Trigonometrics become Logarithmics

$I = \large \int\limits_{0}^{\infty} 2017 \sin 2016x \sin 2015x \frac {dx}{x}$

The integrand above may be transformed using this identity:

$\sin A \sin B= \frac{1}{2} [ \cos (A-B) -\cos (A+B) ]$

by doing so we get

$\large I = \frac{2017}{2} \int\limits_{0}^{\infty} \frac{\cos x - \cos 4031x }{x}dx$

which follows the structure of Frullani's Integral . That is, for an everywhere continuous and differentiable function $f(x)$ that is bounded at infinity and whose derivative is also continuous, the following equation holds.

$\int\limits_0^{\infty} \frac{ f(ax) - f(bx) }{x} dx = [ f(0) - f(\infty) ] \ln \frac{b}{a}$

So, by inspection, we find that $f(x) = \cos x$ . And that simplifies our integral to

$\frac{2017}{2} [ \cos(0) - \cos(\infty)] \ln -\frac{4031}{1}$

However, we don't have $\cos(\infty)$ . So why don't we consider multiplying $f(x)$ by an arbitrary factor $e^{-\epsilon x}$ , where $\epsilon$ is an infinitesimal parameter, and calculate its limit as $\epsilon \to 0$ .

So from there we get the following:

$\lim\limits_{\epsilon \to 0} f(0) = 1$

$\lim\limits_{\epsilon \to 0} f(\infty) = 0$

And that makes it

$I = \frac{2017}{2} (1 - 0) \ln \frac{4031}{1}$

$I = \frac{2017}{2} \ln (4031)$

and that's just about it. $\boxed{2017+4031+2=6050}$ .