How Do You Solve the Infinite Integral in Quantum Mechanics Demystified?

[Pollywoggy]

I found an example on p. 14 of Quantum Mechanics Demystified (McMahon) which I don't understand. The part I do not understand is how they got the part on the right side of the equal sign from the part on the left.

$$\int^\infty_0 e^{-2x^2}\,dx =\sqrt{\pi/8}$$

I looked at several books but could not find an explanation that applies here.

thanks

 

[Rainbow Child]

It is based on

$$I=\int_0^{+\infty}e^{-u^2}\,d\,u=\frac{\sqrt{\pi}}{2}$$

which is proved by several methods. If you know double integrals there is an easy way to show it.

 

[Pollywoggy]

It is based on

$$I=\int_0^{+\infty}e^{-u^2}\,d\,u=\frac{\sqrt{\pi}}{2}$$

which is proved by several methods. If you know double integrals there is an easy way to show it.

Thanks, you have answered my question. I saw a *similar* equation in another book explained with double integrals and it was stated that the equation could not be solved by the methods shown up to that point.

 

[Kaan]

It will basically work if you know how to switch to polar cordinates.