Asymptotic expansion of a function $\frac{4}{\sqrt \pi} \int_0^\infty \frac{x^2}{1 + z^{-1} e^{x^2}}dx$ How to find the asymptotic expansion of the following function for large values of $z$.
$$f_{3/2}(z) = \frac{4}{\sqrt \pi} \int_0^\infty \frac{x^2}{1 + z^{-1} e^{x^2}}dx $$
I have to get something like (in the book)
$$ f_{3/2}(z) = \frac{4}{3 \sqrt \pi} \left[ (\ln z)^{3/2} + \frac{\pi^2}{8} (\ln z)^{-1/2} + \dots \right ] $$
the problem comes from physics in determining the chemical potential for Fermi gas.
 A: For simplicity let's write $\ln z = \lambda$.  The integral we're trying to approximate is
$$
I(\lambda) = \int_0^\infty \frac{x^2}{1+\exp(x^2-\lambda)}\,dx
$$
as $\lambda \to \infty$.
Begin by making the change of variables $x^2-\lambda = y$ to get
$$
\begin{align}
I(\lambda) &= \frac{1}{2} \int_{-\lambda}^\infty \frac{\sqrt{y+\lambda}}{1+e^y}\,dy \\
&= \frac{\sqrt{\lambda}}{2} \left( \int_{-\lambda}^0 \frac{\sqrt{1+y/\lambda}}{1+e^y}\,dy + \int_0^\infty \frac{\sqrt{1+y/\lambda}}{1+e^y}\,dy \right). \tag{1}
\end{align}
$$
The asymptotics for the second integral can be found by simply expanding the square root in a power series
$$
\sqrt{1+y/\lambda} = 1 + \frac{y}{2\lambda} - \frac{y^2}{8\lambda^2} + \cdots
$$
and integrating term-by-term to get
$$
\int_0^\infty \frac{\sqrt{1+y/\lambda}}{1+e^y}\,dy = \ln 2 + \frac{\pi^2}{24\lambda} + O\left(\frac{1}{\lambda^2}\right). \tag{2}
$$
Getting the asymptotics for the first integral is a little more tricky.  Making the change of variables $y = -\lambda u$ yields
$$
\int_{-\lambda}^0 \frac{\sqrt{1+y/\lambda}}{1+e^y}\,dy = \lambda \int_0^1 \frac{\sqrt{1-u}}{1+e^{-\lambda u}}\,du. \tag{3}
$$
By invoking the dominated convergence theorem we may expand the denominator as a geometric series in $e^{-\lambda u}$ and interchange the order of integration and summation to get
$$
\begin{align}
\lambda \int_0^1 \frac{\sqrt{1-u}}{1+e^{-\lambda u}}\,du &= \lambda \sum_{n=0}^{\infty} (-1)^n \int_0^1 \sqrt{1-u} \ e^{-n\lambda u}\,du \\
&= \frac{2\lambda}{3} + \lambda \sum_{n=1}^{\infty} (-1)^n \int_0^1 \sqrt{1-u} \ e^{-n\lambda u}\,du.
\end{align}
$$
By Watson's lemma we have
$$
\int_0^1 \sqrt{1-u} \ e^{-n\lambda u}\,du = \frac{1}{n\lambda} - \frac{1}{2n^2\lambda^2} + O\left(\frac{1}{n^3 \lambda^3}\right), \tag{4}
$$
and upon substituting this into the above sum we find that
$$
\begin{align}
\lambda \int_0^1 \frac{\sqrt{1-u}}{1+e^{-\lambda u}}\,du &= \frac{2\lambda}{3} + \lambda \sum_{n=1}^{\infty} (-1)^n \left[
\frac{1}{n\lambda} - \frac{1}{2n^2\lambda^2} + O\left(\frac{1}{n^3 \lambda^3}\right)\right] \\
&= \frac{2\lambda}{3} + \sum_{n=1}^{\infty} \frac{(-1)^n}{n} - \frac{1}{2\lambda} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} + O\left(\frac{1}{\lambda^2}\right) \\
&= \frac{2\lambda}{3} - \ln 2 + \frac{\pi^2}{24\lambda} + O\left(\frac{1}{\lambda^2}\right). \tag{5}
\end{align}
$$
Recalling from $(3)$ that this is equal to the first integral in $(1)$, we substitute this and $(2)$ into $(1)$ to get
$$
\begin{align}
I(\lambda) &= \frac{\sqrt{\lambda}}{2} \left[ \frac{2\lambda}{3} - \ln 2 + \frac{\pi^2}{24\lambda} + O\left(\frac{1}{\lambda^2}\right) + \ln 2 + \frac{\pi^2}{24\lambda} + O\left(\frac{1}{\lambda^2}\right) \right] \\
&= \frac{1}{3} \lambda^{3/2} + \frac{\pi^2}{24} \lambda^{-1/2} + O\left(\lambda^{-3/2}\right).
\end{align}
$$
This concludes the answer since
$$
f_{3/2}(z) = \frac{4}{\sqrt{\pi}} \,I(\ln z).
$$
A: Here is a way a physicist makes a rough quick estimation:  
After the variable change $t=x^2$ we get:  
$$f_{3/2}(z) = \frac{2}{\sqrt \pi} \int_0^\infty \frac{\sqrt{t}}{1 + z^{-1} e^{t}}dt$$  
Now, the trick is to replace the upper limit $\infty$ with $\ln z$:  
$$f_{3/2}(z) \approx \frac{2}{\sqrt \pi} \int_0^{\ln z} \frac{\sqrt{t}}{1 + z^{-1} e^{t}}dt$$ for large $z$.  
The larger z is, the less $z^{-1}e^t$ will contribute to the integral and we get approximately:  
$$f_{3/2}(z) \approx \frac{2}{\sqrt \pi} \int_0^{\ln z} \sqrt{t}\;dt=\frac{4}{3\sqrt \pi}(\ln z)^{\frac{3}{2}}$$
