My Math Forum Quantum integral

 Calculus Calculus Math Forum

 December 28th, 2017, 12:01 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27 Quantum integral Post your own way of solution $\displaystyle \int _{-\infty}^{\infty} e^{-X^2 }dX$
 December 28th, 2017, 12:40 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,636 Thanks: 569 Math Focus: Yet to find out. $\sqrt{\pi}$ from memory.
 December 28th, 2017, 01:38 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27 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, 03:22 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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 Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Loren Computer Science 4 December 1st, 2016 07:19 AM VisionaryLen Physics 10 September 12th, 2016 04:03 AM BenFRayfield Physics 2 February 6th, 2014 11:45 AM johnny Computer Science 3 July 31st, 2007 06:35 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top