# 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.

• By factoring $z^{-1} e^{x^2}$ out of the denominator, expanding the resulting fraction as a series in powers of $ze^{-x^2}$, and integrating term-by-term, you can show that your integral is equal to the polylogarithm $-\operatorname{Li}_{3/2}(-z)$. The desired asymptotic expansion can probably be deduced from the series representation of the polylog using Euler-Maclaurin summation. That's not to say it can't be derived directly from the integral you have. Aug 24, 2013 at 17:14

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}

$$f_{3/2}(z) = \frac{4}{\sqrt{\pi}} \,I(\ln z).$$

• To note: by taking more terms in the application of Watson's lemma $(4)$ and in the expansion $(2)$ one can recover as many terms as are desired of the final asymptotic expansion. Aug 24, 2013 at 22:39

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:

$$f_{3/2}(z) \approx \frac{2}{\sqrt \pi} \int_0^{\ln z} \sqrt{t}\;dt=\frac{4}{3\sqrt \pi}(\ln z)^{\frac{3}{2}}$$

• thanks, but i need the second term too, the second term contains the info on the chemical potential. Aug 27, 2013 at 7:00