Edit: Why is $$\left| \int^{\infty}_{0} \frac{e^{is\cos \phi} - e^{-s}}{s}ds \right| < 2\log \frac{c}{|\cos \phi|}, \quad \text{ where }c\in (1,\infty]?$$ Reference:
Let $E^n$ be the $n$-dimensional euclidean space. If $P$ and $Q$ are points in $E^n$, $(P-Q)$ will denote either the vector going from $Q$ to $P$, or the point whose coordinates are the components of $(P-Q)$. The length of $(P-Q)$ will be denoted by $|P-Q|$, and $\Sigma$ will stand for the surface of the sphere of radius $1$ with center at the origin of coordinates, $O$. We shall be concerened with kernels of the form $$K(P-Q)= \frac{\Omega (\frac{P-Q}{|P-Q|} )}{|P-Q|^n},$$ where $\Omega(P)$ is a function defined on $\Sigma$ and satisfying the conditions \begin{equation} \int_\Sigma \Omega(P)d\sigma =0, \end{equation} and $$\left| \Omega(P)-\Omega(Q) \right| \leq \omega(|P-Q|),$$ where $\omega$ is an increasing function such that $\omega(t)\ge t$, and $$\int_0^1 \omega(t) \frac{dt}{t} = \int_1^\infty \omega(\frac{1}{t}) \frac{dt}{t} <\infty.$$
More precisely, we shall investigate the convergence of the integral \begin{equation} \tilde f_\lambda (P) = \int_{E^n} K_\lambda(p-Q)f(Q)dQ, \end{equation} where $f(Q)$ is a function of $L^p$, $p\ge 1$ in $E^n$, $dQ$ is the element of volume in $E^n$ and
$$K_\lambda(P-Q) = \begin{cases} K(P-Q) &\quad if |P-Q| \ge 1/\lambda,\\ 0 &\quad otherwise. \end{cases} $$ Using Holder's inequality, or the boundedness of $K_\lambda$, we see that $\tilde{f}_\lambda(P)$ is absolutely convergent for $1<p<\infty$ and $p=1$ respectively.
We begin by proving that in the case when the function $f$ in $\tilde{f}_\lambda(P)$ belongs to $L^2$, $\tilde f_\lambda$ converges in the $L^2$ norm as $\lambda \to \infty$.
Let $$K_{\lambda \mu}(P-Q) = \begin{cases} K(P-Q) &\quad if \mu \ge |P-Q| \ge 1/\lambda,\\ 0 &\quad otherwise. \end{cases} $$
As we shall see, the Fourier transform of $K_{\lambda \mu}$ converges boundedly as $\mu$ and $lambda$ tend to infinity successively, and then the desired result will follow easily.
In polar coordinates we have the following expression for the Fourier transform $\hat{K}_{\lambda \mu}$ of $K_{\lambda \mu}$, $$\hat{K}_{\lambda \mu} (P) = \int_{E^n} K_{\lambda \mu} (Q) e^{ir \varrho \cos \phi} dQ = \int_{1/\lambda}^\mu \varrho^{-1}d\varrho \int_\Sigma \Omega(Q') e^{ir \varrho \cos \phi} d\sigma,$$ where $r=|P-O|,\varrho = |Q-O|, Q'=(Q-O)|Q-O|^{-1}$, and $\phi$ is the angle between the vectors $(P-O)$ and $(Q-O)$. Introducing the variable $s=\varrho r$ we can write $$\hat K_{\lambda \mu}(P) = \int_{r/\lambda}^{r\mu} \frac{ds}{s} \int_{\Sigma} \Omega(Q') e^{is \cos \phi}d\sigma,$$ and by our assumption $$\int_\Sigma \Omega(Q')d\sigma =0,$$ we also have \begin{align*} \hat K_{\lambda \mu}(P) &= \int_{r/\lambda}^{r\mu} \frac{ds}{s} \int_{\Sigma} \Omega(Q')e^{is \cos \phi} d\sigma -\int_{r/\lambda}^{r\mu} \frac{ds}{s} \int_{\Sigma} \Omega(Q') e^{-s}d\sigma \\ &=\int_{r/\lambda}^{r\mu} \frac{ds}{s} \int_{\Sigma} \Omega(Q')[e^{is \cos \phi} - e^{-s}] d\sigma\\ &= \int_\Sigma \Omega(Q')d\sigma \int^{r\mu}_{r/\lambda} \frac{e^{is\cos \phi} - e^{-s}}{s}ds. \end{align*}
Now, if $\phi \not = \pi /2$, the inner integral in the last expression converges as $\lambda$ and $\mu$ tend to infinity, and it never exceeds $2\log \frac{c}{|\cos \phi|}$ in absolute value, where $c>1$ is a constant.