I want to determine if the following integral converges
$$\intop_{1}^{2}\frac{\sin\left(x\right)}{\log\left(x\right)}dx$$
My intuition tells me it diverges. I thought the way to solve that woudln't be by definition but by finding another function that is smaller than that and use integral comparison test. I also know that $$\lim_{x\rightarrow1}\log=0$$ but I don't see how that helps me
It suffices to check that for $x\to 1$, we have $$\frac{\sin x}{\ln x}\sim \frac{c}{x-1},$$ where $c=\sin(1)$.
Hence, your integral converges iff. $$\int_1^2\frac{c}{x-1}\mathrm d x=c\int_0^1\frac{1}{x}\mathrm d x$$ converges.
Does it?
Use the fact that $logx \leq x-1$. So $\frac{sinx}{lnx} \geq \frac{sin1}{x-1}$ for $x \in (1,2]$, and $\intop_{1}^{2}\frac{\sin\left(1\right)}{x-1}$ diverges. So $\intop_{1}^{2}\frac{\sin\left(x\right)}{\log\left(x\right)}$ divergent.
$f: x\mapsto \frac{\sin(x)}{\ln(x)} $ is continuous at $ (1,\frac{\pi}{2} ]$ and positive.
in the other hand,
$$\ln(x)=\ln(1+(x-1))$$ thus $$f(x)\sim \frac{\sin(1)}{x-1} \;\;\;(x\to 1^+)$$
and
$$\int_1^{\frac{\pi}{2}}\frac{dx}{x-1}$$ is divergent. So, your integral is Divergent.
Notice that $e^x \ge 1+x$ for $x\in(-1, \infty)$. This means that $\log(1+x) \le x$ for $x\in(-1,\infty)$. Hence, $\log(x)\le x-1$ and $1/\log x \ge (x-1)^{-1} $ for $x\in(0, \infty)$. Now, you know that $|\sin x/\log(x)| \ge |\sin(x)|/(x-1)$ on $[1,2]$. The norm of your integral is thus bounded below by $$\int_1^2\frac{|\sin x|}{x-1}\,dx,$$ which diverges because of the singularity at $x=1$. Conclusion: your integral also diverges.
We have $$\frac{\sin x}{\log x}>\frac1{2\log x}\quad\forall x\in(1,2]$$ and as such, by finding that $$\int_1^2\frac1{2\log x}\,dx=\frac12\left(\text{Li}(2)-\text{Li}(1)\right)$$ is divergent, we can conclude that $$\int_1^2\frac{\sin x}{\log x}\,dx$$ is divergent as well.
