Integral $\int_0^\infty \frac{|\sin\sqrt{qx}|-|\sin\sqrt{px}|}{x}dx$

Prove that $$ \int_{0}^{\infty} \frac{\left\vert\sin\left(\sqrt{qx}\right)\right\vert- \left\vert\sin\left(\sqrt{px}\right)\right\vert}{x}\,\mathrm{d}x = \frac{2}{\pi}\,\log\left(\frac{q}{p}\right) $$

This is a Frullani integral, but I am not sure if it converges. Anyway, I investigated it in my article on "fascinating integrals" (see here) if you are interested about how I came to that result.

My interest in this integral is because I solved it using non-traditional methods ( purely based on statistical analysis ), Wolfram Alpha is unable to compute it ( thought it provides the exact value of other Frullani integrals ), and I want to make sure my answer is correct or makes sense, maybe not in the context of Rienman integrals, but some other types of integrals.

Solution 1:

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.

Solution 2:

I also posted the question on Quora, and Joel Campbell proved the result, see the answer on Quora. In short, it starts with a change of variable $x=y^2$, uses the Fourier series $|\sin u|=\frac{2}{\pi} - \frac{4}{\pi}\sum_{n=1}^\infty \frac{\cos(2nu)}{4n^2-1}$, and the fact that $\int_0^\infty \frac{\cos(ax)-\cos(bx)}{x}dx = \log\frac{b}{a}$.