How “to take” this integral?

Question

When I learned anharmonic model of crystal, I read that considering anharmonic oscillations and Boltzmann distribution for the "atoms" of crystal we can get the dependence of distance between the "atoms" from a temperature as

$$ \langle r \rangle = r_{0} + \alpha T. $$

As I understood the words below, it's like

$$ \langle r \rangle = \frac{\int \limits_{0}^{\infty}re^{-\frac{U}{kT}}dr}{\int \limits_{0}^{\infty}e^{-\frac{U}{kT}}dr} \approx \frac{\int \limits_{0}^{\infty}re^{-\frac{U_{0} + a(r - r_{0})^{2} + b(r - r_{0})^{3} }{kT}}dr}{\int \limits_{0}^{\infty}e^{-\frac{U_{0} + a(r - r_{0})^{2} + b(r - r_{0})^{3} }{kT}}dr} = |x = r - r_{0}| = \frac{\int \limits_{0}^{\infty}(x + r_{0})e^{-\frac{ax^{2} + bx^{3}}{kT}}dr}{\int \limits_{0}^{\infty}e^{-\frac{ax^{2} + bx^{3}}{kT}}dr}, $$

and then - $$ \langle r \rangle \approx r_{0} + \frac{\int \limits_{0}^{\infty}xe^{-\frac{ax^{2} + bx^{3}}{kT}}dr}{\int \limits_{0}^{\infty}e^{-\frac{ax^{2} + bx^{3}}{kT}}dr}. $$

What can I do on the next step?

Answer 1

You expand the top and bottom integral in a power series in b, and keep the lowest order term in $b$:

For the top integral:

$$ \int x e^{-ax^2} (1 + bx^3) = b\int x^4 e^{-ax^2} = b {d^2\over da^2} \int e^{-ax^2} = b{d^2\over da^2} \sqrt{\pi\over a} = -\sqrt{\pi\over a} {3b\over 4a^2} $$

For the bottom integral

$$ \int e^{-ax^2} (1 + b x^3) = \sqrt{\pi \over a} $$

So the quotient is

$$ {3b\over 4a^2}$$

The coefficients $a$ and $b$ absorb the $kT$ on the bottom in your expression, so replace $a$ by $a\over kT$ and b by $b\over kT$. The above becomes:

$$ kT {3b\over 4a^2} $$

and it is linearly proportional to T, as you expected.