# How do I calculate this integral?

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1. Dec 24, 2016

### bwest121

1. The problem statement, all variables and given/known data
We're given the gaussian distribution: $$\rho(x) = Ae^{-\lambda(x-a)^2}$$ where A, a, and $\lambda$ are positive real constants. We use the normalization condition $$\int_{-\infty}^{\infty} Ae^{-\lambda(x-a)^2} \,dx = 1$$ to find: $$A = \sqrt \frac \lambda \pi$$ What I want to find is $\langle x^2 \rangle$.

2. Relevant equations
$$\langle x^2 \rangle = \int_{-\infty}^{\infty} x^2Ae^{-\lambda(x-a)^2} \, dx$$
Hence, I need to solve the RHS integral.

3. The attempt at a solution
I'm really not sure how to solve this integral. Converting to polar seems to produce a very nasty integral. Integration by parts also produces very nasty integrals. I don't think this function is even or odd so the symmetric integration interval won't simplify things. I'm not sure what to do.

I would appreciate any hints, friends. :) Thank you very much.

2. Dec 24, 2016

### BvU

Helo BWest,

The hint is: integration by parts (in spite of the nastiness you experienced -- perhaps you can post your steps ?).

3. Dec 24, 2016

### vela

Staff Emeritus
Try calculating $\langle (x-a)^2 \rangle$.

4. Dec 24, 2016

### Ssnow

Integration by parts is a good idea, but you must choose well what integrate and what derive in order to simplify the expression ...

5. Dec 24, 2016

### bwest121

Thank you everyone. I understand how to do this now. I truly appreciate all the responses!