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$ \displaystyle\int_{-\infty }^\infty e^{-\frac{1}{2} x^2} \; dx $ and $ \displaystyle\int_{-\infty }^\infty x^{2}e^{-\frac{1}{2}x^2} \; dx $

how i compute these integrals via Gauss Integral?

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An interesting alternate way to evaluate the second integral $\displaystyle\int_{-\infty }^\infty x^{2}e^{-\frac{1}{2}x^2} \; dx$ 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 $ \forall \alpha > 0 \in \mathbb{R} $,

$\displaystyle\int_{-\infty }^\infty e^{-\alpha x^2} \; dx = {\dfrac{\sqrt{\pi}}{\sqrt{\alpha}}} $.

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

$ \displaystyle\int_{-\infty }^\infty -x^2 e^{-\alpha x^2} \; dx$ = $-\dfrac{\sqrt{\pi}}{2\alpha\sqrt{\alpha}}$

Factoring out the $-1$ from both sides and plugging in $\alpha = 1/2$ gives us our answer:

$\displaystyle\int_{-\infty }^\infty x^{2}e^{-\frac{1}{2}x^2} \; dx = \dfrac{\sqrt{\pi}}{\sqrt{1/2}} = \sqrt{2\pi}.$

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Make the substitution $x = \sqrt{2}\ t$ in both, and use integration by parts in the second, by writing $$2t^2 e^{-t^2} = -t \times \frac{d e^{-t^2}}{dt}$$

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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 $$ \int e^{-w^2}\;dw=\sqrt{\pi}, $$ then you can write $x=\sqrt{2}\;w$ and $dx=\sqrt{2}\; dw$. As $w$ goes from $-\infty$ to $\infty$, so does $x$. So $$ \int_{-\infty}^\infty e^{-\frac 12 x^2}\; dx = \int_{-\infty}^\infty e^{-w^2} \sqrt{2} \; dw = \sqrt{2\pi\;{}}. $$

For $\displaystyle\int_{-\infty}^\infty e^{-\frac 1 2 x^2} \; dx,$ begin by observing that it's an even function integrated over an interval that is symmetric about $0$, so it's equal to $$ 2 \int_0^\infty x^2 e^{-\frac 1 2 x^2} \; dx. $$ Then substitute $u=\frac 1 2 x^2$, $du = x\;dx = \sqrt{2u} \; dx$, so $\dfrac{du}{\sqrt{2u}} =dx$. The integral becomes $$ 2\int_0^\infty 2u e^{-u} \frac{du}{\sqrt{2u}} = \frac{4}{\sqrt{2}} \int_0^\infty u^{\frac 3 2 - 1} e^{-u}\;du = 2\sqrt{2}\Gamma\left(\frac 3 2 \right) = 2\sqrt{2}\cdot \frac 1 2 \Gamma\left( \frac 1 2 \right) = \sqrt{2\pi}. $$

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