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December 28th, 2017, 01:01 AM   #1
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Quantum integral

Post your own way of solution
$\displaystyle \int _{-\infty}^{\infty} e^{-X^2 }dX $
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December 28th, 2017, 01:40 AM   #2
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$\sqrt{\pi}$ from memory.
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December 28th, 2017, 02:38 AM   #3
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First use $\displaystyle \int _C y_sds = \int _C y_z dz$
so $\displaystyle \int _{-\infty}^{\infty} e^{-s^2} dx \int _{-\infty}^{\infty} e^{-z^2} dz =\int _{-\infty}^{\infty} \int _{-\infty}^{\infty} e^{-s^2-z^2 } ds dz =(\int _{-\infty}^{\infty} e^{-x^2} dx)^2 $
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December 28th, 2017, 04:22 AM   #4
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Standard method in any text book:
$\int_{-\infty}^\infty e^{-x^2}dx= 2\int_0^\infty e^{-x^2}dx$

Let $I= \int_0^\infty e^{-x^2}dx$. Then $I^2= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy\right)$$= \int_0^\infty \int_0^\infty e^{-x^2- y^2}dydx$.

Now convert to polar coordinates. $-x^2- y^2= -r^2$ and $dydx= r drd\theta$. To cover the first quadrant, r goes from 0 to infinity and $\theta$ goes from 0 to $\frac{\pi}{2}$:
$I^2= \int_0^\infty \int_0^{\pi/2} e^{-r^2} r dr d\theta= \left(\int_0^{\pi/2}d\theta\right)\left(\int_0^\infty e^{-r^2} r dr\right)= \frac{\pi}{2}\int_0^\infty e^{-r^2} r dr$

And now that "r" in the integrand allows us to use the substitution $u= r^2$, $du= 2rdr$, $dr= \frac{1}{2}du$. When r= 0, u= 0, when $r= \infty$, $u= \infty$.

$I^2= \frac{\pi}{4}\int_0^\infty e^{-u}du= \frac{\pi}{4}\left[-e^{-u}\right]_0^\infty$. $e^0= 1$ and as u goes to $\infty$, $e^{-u}$ goes to 0.

$I^2= \frac{\pi}{4}\left[0- (-1)\right]= \frac{\pi}{4}$ so $I= \frac{\sqrt{\pi}}{2}$.

$\int_{-\infty}^\infty e^{-x^2}dx= 2I= \sqrt{\pi}$.
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