Following @Nebulae's comment:
Let the standard integral $$\int_0^{\infty} e^{-u^2}\, du = \frac{\sqrt{\pi}}2$$
Next, let $u = \frac{x}{\sqrt2}$, then $du = \frac{dx}{\sqrt2}$
The integrals will be equal:
$$\int_0^{\infty} e^{-u^2}\, du = \int_0^{\infty} \frac1{\sqrt2}e^{\frac{-x^2}2}\, dx = \frac{\sqrt{\pi}}2$$
Lastly, $\int_0^{\infty} \frac1{\sqrt2}e^{\frac{-x^2}2}\, dx = \frac{\sqrt{\pi}}2$
Multiply both sides by root 2
$$\int_0^{\infty} e^{\frac{-x^2}2}\, dx = \sqrt{\frac\pi2}$$
Next one! $$\int_0^\infty x^2 e^{-x^2}\, dx$$
This one is a really cool by-parts trick. One would think to separate it by the $x^2$ term and the $e^{[::]}$ term, but no...
Let $$u = x,\ \ du = dx,\ \ dv = xe^{-x^2},\ \ v = -\frac12e^{-x^2}$$
Now use $\int u\, dv = uv - \int v\, du$
Then we have $$-\frac{x}2e^{-x^2} \Bigr|_0^\infty+ \int_0^\infty \frac12e^{-x^2}\, dx$$
$$=\frac12 \int_0^\infty e^{-x^2}\, dx = \frac{\sqrt{\pi}}4$$
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
Here is an approach.
You may write
as desired.