Consider the integral
$$I(t) = \int_0^{1/2} \frac{t dx}{x^{-t} - x^t}$$
in the vicinity of $t=0$.
When $t$ is viewed as a real variable, $I(t)$ is very well-behaved. For example, it's infinitely differentiable.
On the other hand, when $t$ is on the imaginary axis, the integrand is very poorly behaved as a function of $x$ close to zero. The denominator becomes a rapidly oscillating function that repeatedly passes through zero. This is easiest to see by writing $x^{-t} - x^t = 2\sinh(t \log(1/x))$ which is $2i\sin(\Im(t) \log(1/x))$ for imaginary $t$.
Thus $I(t)$ is infinitely-differentiable for real $t$, but is non-analytic in the complex plane about $t=0$.
Note that I expect that $I(t)$ will be asymptotic to some power series in $t$ for small real $t$, $I(t) \sim \sum_{n=0}^\infty c_n t^n$, but that is not quite what I'm looking for. Rather, I'm interested in the leading "non-analytic" contribution to $I(t)$ at small real $t$. For example, one might guess such a piece to go something like $e^{-1/t^2}$.
Is there a systematic way to extract the leading "non-analytic" piece of $I(t)$ for small real $t$?
Long Comment.
Part 1. We first show that $I(t)$ extends to an analytic function on $\Omega = \mathbb{C}\setminus B$, where $B$ is the union of branch cuts given by
$$ B = \left\{ \sigma + \frac{2\pi i k}{\log 2} : \sigma \in (-\infty, 0] \text{ and } k \in \mathbb{Z} \right\}. $$
Below is a complex plot of $I(t)$:
Indeed, assume for a moment that $t > 0$. Substituting $x = \frac{1}{2}u^{\frac{1}{2t}}$, $I(t)$ can be written as
$$ I(t) = \int_{0}^{\frac{1}{2}} \frac{t x^t}{1 - x^{2t}} \, \mathrm{d}t = 2^{-t-2} \int_{0}^{1} \frac{u^{\frac{1}{2t} - \frac{1}{2}}}{1 - 4^{-t}u} \, \mathrm{d}u. \tag{1} $$
Note that the last integral can be expressed in terms of the hypergeometric function, which is useful for computing $I(t)$ numerically.
Then for any integer $N \geq 1$, invoking the identity $\frac{1}{1-a} = 1 + a + \cdots + a^{N-1} + \frac{a^N}{1-a}$ yields
\begin{align*} I(t) &= 2^{-t-2} \int_{0}^{1} u^{\frac{1}{2t} - \frac{1}{2}} \left( \sum_{k=0}^{N-1} (4^{-t}u)^k + \frac{(4^{-t}u)^N}{1 - 4^{-t}u} \right) \, \mathrm{d}u \\ &= \sum_{k=0}^{N-1} \frac{t 2^{-(2k+1)t-1}}{(2k+1)t + 1} + 2^{-(2N+1)t-2} \int_{0}^{1} \frac{u^{N - \frac{1}{2} + \frac{1}{2t}}}{1 - 4^{-t}u} \, \mathrm{d}u \tag{2} \end{align*}
Note that the last line defines an analytic function of $t$ in the region $\Omega_N$ defined by
$$ 4^t \notin [0, 1] \qquad\text{and}\qquad \operatorname{Re}\left(N-\frac{1}{2}+\frac{1}{2t}\right) > -1, $$
or equivalently,
$$ t \notin B \qquad \text{and}\qquad \left| t + \frac{1}{4N+2}\right| > \frac{1}{4N+2}. $$
This shows that $\Omega_N$ is increasing and $\bigcup_{N\geq 1}\Omega_N = \Omega$. Hence, $I(t)$ is defined as an analytic function on all of $\Omega$ and $I(t)$ can be expanded as $\text{(2)}$ on $\Omega_N$.
Part 2. Using the above representation, I can prove that
$$ I(t) \to A := -\frac{1}{2} \int_{0}^{\frac{1}{2}} \frac{\mathrm{d}x}{\log x} \approx 0.189336 $$
as $ t \to 0$ in the sector $\left| \arg t \right| < \frac{\pi}{2} - \delta$ for any $\delta > 0$. Let me only provide a sketch of the proof. In each of these sectors, we can show that
$$ I(t) = \sum_{k=0}^{\infty} \frac{t 2^{-(2k+1)t-1}}{(2k+1)t + 1} $$
and that this sum behaves like a Riemann sum for the integral $A = \int_{0}^{\infty} \frac{2^{-x-1}}{x+1} \, \frac{\mathrm{d}x}{2}$.
However, the plot of $u \mapsto I(1/u)$ shows that the above convergence probably manifests in Stolz angle (i.e., the above convergence holds in the sector $\left| \arg t \right| < \pi - \delta$ for any $\delta > 0$). This can be glimpsed from the graph of $I(1/u)$ as a function of $u$:
Finally, although I have no idea how the leading non-analytic term would look like, the below conjecture might shed a light on how it should look like.
Indeed, this describes the limiting shape of the "poles" we can see from Figure 2 above. I obtained this using heuristic computations and numerical experiments altogether.