Question

Prove that b is irrelevant in this

Answer 1

\(\large{ \int\limits _{-\infty }^{\infty }e^{-a(x+b)^{2}} ~ dx={\sqrt {\frac {\pi }{a}}}}\)

it's a dumb question :)

Answer 2

I'm a little pressed for time but i believe it has something to go with u-substitution/series expansion, you can redefine the index so that the entire exponent is equal to a constant

Answer 3

am I on the right track on this? this question has been bugging me for a while

Answer 4

Yes, I think you are totally right

I shoulda said this as clarification,, this is the Gaussian integral which is very commonly presented as: \( \int\limits _{-\infty }^{\infty } ~ dx ~ e^{-x^{2}} ={\sqrt \pi}\).

And there's now \(I(x, a, b) = \int\limits _{x = -\infty }^{\infty } ~ dx ~ e^{-a(x+b)^2}\)

I thought there was a Feynman wheeze here - differentiation under the integral wrt b, ie \(\frac{\partial I}{\partial b}\), but it's not working...for me at least. So if we go with the sub idea: \( X^2 = a(x+ b)^2 \implies 2X ~ dX = 2 a (x + b) ~ dx \), we have:

\(I = \int\limits_{-\infty }^{\infty } \dfrac{2X ~ dX}{ 2a (x + b)} ~ e^{- X^{2}} = \int\limits_{-\infty }^{\infty } ~ dX ~ \dfrac{X }{ \sqrt{a} X} ~ e^{X^{2}} = \frac{1}{\sqrt{a}} \int\limits_{-\infty }^{\infty } ~ dX ~ e^{X^{2}} = \sqrt{\dfrac{\pi}{a}}\)


The b terms mysteriously disappears :(


Which is because x is going to infinity, so it's understandable

but it is seriously impressive that the math gets that ...and the scaling effect of a

Answer 5

correction:

\(I = \int\limits_{-\infty }^{\infty } \dfrac{2X ~ dX}{ 2a (x + b)} ~ e^{- X^{2}} = \int\limits_{-\infty }^{\infty } ~ dX ~ \dfrac{X }{ \sqrt{a} X} ~ e^{-X^{2}} = \frac{1}{\sqrt{a}} \int\limits_{-\infty }^{\infty } ~ dX ~ e^{-X^{2}} = \sqrt{\dfrac{\pi}{a}}\)