How does integrating this work?

[vorcil]

question:

Consider the gaussian distribution:

p(x) = Ae^(-\lambda (x-a)^2)

(a) use the equation, 1={\int_{-\infty}^{\infty}} p(x)dx

(b) find <x>, <x^2> and \sigma

------------------------------------------------------

a) if i take (x-a) to be u,

1=\int_{-\infty}^\infty Ae^(-\lambda(x-a)^2)dx
=
\int_{-\infty}^\infty Ae^(-\lambda(u)^2)dx (not sure if this is right, when i substitute in u, is dx, supposed to be replaced by du?

why? because of the integration limits? idk understand why please explain,

- after i get told why that occurs,

i need help integrating this
{\int_{-\infty}^{\infty}} Ae^(-\lambda(u)^2)du

 

[Char. Limit]

question:

Consider the gaussian distribution:

p(x) = Ae^(-\lambda (x-a)^2)

(a) use the equation, 1={\int_{-\infty}^{\infty}} p(x)dx

(b) find <x>, <x^2> and \sigma

------------------------------------------------------

a) if i take (x-a) to be u,

1=\int_{-\infty}^\infty Ae^(-\lambda(x-a)^2)dx
=
\int_{-\infty}^\infty Ae^(-\lambda(u)^2)dx (not sure if this is right, when i substitute in u, is dx, supposed to be replaced by du?

why? because of the integration limits? idk understand why please explain,

- after i get told why that occurs,

i need help integrating this
{\int_{-\infty}^{\infty}} Ae^(-\lambda(u)^2)du

Assuming that a is a constant...

Well, start with your replacement...

u=x-a

then differentiate it with respect to x.

\frac{du}{dx}=1

And multiply both sides by dx (you can do this).

du=dx

So du=dx!

 

[vorcil]

my first question is how to solve for A and a

I've been given that i need to figure out A first, before a

helllp

 

[Char. Limit]

my first question is how to solve for A and a

I've been given that i need to figure out A first, before a

helllp

Are you given a value for lambda? That would really help if you were trying to solve the integral...

However, first thing I'd do, assuming A is constant, is factor it out of the integral. After all, the resulting integral is solvable, and would be even more solvable if we had a value lambda (and I hope it's a square number too).

 

[HallsofIvy]

By the way, with LaTex, put everything you want grouped together, such as an exponent, in { }.

[ tex ]e^{-\lambda (x- a)^2} [ tex ]
gives
e^{-\lamda (x- a)^2}

 

[PhaseShifter]

question:i need help integrating this
{\int_{-\infty}^{\infty}} Ae^{-\lambda(u)^2}du

Don't worry about integrating that one.

Instead, try integrating
{\int_{-\infty}^{\infty}} (u+a)Ae^{-\lambda(u)^2}du
and
{\int_{-\infty}^{\infty}} (u+a)^{2}Ae^{-\lambda(u)^2}du
If you integrate by parts (and you may need to do this more than once, to get rid of the "u+a" coefficients), you'll get (some expression)*(the integral you can't do, which is equal to 1)