1
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ 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. $\endgroup$ – Ian Jan 19 at 0:00
  • $\begingroup$ 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. $\endgroup$ – Semiclassical Jan 19 at 0:02
  • $\begingroup$ @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. $\endgroup$ – Martin Jan 19 at 0:29
  • 1
    $\begingroup$ @Martin The dilogarithm has a series expansion...but the non-uniformity I mentioned will come back to bite you eventually if you try this. $\endgroup$ – Ian Jan 19 at 0:37
  • 1
    $\begingroup$ @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 $\endgroup$ – Martin Jan 19 at 0:40
4
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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 $\Gamma(3/2)=\Gamma(1/2)\cdot 1/2=\sqrt{\pi}/2$. $\endgroup$ – Semiclassical Jan 19 at 0:41
  • 1
    $\begingroup$ 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. $\endgroup$ – Ian Jan 19 at 0:53
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.