Not so standard

Let the required integral be $I$ . Substituting $\displaystyle f(x) = \frac{\sin x}{x}$ , in the integral, we obtain, $I = \int \limits_0^\infty \frac{(2x + \sin x)(x - \sin x )}{x^4} \text{d}x$ $\displaystyle \Rightarrow I = \int \limits_0^\infty \frac{2x^2 -x \sin x - \sin ^2 x }{x^4} \text{d}x$ $\displaystyle \Rightarrow I = \int \limits_0^\infty \frac{x^2 -x \sin x + x^2 - \sin ^2 x }{x^4} \text{d}x$ $\displaystyle \Rightarrow I = \int \limits_0^\infty \frac{x -\sin x}{x^3} \text{d}x + \int \limits_0^\infty \frac{x^2 -\sin^2 x}{x^4} \text{d}x \Rightarrow I = A + B$
where, $\displaystyle A= \int \limits_0^\infty \frac{x -\sin x}{x^3} \text{d}x$ and $\displaystyle B= \int \limits_0^\infty \frac{x^2 -\sin^2 x}{x^4} \text{d}x$ .

I will evaluate these integrals not by any theorem but by a nice substitution. For $A : x \to 3y$ and for $B: x \to 2z$ .

Thus, now we have,
$\displaystyle A = \int \limits_0^\infty \frac{3y -\sin 3y}{27y^3} (3 \text{d}y)$ $\displaystyle \Rightarrow 9A = \int \limits_0^\infty \frac{3y -3 \sin y + 4\sin ^3 y}{y^3} \text{d}y$ $\displaystyle \Rightarrow 9A =3 \int \limits_0^\infty \frac{y -\sin y}{y^3} \text{d}y + 4\int \limits_0^\infty \bigg( \frac{ \sin y}{y} \bigg)^3 \text{d}y$ $\displaystyle \Rightarrow 9A = 3A + 4\bigg( \frac{3 \pi }{8} \bigg) \Rightarrow A= \frac{\pi }{4}$

Also,
$\displaystyle B = \int \limits_0^\infty \frac{4z^2 - \sin^2 2z}{16z^4} (2\text{d}z)$ $\displaystyle \Rightarrow B = \frac{4}{8} \int \limits_0^\infty \frac{z^2 - \sin^2 z \cos^2 z}{z^4} \text{d}z$ $\displaystyle \Rightarrow 2B = \int \limits_0^\infty \frac{z^2 - \sin^2 z + \sin^4 z}{z^4} \text{d}z$ $\displaystyle \Rightarrow 2B = \int \limits_0^\infty \frac{z^2 - \sin^2 z}{z^4} \text{d}z + \int \limits_0^\infty \bigg( \frac{ \sin z }{z} \bigg)^4 \text{d}z$ $\displaystyle \Rightarrow 2B = B + \frac{\pi }{3} \Rightarrow B = \frac{\pi }{3}$

Hence,
$\displaystyle I = A + B = \frac{\pi }{4} + \frac{\pi }{3} = \frac{7 \pi }{12}$ .
Thus, $a+b+c = \boxed{20}$ .