Sam Longoria

2022-01-03

how i compute these integrals via Gauss Integral?

Jonathan Burroughs

Beginner2022-01-04Added 37 answers

Make the substitution $x=\sqrt{2}t$ in both, and use integration by parts in the second, by writing

$2{t}^{2}{e}^{-{t}^{2}}=-t\times \frac{d{e}^{-{t}^{2}}}{dt}$

Stuart Rountree

Beginner2022-01-05Added 29 answers

An interesting alternate way to evaluate the second integral ${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{x}^{2}{e}^{-\frac{1}{2}{x}^{2}}dx$ using the results form the first one is to utilize a technique called ''differentiation under the integral sign''.

From a substitution and our first integral's value in hand, we can conclude that$\mathrm{\forall}a>0\in \mathbb{R}$

$\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-a{x}^{2}}dx=\frac{\sqrt{\pi}}{\sqrt{a}$

Now we differentiate both sides with respect to a (under the integral sign on the left), and we obtain

$\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}-{x}^{2}{e}^{-a{x}^{2}}dx=-\frac{\sqrt{\pi}}{2a\sqrt{a}$

Factoring out the$-1$ from both sides and plugging in $a=\frac{1}{2}$ gives us our answer:

$\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{x}^{2}{e}^{-\frac{1}{2}{x}^{2}}dx=\frac{\sqrt{\pi}}{\sqrt{\frac{1}{2}}}=\sqrt{2\pi$

From a substitution and our first integral's value in hand, we can conclude that

Now we differentiate both sides with respect to a (under the integral sign on the left), and we obtain

Factoring out the

karton

Expert2022-01-11Added 613 answers

The one on the left is what I usually think of when I hear the term ''Gaussian integral''. 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

then you can write

For

Then substitute

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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