What is the integral of this Gaussian distribution?

[Boltzman Oscillation]

Homework Statement

Find A in

p(x) = Aexp(-λ(x-a)^2)
by using the equation 1 = ∫ p(x)dx

Homework Equations

1 = ∫p(x)dx

The Attempt at a Solution

I expand the power of the exponential and then extract the constant exponential to get:

Aexp(λa^2) ∫exp(-λx^2)exp(2aλx)dx

I don't know how to proceed now.

 

[Dr Transport]

I'd just let $\alpha = x - a$ and integrate it out. $A$ is the normalization constant. it is a pretty simple analytic integral.

 

[Hiero]

it is a pretty simple analytic integral.

Pretty simple if you know the trick! But the trick is not obvious if you’ve never seen it before.

@Boltzman Oscillation
I would not expand the exponent like you did, I would instead make the substitution that Dr Transport suggested.

The trick behind solving the integral (the only way I know) is to multiply the integral with itself but with another dummy variable, then combine and convert to polar coordinates and integrate over the plane, then take the square root at the end.

I think it’s unclear in words so I’ll show the first few steps without the constants:
$$[\int_{-∞}^∞ e^{-x^2}dx]^2 = [\int_{-∞}^∞ e^{-x^2}dx]⋅ [\int_{-∞}^∞ e^{-y^2}dy] = \int_{-∞}^∞\int_{-∞}^∞ e^{-x^2}e^{-y^2}dxdy = \int_{-∞}^∞\int_{-∞}^∞ e^{-(x^2+y^2)}dxdy$$

 

[Ray Vickson]

Homework Statement

Find A in

p(x) = Aexp(-λ(x-a)^2)
by using the equation 1 = ∫ p(x)dx

Homework Equations

1 = ∫p(x)dx

The Attempt at a Solution

I expand the power of the exponential and then extract the constant exponential to get:

Aexp(λa^2) ∫exp(-λx^2)exp(2aλx)dx

I don't know how to proceed now.

Is this supposed to be an integral from $-\infty$ to $+\infty?$ If so, just put $u = \sqrt{\lambda}(x-a),$ to get a standard integral which is easily accessed on-line, or even in good old-fashioned books.

Note added in edit: I see that Hiero beat me to it in post #3, but that post did not appear on my screen until after I had pressed the enter key for my contribution.

 

[Boltzman Oscillation]

Pretty simple if you know the trick! But the trick is not obvious if you’ve never seen it before.

@Boltzman Oscillation
I would not expand the exponent like you did, I would instead make the substitution that Dr Transport suggested.

The trick behind solving the integral (the only way I know) is to multiply the integral with itself but with another dummy variable, then combine and convert to polar coordinates and integrate over the plane, then take the square root at the end.

I think it’s unclear in words so I’ll show the first few steps without the constants:
$$[\int_{-∞}^∞ e^{-x^2}dx]^2 = [\int_{-∞}^∞ e^{-x^2}dx]⋅ [\int_{-∞}^∞ e^{-y^2}dy] = \int_{-∞}^∞\int_{-∞}^∞ e^{-x^2}e^{-y^2}dxdy = \int_{-∞}^∞\int_{-∞}^∞ e^{-(x^2+y^2)}dxdy$$

Interesting ill use both yours and the previous method. Whats the name of the method you ised?

 

[Boltzman Oscillation]

Pretty simple if you know the trick! But the trick is not obvious if you’ve never seen it before.

@Boltzman Oscillation
I would not expand the exponent like you did, I would instead make the substitution that Dr Transport suggested.

The trick behind solving the integral (the only way I know) is to multiply the integral with itself but with another dummy variable, then combine and convert to polar coordinates and integrate over the plane, then take the square root at the end.

I think it’s unclear in words so I’ll show the first few steps without the constants:
$$[\int_{-∞}^∞ e^{-x^2}dx]^2 = [\int_{-∞}^∞ e^{-x^2}dx]⋅ [\int_{-∞}^∞ e^{-y^2}dy] = \int_{-∞}^∞\int_{-∞}^∞ e^{-x^2}e^{-y^2}dxdy = \int_{-∞}^∞\int_{-∞}^∞ e^{-(x^2+y^2)}dxdy$$

Alright so when i substitute polar coordnates for my eauation i have to find the integral of:

(-A/sqrt(λ))∫∫exp(r^2)dφdr
I can integrate dφ easily but how do i integrate dr?

 

[Ray Vickson]

Alright so when i substitute polar coordnates for my eauation i have to find the integral of:

(-A/sqrt(λ))∫∫exp(r^2)dφdr
I can integrate dφ easily but how do i integrate dr?

The polar-coordinate area element is $dA = r dr d\phi,$ NOT the $dr d\phi$ that you wrote. That is actually the whole point of the "trick".

 

[Hiero]

Interesting ill use both yours and the previous method. Whats the name of the method you ised?

I don’t think it has a name. In fact I’ve never seen it used for anything but this exact problem!

I will always remember it though because of how clever it is, it left a strong impression when I first saw it.

 

[haruspex]

According to https://www.york.ac.uk/depts/maths/histstat/normal_history.pdf, this trick was discovered by Poisson. The integral had already been evaluated by several more complicated methods, so it was known to take the form of a square root. I would guess that gave M. Poisson the idea to consider squaring the integral.