Let us consider the following integral
$$
I(x) = \int_{0}^{\infty} e^{-x \cosh{\theta}} d\theta
$$
where $x \in \mathbb{R}$.
I am trying to get an approximation to this integral for the regime where $x \ll 1$. Since the argument of the exponential is $x \cosh{\theta}$, I cannot just consider a Taylor expansion of the integrand. What would be the smartest way to approximate this integral? In particular, I would like to know an asymptotic expression of $I(x)$ around $x\sim 0$.
I was thinking about considering the following approximation
$$
\tilde{I}(x) = \int_{0}^{\hat\theta} d\theta
$$
where $\hat{\theta}$ is determined such that $x \cosh{\theta}\sim O(1)$. However, for $x\sim 0$, $\hat{\theta}$ is very big which would signal that the integral $I(x)$ cannot be expanded around $x\sim 0$. Indeed, for $x=0$ the integral diverges. Is there a way to interpret this integral, maybe in the sense of distributions?
Can I still find a good approximation of $I(x)$ for small $x$?
Best Answer
Substitute $\theta=\operatorname{arcosh}(s)$. Then
\begin{align*} I(x) &= \int_{1}^{\infty} \frac{e^{-xs}}{\sqrt{s^2-1}} \, \mathrm{d}s \\ &= \int_{1}^{\infty} \frac{e^{-xs}}{s} \, \mathrm{d}s - \int_{1}^{\infty} \left( \frac{1}{s} - \frac{1}{\sqrt{s^2-1}} \right) e^{-xs} \, \mathrm{d}s. \end{align*}
As for the first integral, by substituting $u = xs$ and noting that $-\gamma = \int_{0}^{\infty} e^{-u}\log u \, \mathrm{d}u$,
\begin{align*} \int_{1}^{\infty} \frac{e^{-xs}}{s} \, \mathrm{d}s &= \int_{x}^{\infty} \frac{e^{-u}}{u} \, \mathrm{d}u \tag{$u=xs$}\\ &= \left[e^{-u}\log u\right]_{x}^{\infty} + \int_{x}^{\infty} e^{-u}\log u \, \mathrm{d}u \\ &= -e^{-x}\log x - \gamma - \int_{0}^{x} e^{-u}\log u \, \mathrm{d}u \\ &= -\log x - \gamma + \mathcal{O}(x\log x). \end{align*}
As for the second integral, by the dominated convergence theorem we get
\begin{align*} \lim_{x \to 0^+} \int_{1}^{\infty} \left( \frac{1}{s} - \frac{1}{\sqrt{s^2-1}} \right) e^{-xs} \, \mathrm{d}s &= \int_{1}^{\infty} \left( \frac{1}{s} - \frac{1}{\sqrt{s^2-1}} \right) \, \mathrm{d}s \\ &= \lim_{R\to\infty} \int_{1}^{R} \left( \frac{1}{s} - \frac{1}{\sqrt{s^2-1}} \right) \, \mathrm{d}s \\ &= \lim_{R\to\infty} \left[ \log s - \log\left(s + \sqrt{s^2-1}\right) \right]_{1}^{R} \\ &= -\log 2. \end{align*}
Combining altogether, we get
$$ I(x) = -\log(x/2) - \gamma + o(1) $$
as $x \to 0^+$.
According to the formula 10.31.1 in DLMF,
$$ K_0(z) = \sum_{n=0}^{\infty} \frac{- \log(z/2) - \gamma + H_n}{(n!)^2} \left(\frac{z}{2}\right)^{2n} $$