$$\begin{aligned}
\label{eq:first deformation}
\int_{\mathbb R}\exp(ix\xi) m(\xi)\, {{\mathrm d}}\xi&= 
\int_{\mathbb R}\exp(ix(\xi+is_0))m(\xi+is_0)\, {{\mathrm d}}\xi\\
&
=\exp(-xs_0) \int_{\mathbb R}\exp(ix\xi)m(\xi+is_0)\, {{\mathrm d}}\xi.
\end{aligned}$$ Integrating by parts and using the estimate $|\partial_\xi m(\xi+is_0)|=\mathcal{O}(|\xi|^{-3/2})$ as $|\xi|\to \infty$, we obtain the desired exponential decay of $K$.

In order to estimate the derivatives of $K$, we note that $|\operatorname{{\mathcal F}}(x^{n} \operatorname{D}_x^n K)(\xi)|= |\operatorname{D}_\xi^{n} (\xi^n m(\xi))|$, for any $n\ge 0$, where $\operatorname{D}_\xi^{n} (\xi^n m(\xi))$ extends analytically to the strip $0\le \operatorname{Im}\zeta <\pi/2$ and satisfies the estimate $|D_\xi^{n+k} ((\xi+is_0)^n m(\xi+is_0))|=\mathcal{O}(|\xi|^{-1/2-k})$ as $|\xi|\to \infty$, for any $k\ge 0$ and $s_0\in (0, \pi/2)$. Repeating the above argument, we obtain exponential decay for $x^n \operatorname{D}_x^n K(x)$ and hence also for $\operatorname{D}_x^n K$.

Due to the fact that $m\not \in L^1({\mathbb R})$, it follows that $K$ is singular at the origin. We can give a precise description of this singularity as follows.

\[prop:kernel decomposition\] The Whitham kernel satisfies $$K(x)=\frac{1}{\sqrt{2\pi|x|}}+K_\text{reg}(x),$$ where $K_\text{reg}$ is real analytic on ${\mathbb R}$.

Write $$\sqrt{\frac{\tanh \xi}{\xi}}=\frac{1}{\sqrt{|\xi|}}+\frac{\sqrt{\tanh |\xi|}-1}{\sqrt{|\xi|}}.$$ The first term has inverse Fourier transform $1/\sqrt{2\pi|x|}$, while the second term is integrable and exponentially decaying and hence has a real-analytic transform.

Positivity and monotonicity properties: general theory {#positivity-and-monotonicity-properties-general-theory .unnumbered}
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Our next aim is to show certain positivity and monotonicity properties 