Consider the integral
$$ \int_0^1 \frac{dx}{x(3+\sin(\log(x))}. $$ Since $$\int_0^1 \frac{dx}{x(3+\sin(\log(x))}>\int_0^1 \frac{dx}{4x}$$.
The considered integral is infinite.
However, if we compute the anti-derivative of the function, we get,
$$ \int_t^1 \frac{dx}{x(3+\sin(\log(x))}=\frac{1}{\sqrt 2}\tan^{-1}(\frac{1}{2\sqrt2})-\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{3\tan\left(\frac{\log(t)}{2}\right)+1}{2\sqrt2}\right). $$
Clearly, if we take $t\rightarrow 0$, RHS will not diverge to infinity due to the $\tan^{-1}$ term. What am I missing here? I think it has to be something to do with the branch of $\tan^{-1}$.
The problem is the term $\tan\left(\frac{\log(t)}{2}\right)$. Note that the domain of $\tan (x)$ is $x\neq n\pi+\frac\pi2$ for $n\in \mathrm Z$. So when $t\to 0^+$, $\log(t)$ passes through $n\pi+\frac\pi2$ for $n\in \mathrm Z^-$ and hence $\tan\left(\frac{\log(t)}{2}\right)$ changes from $-\infty$ to $\infty$ or from $\infty$ to $-\infty$ infinitely times. Thus $$ \lim_{t\to0^+}\tan^{-1}\left(\frac{3\tan\left(\frac{\log(t)}{2}\right)+1}{2\sqrt2}\right) $$ does not make sense.