More fun in 2015, Part 19

Let $u=x^{2015}$ , then $du=2015x^{2014}dx\implies\frac{du}{2015x^{2015}}=\frac{dx}{x}\implies\frac{du}{2015u}=\frac{dx}{x}$ :

Substituting into $\left(\frac{1}{\pi}\int_{0}^{\infty}\frac{\sin(x^{2015})}{{x}}dx\right)^{-1}=\left(\frac{1}{\pi}\int_{0}^{\infty}\sin(x^{2015})\frac{dx}{x}\right)^{-1}$

we get

$\left(\frac{1}{\pi}\int_{0}^{\infty}\frac{\sin(u)}{2015u}du\right)^{-1}=\left(\frac{1}{\pi}\cdot\frac{1}{2015}\cdot\frac{\pi}{2}\right)^{-1}=\boxed{4030}$