I want to calculate the improper integral $$\int_0^{\to \infty} \frac{\sin(ax)\sin(bx)}{x} dx$$ where $a,b > 0$ and $a \neq b$.
One of my guess was to use the identity $$\sin(ax)\sin(bx) = \frac{1}{2}\left(\cos((a-b)x) – \cos((a+b)x)\right)$$ and then use the Frullani's theorem, but $\cos(kx)$ does not admit any limit when $x \to \infty$. Also, I know that $$x \mapsto \frac{2\sin(ax)}{x}$$ is the Fourier transform$^1$ of the function $x \mapsto \mathbf{1}_{[-a,a]}(x)$, but I can't figure out how to use this identity. Therefore, I tried to integrate by parts:
(1) taking $v'(x) = \sin(ax)\sin(bx)$ and $u(x) = \frac{1}{x}$ leads to an integral of the form $$\int_0^{\to \infty} \frac{\frac{1}{a-b} \sin((a-b)x) – \frac{1}{a+b} \sin((a+b)x)}{x^2} dx$$ and I can't go further in the computation ;
(2) taking $v'(x) = \sin(bx)$ and $u(x) = \frac{\sin(ax)}{x}$ leads to an integral of the form $$\int_0^{\to \infty} \frac{ax\cos(ax)\cos(bx) – \sin(ax)\cos(bx)}{x^2} dx$$ and I fail to do the calculation.
For context, this problem is part of an introduction to harmonic analysis course, so we can use Fourier transform and its properties.
$1.$ The Fourier transform of $f$ is defined by $\hat{f}(y) = \int_{\mathbb{R}} e^{ixy} f(x)dx$
Best Answer
The Frullani formula does not require existence of the limit at infinity provided that the integrals $$\ \int\limits_1^\infty {f(x)\over x}\,dx \qquad \int\limits_0^1{f(x)-f(0)\over x}\,dx \qquad \qquad \qquad (*)$$ are convergent. Namely, if $f$ is continuous on $[0,\infty),$ then $$\int\limits_0^\infty {f(\alpha x)-f(\beta x)\over x}\,dx =f(0)\,\log {\beta\over \alpha},\quad \alpha,\beta>0\qquad (**)$$ In your question $\alpha=|a-b|$ and $\beta =a+b.$
Here is a proof of $(**)$.
Assume $0<\alpha<\beta.$ Then $$\int\limits_0^\infty {f(\alpha x)-f(\beta x)\over x}\,dx =\lim_{N\to \infty }\left [ \int\limits_0^N{f(\alpha x)-f(0)\over x}\,dx -\int\limits_0^N{f(\beta x)-f(0)\over x}\,dx\right ]$$ $$ =\lim_{N\to \infty }\left [ \int\limits_0^{\alpha N}{f(y)-f(0)\over y}\,dy -\int\limits_0^{\beta N}{f(y)-f(0)\over y}\,dy\right ]= -\lim_{N\to \infty } \,\int\limits_{\alpha N}^{\beta N}{f(y)-f(0)\over y}\,dy $$ $$= f(0)\,\log{\beta\over \alpha}-\lim_{N\to \infty}\,\int\limits_{\alpha N}^{\beta N} {f(y)\over y}\,dy= f(0)\,\log{\beta\over \alpha}$$