Statistical Mechanics of interacting Particles. Quantized Fields. Solving Integral?

Question

Hi everyone How we can analytically without using a software solve below integral . Chapter 11 of Pathria (edition 1). and x is dimensionless.

Answer 1

Building off of @SuperCiocia 's answer, parts 1 and 3 give you $\frac{x}{2}-\frac{x^3}{3}-\frac{x^5}{5}$. As you noted, these diverge; but the 2nd term comes to our rescue.

The second term will give you $\frac{1}{15}(x^2+2)^{\frac{3}{2}}(3x^2-4)$. Evaluating this at zero is easy (and will give you the answer you seek). The other part is evaluating it at infinity. So write it as $$\frac{x^3}{15}(1+\frac{2}{x^2})^{\frac{3}{2}}(3x^2-4).$$ Do a binomial expansion (which you can do since $\frac{2}{x^2}$ is small). The leading terms will be $-\frac{x}{2}+\frac{x^3}{3}+\frac{x^5}{5}$, which cancel out the other problem terms.

Answer 2

We sure can.

You have 3 terms in the intergrand:

1) $\frac{1}{2} \rightarrow $ trivial to integrate

2) $x^3\sqrt{x^2+2}$:

write this as $\int \frac{x^2}{2} \, 2x \, \sqrt{x^2+2}$, so it is in the form $\int f(x) g'(x)$ where $g(x) = \frac{2}{3}(x^2+2)^{3/2}$.

You can then proceed to integrate by parts.

3) $-x^2 -x^4 \rightarrow$ trivial to integrate