Gaussian function integration - Application to statistical mechanics

Question

I am struggling solving a integral. I am making a slight mistake but do not know where... Could you help me solve my problem please :)?

$$ \int_{-\infty}^{+\infty} N^2 e ^{-\frac{\beta\gamma}{2}(N-N^{*})^2} dN $$

Where $ \beta, \gamma, N^{*}$ are constants.

When doing the integration, I am finding $ \frac{1}{\beta \gamma} \times \sqrt{\frac{2\pi}{\beta \gamma}}$. But I am pretty sure that I should also have something with a $N^*$.

Thanks for your help and sorry for the dummy question !

Léonard

Note : Obviously, to solve the integral one should use

$$\int_{-\infty}^{+\infty} x^n e^{-\alpha x^2} dx = \frac{n-1}{2\alpha} \int_{-\infty}^{+\infty} x^{n-2} e^{- \alpha x^2} dx$$

and

$$\int_{-\infty}^{+\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}$$

Answer 1

Hint:

You can shift the variable by $N:=M+N^*$ so that you get three terms in $M^2,2MN^*$ and $N^{*2}$. The second term will vanish by symmetry and you'll end up with

$$a+bN^{*2}.$$

Answer 2

So I just find the answer! Yves is right :)

By shifting the variable, you find the solution easily !

$$ \int_{-\infty}^{+\infty} N^2 e ^{-\frac{\beta\gamma}{2}(N-N^{*})^2} dN $$

By doing $ X = N - N^{*}$ (thus $N = X + N^{*}$ and $ dX = dN + dN^{*} = dN$ , $N^{*}$ being a constant).

We now have

$$ \int_{-\infty}^{+\infty} (X+N^{*})^2 e ^{-\frac{\beta\gamma}{2}X^2} dX $$

$$ \int_{-\infty}^{+\infty} X^2 e ^{-\frac{\beta\gamma}{2}X^2} dX + 2 N^{*}\int_{-\infty}^{+\infty} X e ^{-\frac{\beta\gamma}{2}X^2} dX + \int_{-\infty}^{+\infty} N^{*2} e ^{-\frac{\beta\gamma}{2}X^2} dX $$

Solving those integrals individually, one gets

• For the first one

$$ \int_{-\infty}^{+\infty} X^2 e ^{-\frac{\beta\gamma}{2}X^2} dX = \frac{1}{\beta \gamma} \times \int_{-\infty}^{+\infty} e ^{-\frac{\beta\gamma}{2}X^2} dX = \frac{1}{\beta \gamma} * \sqrt{\frac{2 \pi}{\beta \gamma}} $$

• For the second one

$$ 2 N^{*}\int_{-\infty}^{+\infty} X e ^{-\frac{\beta\gamma}{2}X^2} dX = 0 $$

For parity reason, the integral is equal to zero (the integral is odd on $ ] - \infty ; + \infty [ $, thus is equal to zero when integrating on this ensemble)

• For the last one

$$ \int_{-\infty}^{+\infty} N^{*2} e ^{-\frac{\beta\gamma}{2}X^2} dX = N^{*2} \int_{-\infty}^{+\infty} e ^{-\frac{\beta\gamma}{2}X^2} dX = N^{*2} \times \sqrt{\frac{2 \pi}{ \beta \gamma}} $$

In conclusion,

$$ \int_{-\infty}^{+\infty} N^2 e ^{-\frac{\beta\gamma}{2}(N-N^{*})^2} dN = \sqrt{\frac{2 \pi}{\beta \gamma}} \times (N^{*2} + \frac{1}{\beta \gamma}) $$