Let $F(p,q)$ be given by the integral
$$\begin{align}
F(p,q)&=\int_0^\infty \frac{|\sin(\sqrt{qx})|-|\sin(\sqrt{px})|}{x}\,dx\\\\
&\overbrace{=}^{x\mapsto x^2}2\int_0^\infty \frac{|\sin(\sqrt{q}x)|-|\sin(\sqrt{p}x)|}{x}\,dx\tag1
\end{align}$$
We will evaluate the integral on the right-hand side of $(1)$ using two distinct approaches. In the first approach, we begin with a common way of evaluating a standard Frullani integral and finish with an heuristic evaluation. In the second, we simply integrate by parts and reduce the integral in $(1)$ to a standard Frullani integral. To that end, we now proceed.
METHODOLOGY $1$:
We proceed by writing the improper integral on the right-hand side of $(1)$ as the limit
$$\begin{align}
\int_0^\infty \frac{|\sin(\sqrt{q}x)|-|\sin(\sqrt{p}x)|}{x}\,dx&=\lim_{L\to\infty}\int_0^L \frac{|\sin(\sqrt{q}x)|-|\sin(\sqrt{p}x)|}{x}\,dx\tag2
\end{align}$$
Next, writing the integral in $(2)$ as the difference of integrals, enforcing substitutions $\sqrt{q}x\mapsto x$ and $\sqrt{p}x\mapsto x$, and adding the resulting integrals reveals
$$\begin{align}
F(p,q)&=2\lim_{L\to\infty}\int_0^L \frac{|\sin(\sqrt{q}x)|-|\sin(\sqrt{p}x)|}{x}\,dx\\\\
&=\lim_{L\to\infty}2\int_0^L \frac{|\sin(\sqrt{q}x)|}{x}\,dx-2\lim_{L\to\infty}\int_0^L \frac{|\sin(\sqrt{p}x)|}{x}\,dx\\\\
&=2\lim_{L\to\infty}\int_0^{\sqrt{qL}} \frac{|\sin(x)|}{x}\,dx-2\lim_{L\to\infty}\int_0^{\sqrt{pL}} \frac{|\sin(x)|}{x}\,dx\\\\
&=2\lim_{L\to\infty}\int_{\sqrt{pL}}^{\sqrt{qL}} \frac{|\sin(x)|}{x}\,dx\tag3
\end{align}$$
The following heuristic analysis can be made rigorous and we leave the details to the reader. We break the integral in $(3)$ into a sum of integrals over intervals $[k\pi,(k+1)\pi]$ and write (for "large" $L$)
$$\begin{align}
\int_{\sqrt{pL}}^{\sqrt{qL}} \frac{|\sin(x)|}{x}\,dx&\approx\sum_{k=\lfloor \sqrt{pL}/\pi\rfloor}^{\lfloor\sqrt{qL}/\pi\rfloor}\int_{k\pi}^{(k+1)\pi}\frac{|\sin(x)|}{x}\,dx\\\\
&\approx \sum_{k=\lfloor \sqrt{pL}/\pi\rfloor}^{\lfloor\sqrt{qL}/\pi\rfloor}\frac{2}{k\pi}\\\\
&\approx \frac2\pi \left(\log\left(\frac{\lfloor\sqrt{qL}/\pi\rfloor}{\lfloor\sqrt{pL}/\pi\rfloor}\right)\right) \\\\
&\approx \frac1\pi \log(q/p)\tag4
\end{align}$$
Using $(4)$ into $(3)$ yields
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{|\sin(\sqrt{qx})|-|\sin(\sqrt{px})|}{x}\,dx=\frac2\pi \log(q/p)}$$
as was to be shown!
METHODOLOGY $2$: Integrating by Parts
Let $\bar S(x)$ denote that average value the absolute value of the sine function on $[0,x]$. That is,
$$\bar S(x) =\frac1x\int_0^x |\sin(t)|\,dt$$
It is easy to see that the following limits hold:
$$\begin{align}
\lim_{x\to0^+}\bar S(x)&=0\tag5\\\\
\lim_{x\to\infty}\bar S(x)&=\frac2\pi\tag6
\end{align}$$
Integrating by parts the integral on the right-hand side of $(1)$ with $u=\frac1x$ and $v=\int_0^x \left(|\sin(\sqrt{q}t)|-|\sin(\sqrt{p}t)|\right)\,dt$ reveals
$$F(p,q)=2\int_0^\infty \frac{\bar S(\sqrt{q}x)-\bar S(\sqrt{p}x)}{x}\,dx\tag7$$
The integral in $(7)$ is a Frullani integral. Therefore, using $(5)$ and $(6)$ in $(7)$ yields to coveted result
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{|\sin(\sqrt{qx})|-|\sin(\sqrt{px})|}{x}\,dx=\frac2\pi \log(q/p)}$$
NOTE: The approach in the Methodology $2$ is generalized in This Answer.
Best Answer
I'm going to assume that $f$ does not go to 0 on any other point of the interval, so we just have to worry of what happens near 0. On both cases use substitution $u = \sqrt{x}$ then the integral we are interested in becomes
$ 2\int_{0}^{1} \frac{u}{f\left(u\right)} du. $
Now, for item a, you get
$ \lim_{u \to 0} \frac{f\left(u\right)}{u} = f'\left(0\right) \neq 0 $
So you can find a lower bound for $\left|\frac{f\left(u\right)}{u}\right|$ in a neighboorhood of $0$, which gives you an upper bound for $\left|\frac{u}{f\left(u\right)}\right|$ and from this you can conclude.
For item b, you get using L'Hopital
$ \lim_{u \to 0} \frac{f\left(u\right)}{u^{2}} = \lim_{u \to 0} \frac{f'\left(u\right)}{2u} = f''\left(0\right) \neq 0 $
Now we have an upper bound (because the limit exists) and an lower bound, but for $\left|\frac{f\left(u\right)}{u^{2}}\right|$. I am going to assume this is positive, the negative case is analogous, so there are $\epsilon,c,C >0$ such that
$ c > \frac{f\left(u\right)}{u^{2}} > C \ \forall u \in \left[0,\epsilon\right], $
hence
$ 2\int_{0}^{1} \frac{u}{f\left(u\right)}du > 2\int_{0}^{\epsilon} \frac{u}{f\left(u\right)}du > 2\int_{0}^{\epsilon} \frac{u}{cu^{2}} du = 2\int_{0}^{\epsilon} \frac{du}{cu} = +\infty $