Ytukyg

Edit:
Why is $$left| int^{infty}_{0} frac{e^{iscos phi} - e^{-s}}{s}ds right| < 2log frac{c}{|cos phi|}, quad text{ where }cin (1,infty]?$$
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)dsigma =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 $L^p$, $pge 1$ 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 $K_{lambda mu}$,
$$hat{K}_{lambda mu} (P) = int_{E^n} K_{lambda mu} (Q) e^{ir varrho cos phi} dQ = int_{1/lambda}^mu varrho^{-1}dvarrho int_Sigma Omega(Q') e^{ir varrho cos phi} dsigma,$$
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
$$hat K_{lambda mu}(P) = int_{r/lambda}^{rmu} frac{ds}{s} int_{Sigma} Omega(Q') e^{is cos phi}dsigma,$$
and by our assumption
$$int_Sigma Omega(Q')dsigma =0,$$
we also have
begin{align*}
hat K_{lambda mu}(P) &= int_{r/lambda}^{rmu} frac{ds}{s} int_{Sigma} Omega(Q')e^{is cos phi} dsigma -int_{r/lambda}^{rmu} frac{ds}{s} int_{Sigma} Omega(Q') e^{-s}dsigma \
&=int_{r/lambda}^{rmu} frac{ds}{s} int_{Sigma} Omega(Q')[e^{is cos phi} - e^{-s}] dsigma\
&= int_Sigma Omega(Q')dsigma int^{rmu}_{r/lambda} frac{e^{iscos 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 $2log frac{c}{|cos phi|}$ in absolute value, where $c>1$ is a constant.