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 December 28th, 2017, 01:01 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 831 Thanks: 113 Math Focus: Elementary Math Quantum integral Post your own way of solution $\displaystyle \int _{-\infty}^{\infty} e^{-X^2 }dX$ December 28th, 2017, 01:40 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,847 Thanks: 661 Math Focus: Yet to find out. $\sqrt{\pi}$ from memory. December 28th, 2017, 02:38 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 831 Thanks: 113 Math Focus: Elementary Math First use $\displaystyle \int _C y_sds = \int _C y_z dz$ so $\displaystyle \int _{-\infty}^{\infty} e^{-s^2} dx \int _{-\infty}^{\infty} e^{-z^2} dz =\int _{-\infty}^{\infty} \int _{-\infty}^{\infty} e^{-s^2-z^2 } ds dz =(\int _{-\infty}^{\infty} e^{-x^2} dx)^2$ December 28th, 2017, 04:22 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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 $I= \frac{\sqrt{\pi}}{2}$. $\int_{-\infty}^\infty e^{-x^2}dx= 2I= \sqrt{\pi}$. Tags integral, quantum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Loren Computer Science 4 December 1st, 2016 08:19 AM VisionaryLen Physics 10 September 12th, 2016 05:03 AM BenFRayfield Physics 2 February 6th, 2014 12:45 PM johnny Computer Science 3 July 31st, 2007 07:35 AM

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