Calculate $\int_0^{\infty} \frac{\sin(ax)\sin(bx)}{x} dx$

Question

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.

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$1.$ The Fourier transform of $f$ is defined by $\hat{f}(y) = \int_{\mathbb{R}} e^{ixy} f(x)dx$

Answer 1

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 $$\int\limits_0^\infty {f(\alpha x)-f(\beta x)\over x}\,dx =\lim_{N\to \infty }\left [ \int\limits_0^N{f(\alpha x)-f(0)\over x}\,dx -\int\limits_0^N{f(\beta x)-f(0)\over x}\,dx\right ]$$ $$ =\lim_{N\to \infty }\left [ \int\limits_0^{\alpha N}{f(y)-f(0)\over y}\,dy -\int\limits_0^{\beta N}{f(y)-f(0)\over y}\,dy\right ]= -\lim_{N\to \infty } \,\int\limits_{\alpha N}^{\beta N}{f(y)-f(0)\over y}\,dy $$ $$= f(0)\,\log{\beta\over \alpha}-\lim_{N\to \infty}\,\int\limits_{\alpha N}^{\beta N} {f(y)\over y}\,dy= f(0)\,\log{\beta\over \alpha}$$

Answer 2

HINT: Using the complex version of Frullani's theorem we get for $a,b>0$ and $\not=b$, \begin{align*} \int_{0}^{+\infty}\frac{\sin ax \sin bx}{x}\, {\rm d}x&=\frac{1}{4}\int_{0}^{+\infty}\frac{e^{-iz(b-a)}-e^{iz(b+a)}-\left(e^{-iz(b+a)}-e^{iz(b-a)}\right)}{z}\, {\rm d}z=\frac{1}{2}\log\left(\frac{a+b}{a-b}\right) \end{align*} You can now complete the details. Then, a solution using some property of the Fourier transform should be possible. However, for now I know that this method works well and of course Laplace transform too.

Answer 3

Probably, the straightforward way to evaluate this integral is to use the Frullani's integral :

$$\int_0^\infty\frac{f(ax)-f(bx)}{x}dx=\big(f(\infty)-f(0)\big)\ln\frac{a}{b}$$ In our case $$I=\frac{1}{2}\int_0^\infty\frac{\cos(a-b)x - \cos(a+b)x}{x}dx; \quad f(x)=\cos x$$ Let $a>b>0$. The problem is that $f(\infty)$ is not defined. To cope with this problem we can, as an option, to prove first that $J=\int_0^\infty\big(\frac{\cos t}{t}-\frac{e^{-t}}{t}\big)dt=0$.

Indeed, $J=\Re\int_0^\infty\big(\frac{e^{it}}{t}-\frac{e^{-t}}{t}\big)dt$. Now, we go in the complex plane and consider the closed contour $z=0\,\to\,z=R$ along the axix $X$, then along a quarter-circle, counterclockwise, $z=R\,\to\, z=Re^{\pi i/2}$, and then to $z=0$ along the axis $Y$.

There is no singularity at $z=0$; the integral along the quarter circle can be evaluated by means of Jordan's lemma and $\to0$ at $R\to\infty$. There are no singularities inside the contour $\,\,\Rightarrow\,\oint\big(\frac{e^{iz}}{z}-\frac{e^{-z}}{z}\big)dz=2\pi i\sum Res=0$.

Therefore ($t\to ix$), $$\int_0^\infty\big(\frac{e^{it}}{t}-\frac{e^{-t}}{t}\big)dt+\int_\infty^0\big(\frac{e^{-x}}{x}-\frac{e^{-ix}}{x}\big)dx=0$$ $$J=\Re\int_0^\infty\big(\frac{e^{it}}{t}-\frac{e^{-t}}{t}\big)dt=-\Re\int_0^\infty\big(\frac{e^{-it}}{t}-\frac{e^{-t}}{t}\big)dt=-J\,\Rightarrow\,J=0$$ .. and the desired integral $$I=\frac{1}{2}\int_0^\infty\frac{e^{-(a-b)x} - e^{-(a+b)x}}{x}dx=\frac{1}{2}\ln\frac{a+b}{a-b}$$