I am trying to study the following function defined in terms of an integral

$$f(x) = -\int_{x}^{\infty} \sqrt{1-\frac{x^2}{k^2}}\log\left( 1- e^{-k}\right) dk$$

with $x\geq 0$. So far I was able to compute: $f(0)= \frac{\pi^2}{6}$. Using a saddle point approximation I was able to determine the asymptotic form $f(x) \approx c \frac{e^{-x}}{x^{3/2}}$ for large $x$ with $c$ a numerical constant. Moreover, I can evaluate numerically the integral and see that the function is monotonically decreasing and always positive.

I add the numerical plot between 0 and 1 for completeness

Plot of f(x) between 0 and 1

After a week of tinkering, I am tempted to give up. I am missing: 1) a form for $f$ in terms of known functions (probably this is not possible) 2) at least the series expansion of $f(x)$ in a neighborhood of $0$ a.k.a. compute $f'(0)$ and $f''(0)$.

Thank you very much for your help!


Some integral representations

For $x > 0$ we have \begin{align} f(x) &= \int \limits_x^\infty - \log \left(1 - \mathrm{e}^{-k}\right) \sqrt{1 - \frac{x^2}{k^2}} \, \mathrm{d}k \tag{1} \\ &\!\!\!\!~\stackrel{k = x u}{=} x \int \limits_1^\infty - \log \left(1 - \mathrm{e}^{-x u}\right) \sqrt{1 - \frac{1}{u^2}} \, \mathrm{d}u \tag{2} \\ &\!\!\stackrel{\text{i.b.p.}}{=} \int \limits_1^\infty \frac{\operatorname{Li}_2\left(\mathrm{e}^{-x u}\right)}{u^2 \sqrt{u^2-1}} \, \mathrm{d} u \tag{3} \\ &\!\!\!\!\!\!\stackrel{u = \cosh(s)}{=} \int \limits_0^\infty \frac{\operatorname{Li}_2\left(\mathrm{e}^{-x \cosh(s)}\right)}{\cosh^2(s)} \, \mathrm{d} s \tag{4} \\ &\!\!\!\!\!\stackrel{\tanh(s) = \tau}{=} \int \limits_0^1 \operatorname{Li}_2\left(\mathrm{e}^{-\frac{x}{\sqrt{1-\tau^2}}}\right) \, \mathrm{d} \tau \, ,\tag{5} \end{align} where $\operatorname{Li}_2$ is the dilogarithm. Apart from $(2)$ these representations are also valid for $x=0$ and yield $f(0) = \frac{\pi^2}{6}$ as expected. None of them gives much hope for a closed-form expression, but they can be used to study the asymptotic behaviour of $f$.

The expansion for $x \to \infty$

Using $(5)$ and the series representation of the dilogarithm we find $$ f(x) = \sum \limits_{n=1}^\infty \frac{1}{n^2} \int \limits_0^1 \mathrm{e}^{- \frac{n x}{\sqrt{1-\tau^2}}} \, \mathrm{d} \tau \stackrel{\frac{1}{\sqrt{1-\tau^2}} = 1 + t}{=} \frac{1}{\sqrt{2}} \sum \limits_{n=1}^\infty \frac{\mathrm{e}^{-n x}}{n^2} \int \limits_0^\infty \frac{\mathrm{e}^{-n x t}}{\sqrt{t(1+\frac{t}{2})} (1+t)^2} \, \mathrm{d}t \, . $$ We can then employ Watson's lemma with $\lambda = -\frac{1}{2}$ and $g(t) = \frac{1}{\sqrt{1+\frac{t}{2}}(1+t)^2}$ to obtain the asymptotic expansion $$ \int \limits_0^\infty \frac{\mathrm{e}^{-n x t}}{\sqrt{t(1+\frac{t}{2})} (1+t)^2} \, \mathrm{d}t \sim \sum \limits_{k=0}^\infty \frac{g^{(k)}(0) \operatorname{\Gamma}\left(k + \frac{1}{2}\right)}{k! (n x)^{k+\frac{1}{2}}} \, , \, x \to \infty \, , \, n \in \mathbb{N} \, . $$ Simplifying the gamma function we arrive at $$ f(x) \sim \sqrt{\frac{\pi}{2 x}} \sum \limits_{n=1}^\infty \frac{\mathrm{e}^{-n x}}{n^{5/2}} \sum \limits_{k=0}^\infty \frac{{{2k} \choose k} g^{(k)}(0)}{(4nx)^k} \, , \, x \to \infty \, .$$ Clearly, terms with $n > 1$ are exponentially smaller than those with $n = 1$ and may be dropped, so we find the asymptotic series $$ f(x) \sim \sqrt{\frac{\pi}{2x}} \mathrm{e}^{-x} \left[1 - \frac{9}{8x} + \frac{345}{128 x^2} + \sum \limits_{k=3}^\infty \frac{{{2k} \choose k} g^{(k)}(0)}{(4x)^k} \right] \, , \, x \to \infty \, ,$$ which agrees with Robert Israel's result.

The value of $f'(0)$

Using $(3)$, $(4)$ or $(5)$ it is not hard to show that $f$ is smooth on $(0,\infty)$. The derivative at zero, however, does not exist. This can be seen by combining $(1)$ and $(2)$ to write (for $x > 0$) \begin{align} \frac{f(0) - f(x)}{x} &= \int \limits_0^\infty - \log \left(1 - \mathrm{e}^{-x u}\right) \, \mathrm{d} u - \int \limits_1^\infty - \log \left(1 - \mathrm{e}^{-x u}\right) \sqrt{1 - \frac{1}{u^2}} \, \mathrm{d}u \\ &= \int \limits_0^1 - \log \left(1 - \mathrm{e}^{-x u}\right) \, \mathrm{d} u + \int \limits_1^\infty - \log \left(1 - \mathrm{e}^{-x u}\right) \left[1 - \sqrt{1 - \frac{1}{u^2}}\right] \, \mathrm{d}u \\ &\geq \int \limits_0^1 - \log(x u) \, \mathrm{d} u + 0 = 1 - \log(x) \stackrel{x \to 0^+}{\longrightarrow} \infty \, , \end{align} which implies $f'(0) = - \infty$.

An idea for $x \to 0^+$

In the face this result, the expansion of $f$ at zero cannot be a simple Taylor series. Instead, we might want to use the asymptotic expansion $$ \operatorname{Li}_2 \left(\mathrm{e}^{- a}\right) = a [\log(a)-1] + \sum \limits_{k \in \mathbb{N}_0 \setminus \{1\}} \frac{\zeta(2-k)}{k!} (-a)^k = \frac{\pi^2}{6} + a \log(a) - a -\frac{a^2}{4} + \frac{a^3}{72} + \mathcal{O}\left(a^5\right) $$ of the dilogarithm for $0 < a < 2 \pi$. Naively plugging this result into $(5)$ we obtain \begin{align} f(x) &\stackrel{?}{\sim} \int \limits_0^1 \left[\frac{\pi^2}{6} + \frac{x}{\sqrt{1-\tau^2}}\left[\log\left(\frac{x}{\sqrt{1-\tau^2}}\right) -1\right] + \mathcal{O}\left(x^2\right)\right] \mathrm{d} \tau \\ &= \frac{\pi^2}{6} - \frac{\pi}{2} x \left[-\log(2 x) + 1\right] + \mathcal{O}\left(x^2\right) \, , \, x \to 0^+ \, . \end{align} While this approximation appears to be quite good numerically, it seems unlikely that it is entirely correct. The integral $\int_0^1 \frac{\mathrm{d}{\tau}}{1-\tau^2}$, which is the prefactor of the $x^2$-term, is divergent to begin with. This is probably related to the fact that the expansion is only valid for $\tau^2 < 1 - \frac{x^2}{4 \pi^2}$. I do not know (yet?) how to compute or estimate the higher-order terms rigorously, so I'll leave it at that for now.

  • $\begingroup$ Thank you very much! This was way more of what I was hoping to find! $\endgroup$ – Pietro Dona Nov 15 '19 at 10:58

Write $$\ln(1-e^{-k}) = - \sum_{j=1}^\infty e^{-jk}/j$$ and take $k = x + t$ so your integral becomes $$ \sum_{j=1}^\infty e^{-jx} \int_0^\infty \frac{\sqrt{t(t+2x)}}{j(x+t)} e^{-jt} \; dt$$ Now $$ \int_0^\infty \frac{\sqrt{t(t+2x)}}{j(x+t)} e^{-jt}\; dt= \frac{\sqrt{2\pi}}{2 j^{5/2} \sqrt{x}} - \frac{9 \sqrt{2\pi}}{16 j^{7/2} x^{3/2}} + \frac{345\sqrt{2\pi}}{256 j^{9/2} x^{5/2}} + \ldots$$ so it seems to me the asymptotic form should be $$ e^{-x} \left( \frac{\sqrt{2\pi}}{2 \sqrt{x}} - \frac{9 \sqrt{2\pi}}{16 x^{3/2}} + \frac{345\sqrt{2\pi}}{256 x^{5/2}} + \ldots\right)$$

  • $\begingroup$ Thank you very much for your calculation. I clearly made some mistake there. $\endgroup$ – Pietro Dona Nov 14 '19 at 20:48

You can try an asymptotic expansion near $x\rightarrow 0$ :

First, lets rewrite the integral as
$I(x)=x\int_1^\infty du \sqrt({1-{1\over u^2}})\log(1-e^{-xu})$,

next, we split the integral into two ranges: $u\in [1,{1\over \sqrt x}] $ and $u>{1\over \sqrt x}$ $I(x)=I_1(x)+I_2(x)=x\int_1^{1\over \sqrt x} du \sqrt({1-{1\over u^2}})\log(1-e^{-xu})+x\int_{1\over \sqrt x}^\infty du \sqrt({1-{1\over u^2}})\log(1-e^{-xu})$

Note that for $I_1$ $xu$ is always small, so an expansion can be made in $\log(1-e^{-xu})$. For $I_1$ we may consider $1/u^2$ to be small and expand in $\sqrt({1-{1\over u^2}})$.

I believe that this dual expansion will give you the behavior for $x<<1$

  • $\begingroup$ I will look at this as soon as I can. Thank you very much for your help! $\endgroup$ – Pietro Dona Nov 14 '19 at 20:49

the work I believe that the first derivative at zero is zero, my justification is in the attached image. It basically boils down to defining a multivariate function $F(x,y)$ such that $f(x) = F(x,x)$, and computing the partial derivatives of $F$ at (0,0) on the contour x=y.

  • $\begingroup$ That is precisely what was bothering me. I did the same calculation but then I looked at the plot from which the derivative in 0 is clearly non 0. I added the picture to the original post. $\endgroup$ – Pietro Dona Nov 14 '19 at 20:43
  • $\begingroup$ I thought about it a bit, I think the mistake I made was probably in computing the partial with respect to x, I let x go to zero but forgot that to have x=0, y has to go to zero too, which means that the bounds on k include 0, so the ln part explodes. So we can’t make that simplification. $\endgroup$ – Robo300 Nov 14 '19 at 23:09

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