<s>[INST] Evaluating $\int_0^\infty dq\, \frac{\sin(qr)}{r} \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)$ I am trying to evaluate, at least in the limit $|x|\to0$,
$$
F_m(r)\equiv\int \frac{d^3q}{(2\pi)^3}\left(\frac{1}{2\sqrt{\mathbf q^2-m^2}}-\frac{1}{2|\mathbf q|}\right) e^{i\mathbf q \cdot \mathbf x}\,.
$$
Going to spherical coordinates and letting $r=|\mathbf x|$ yields
$$
\frac{1}{(2\pi)^2}\int_0^\infty dq\, \frac{\sin(qr)}{r} \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\,,
$$
which is absolutely integrable since the term between round brackets behaves as $-m^2/2q^2$ for large $q$. 
Since
$$\begin{aligned}
\left|\int_0^\infty dx\, \sin(qr) \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\right|&=
\int_0^\infty dq |\sin(qr)|\left(1-\frac{q}{\sqrt{q^2+m^2}}\right)\\
& \le \int_0^\infty dq \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\\
&=m<\infty
\end{aligned}$$
we can conclude that, by dominated convergence,
$$
\lim_{r\to0}\int_0^\infty dx\, \sin(qr) \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)=0\,;
$$
but unfortunately this does not help since it gives an indefinite expression for 
$$
\lim_{r\to0} F_m(r)= \left[\frac{0}{0}\right]\,.
$$
I tried applying l'Hospital's rule but then we get
$$
\int_0^\infty dq\, \cos(qr) \left(\frac{q^2}{\sqrt{q^2+m^2}}-q\right)\,.
$$
Alternatively, I tried using $\sin(qr)=\Im e^{iq r}$ and integrating by parts formally (we can think that $r$ acquires a small positive imaginary part which we then send to zero), which gives
$$\begin{aligned}
F_m(r)&=\frac{1}{(2\pi)^2} \Re \int_0^\infty dq\, e^{iqr}\left(q-\sqrt{q^2+m^2}\right)
\end{aligned}$$
But still, I'm not getting anywhere. Any suggestion? :)
 [/INST] With $q=mt$ and $x=rm$, 
\begin{align}
 I&=\int_0^\infty dq\, \frac{\sin(qr)}{r} \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\\
&=\frac{m}{r}\int_0^\infty \sin (xt) \left( \frac{t}{\sqrt{1+t^2}}-1 \right)\,dt
\end{align}
We integrate by parts,
\begin{align}
 I&=\frac{m}{r}\left[\left.-\frac{\cos (xt)}{x}\left( \frac{t}{\sqrt{1+t^2}}-1 \right)\right|_0^\infty+\frac{1}{x}\int_0^\infty\frac{\cos (xt)}{\left( 1+t^2 \right)^{\tfrac{3}{2}}} \right]\,dt\\
&=\frac{m}{r}\left[-\frac{1}{x}+K_1(x)\right]
\end{align} 
$K_1(.)$ is the modified Bessel function of the second kind and we used its Basset's integral representation.
Then, 
\begin{equation}
 I=\frac{m}{r}K_1(rm)-\frac{1}{r^2}
\end{equation} 
Its series expansion for $r\to 0^+$ is then
\begin{equation}
 I\sim \frac{m^2}{2}\left( \gamma-\frac{1}{2}+\ln \frac{mr}{2} \right)+\frac{m^4r^2}{16}\left(\gamma-\frac{5}{4}+\ln\frac{mr}{2}  \right)+...
\end{equation} 
</s>