# How to approximate the integral $\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1} dx$ for a constant $\mu \gg 1$

I think this integral has no exact solution, but I've found that you can approximate it for large $$\mu$$ as:

$$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1} dx \approx \sqrt{\frac{\pi \mu}{2}} e^{-\mu}$$

This is used a couple of times in a book I'm reading. I've checked it numerically and it works, but I wanted to get the symbolic expression.

I've tried expanding the square root: $$\sqrt{x^2-\mu^2}=\mu\sqrt{-(1-x^2/\mu^2)}\approx i\mu \left(1-\frac{1}{2}\frac{x^2}{\mu^2}-\frac{1}{8}\frac{x^4}{\mu^4}-...\right)$$

That means that I get an imaginary solution as far as I know, because the first term is a simple logarithmic integral:

$$\int_{\mu}^{\infty} \frac{i\mu}{e^x-1}dx=-i\mu\ln{|e^{-\mu}-1|}$$

And the following terms are integrals that I don't know how to solve. So I guess that this is not the way to proceed.

Any ideas on how to do this approximation, or a more exact solution? Appreciate your help.

• Pulling out $\mu$ is the wrong move there, because it makes the $x^2$ term the bigger term rather than smaller. You want to pull out $x$ instead. – Ian Jan 19 at 0:00
• Out of curiosity, what context does this arise from? It seems like the kind of integral arising from the statistical mechanics of a Bose-Einstein gas. – Semiclassical Jan 19 at 0:02
• @Ian that makes sense... But from there, I get for the first term $$\int_{\mu}^{\infty} \frac{x}{e^x-1} = \sum_{n=1}^{\infty} \left(\frac{\mu e^{-\mu n}}{n} + \frac{e^{-\mu n}}{n^2}\right) = -\mu \ln{(1-e^{-\mu})} + Li_2 (e^{-\mu})$$ I guess I can approximate the logarithm to $0$, but the second serie is a dilogarithm and I don't know how to evaluate it. – Martin Jan 19 at 0:29
• @Martin The dilogarithm has a series expansion...but the non-uniformity I mentioned will come back to bite you eventually if you try this. – Ian Jan 19 at 0:37
• @Semiclassical it appears when computing the Casimir energy in an interval for a non maseless field, arising after using the Abel-Plana formula with a square-root function. Its in the second chapter of the book "Advances in the Casimir Effect" by M. Bordag – Martin Jan 19 at 0:40

With the substitution $$z=x-\mu$$, the integral becomes

$$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1}\,dx=\int_0^\infty \frac{\sqrt{z^2+2\mu z}}{e^{\mu+z}-1}\,dz=\frac{\sqrt{2\mu}}{e^\mu }\int_0^\infty\frac{\sqrt{z+z^2/(2\mu)}}{e^z-e^{-\mu}}\,dz$$

Taking the limit $$\mu\to\infty$$ on the integrand and substituting $$w=\sqrt{z}$$, the integral becomes $$\int_0^\infty \sqrt{z}\,e^{-z}\,dz=\int_0^\infty 2w^2 e^{-w^2}\,dx=\int_{-\infty}^\infty w^2 e^{-w^2}\,dw.$$ This is a standard variation on the Gaussian integral and evaluates to $$\sqrt{\pi}/2$$. Hence $$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1}\,dx\sim \frac{\sqrt{2\mu}}{e^\mu}\frac{\sqrt{\pi}}{2}=\sqrt{\frac{\pi u}{2}}e^{-\mu}$$ in the $$\mu\to\infty$$ limit as claimed.

• Alternatively, one has $\int_0^\infty z^{\alpha-1} e^{-z}\,dz =\Gamma(\alpha)$ as integral representation for the gamma function. So the integral after taking the $\mu\to\infty$ limit is just $\Gamma(3/2)=\Gamma(1/2)\cdot 1/2=\sqrt{\pi}/2$. – Semiclassical Jan 19 at 0:41
• Nice approach with the removing the square root. It is a little bit heuristic at this level of explanation since you need $z/\mu \ll 1$ on the region of significant contribution to integration, so you need to argue that the integral for $z \in [\mu,\infty)$ is insignificant. – Ian Jan 19 at 0:53

Having been stuck with this integral, now using @Semiclassical's elegant approach, I think that we can go further.

$$I=\frac{\sqrt{2\mu}}{e^\mu }\int_0^\infty\frac{\sqrt{z+\frac {z^2}{2\mu}}}{e^z-e^{-\mu}}\,dz\tag1$$

Expanding $$\sqrt{z+\frac {z^2}{2\mu}}$$ as a series, we have

$$\color{blue}{I=2\sqrt{\pi}\,\,\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{3}{2}\right)\,\,\text{Li}_{n+\frac{3}{2}}\left(e^{-\mu }\right)}{(2\mu)^{n-\frac{1}{2}}\,\,\Gamma \left(\frac{3}{2}-n\right)\,\, \Gamma (n+1)} }\tag2$$

and the first term is then $$a_0=\sqrt{\frac{\pi \mu}{2}} \text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right)=\sqrt{\frac{\pi \mu}{2}} \sum_{k=1}^\infty \frac {e^{-k\mu}}{k^{\frac 32}}$$ and then the required approximation using the first term of the last summation. Doing the same (that is to say using $$\text{Li}_{n+\frac{3}{2}}\sim e^{-\mu}~~\forall n$$) a better approximation would be $$I=\sqrt{\frac{\pi\mu}{2}} e^{-\mu } \left(1+t-\frac{5 t^2}{6}+O\left(t^3\right) \right) \qquad \text{with} \qquad t=\frac{3}{8 \mu }$$

Edit

To better see the approximation, we could rewrite $$I=\sqrt{\frac{\pi \mu}{2}} e^{-\mu } \big[e^{\mu } \,\text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right) \big]\Bigg[1+\sum_{n=1} ^\infty B_n\Bigg]$$ with $$B_n=\frac{ \Gamma \left(n+\frac{3}{2}\right)}{(2\mu)^n\,\,\Gamma \left(\frac{3}{2}-n\right)\,\,\Gamma (n+1)}\,\,\frac{\text{Li}_{n+\frac{3}{2}}\left(e^{-\mu }\right) } {\text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right) }$$

$$e^{\mu } \,\text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right)=1+\frac{e^{-\mu }}{2 \sqrt{2}}+\frac{e^{-2 \mu }}{3 \sqrt{3}}+\cdots$$ $$\frac{\text{Li}_{n+\frac{3}{2}}\left(e^{-\mu }\right) } {\text{Li}_{\frac{3}{2}}\left(e^{-\mu }\right) }=1- 2^{-n-\frac{3}{2}} \left(2^n-1\right)e^{-\mu }+\cdots$$