(a) We replace the sum over p p appearing in eqn. (11.3.14) by an integral, viz. 0 { ( p )

(a) We replace the sum over $p$ appearing in eqn. (11.3.14) by an integral, viz.

$${\int }_0^{\infty }\left\{\epsilon \right(p)-\frac{p^2}{2m}-\frac{4\pi a{\hslash }^{2}N}{mV}+{\left(\frac{4\pi a{\hslash }^{2}N}{mV}\right)}^2\frac{m}{p^2}\}\frac{V\cdot 4\pi p^{2}dp}{h^3}$$

Substituting $p={\left(8\pi a{\hslash }^{2}N/V\right)}^{1/2}x$, we get

$${\int }_0^{\infty }\left(\frac{4\pi a{\hslash }^{2}N}{mV}\right)\{x{(x^2+2)}^{1/2}-x^2-1+\frac{1}{2x^2}\}\frac{4\pi V}{h^3}{\left(\frac{8\pi a{\hslash }^{2}N}{V}\right)}^{3/2}{x}^{2}dx$$

which readily leads us from eqn. (11.3.14) to (11.3.15). The resulting integral over $x$ can be done by elementary means, giving

$${\int }_0^{\infty }\{x{(x^2+2)}^{1/2}-x^2-1+\frac{1}{2x^2}\}{x}^{2}dx=\frac{1}{15}(3x^2-4){(x^2+2)}^{3/2}-\frac{1}{5}{x}^5-\frac{1}{3}{x}^3+{\frac{1}{2}x|}_0^{\infty }.$$

For $x\gg 1$,

$$\begin{array}{c}\frac{1}{15}(3x^2-\end{array}$$