Riesel and Gohl show in Some Calculations Related to Riemann's Prime Number Formula that $$\sum_{n=1}^N \frac{\mu (n)}{n}\left(\int_{x^{\frac{1}{n}}}^\infty \frac{\mathrm dt}{t(t^2-1)\ln t}-\ln 2\right)=\frac{1}{2\ln x}\sum_{n=1}^N \mu (n)+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}+\epsilon (x,N)$$ where $$\epsilon (x,N)=-\sum_{n=N+1}^\infty \frac{\mu (n)}{n}\left(\frac{1}{2}\ln\ln x+C\right)+\frac{1}{2}\sum_{n=N+1}^\infty \frac{\mu (n)\ln n}{n}+\sum_{n=N+1}^\infty \frac{\mu (n)}{n}\sum_{k=1}^\infty \frac{1}{\pi k}\arctan\frac{\ln x}{n\pi k}$$ where $$C=\sum_{k=1}^\infty \frac{1}{\pi k}\arctan\frac{1}{\pi k}+\int_1^\infty \frac{\mathrm du}{u(e^{2u}-1)}-\ln 2+\frac{1}{2}.$$ Now, $\epsilon\to 0$ as $N\to\infty$. If $\sum_{n=1}^\infty \mu (n)=-2$, then $$\sum_{n=1}^\infty \frac{\mu (n)}{n}\int_{x^{\frac{1}{n}}}^\infty \frac{\mathrm dt}{t(t^2-1)\ln t}=\frac{1}{\pi}\arctan\frac{\pi}{\ln x}-\frac{1}{\ln x}.$$
But how does one make sense of $\sum_{n=1}^\infty \mu (n)=-2$? Of course, it can be obtained incorrectly by just plugging $s=0$ in $$\frac{1}{\zeta (s)}=\sum_{n=1}^\infty \frac{\mu (n)}{n^s},\, \operatorname{Re}s\gt 1.$$