to ∞, so does x. So ∫−∞∞e−12x2dx=∫−∞∞e−w22dw=2π For ∫−∞∞e−12x2dx begin by observing that it&#039;s an even function integrated over an interval that is symmetric about 0, so it&#039;s equal to 2∫0∞x2e−12x2dx Then substitute u=12x2, du=x dx=2u, so du2u=dx. The integral becomes 2∫0∞2ue−udu2u=42∫0∞u32−1e−udu=22T(32)=22⋅12(12)=2π

\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}dx and \int_{-infty}^\infty x^2e^{-\frac{1}{2}x^2}dx how i compute these integrals via

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New Questions in Calculus 1