I need help spotting the error in my calculation here. Wolfram alpha tells me my integral should work out to $-1$, rather than diverging. I've worked this out as follows, where unless I'm really missing something $\lim\limits_{t\to\infty}\cos(\ln(t))$ should diverge and not equal $0$:
\begin{equation} \int_{e^\pi}^\infty\frac{\sin(\ln(x))dx}{x} = \left.-\lim\limits_{t\to\infty}\cos(\ln(x))\right\rvert_{e^\pi}^t = \left.\lim\limits_{t\to\infty}\cos(\ln(x))\right\rvert_t^{e^\pi} = -1 - \lim\limits_{t\to\infty}\cos(\ln(t)) \end{equation}
As @Sine of the Time suggests, wolfram is incorrect because the integral just oscillates, this can be observed due to the following : Take $M > \pi$, then $$ \int_{e^\pi}^{e^M} \frac{\sin{(\log{x})}}{x}dx \overset{\log{(x)} = t}{=} \int_{\pi}^{M} \sin{(t)} dt = -\cos{(t)}\big|_{\pi}^{M} = -2\cos^2{\left(M/2\right)}. $$ So how do we make sense of Wolfram's result? I am guessing the integral $\int_{\pi}^{\infty} \sin{(t)} dt$ can be "assigned" the value $-1$ as $$ \int_{\pi}^{\infty} \sin{(x)}e^{-\delta x} dx = -\frac{e^{-\pi\delta}}{\delta^2+1},$$
therefore taking small values of delta, the right side converges to $-1$.