Does $\int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx $ converge? Does $ \displaystyle \int_{0}^{\infty} \ \frac{\sin (\tan x)}{x} dx $ converge?
$ \displaystyle \int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin (\tan x)}{x} \ dx + \sum_{n=1}^{\infty} \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx $
The first integral converges since $\displaystyle \frac{\sin (\tan x)}{x}$ has a removable singularity at $x=0$ and is bounded near $ \displaystyle \frac{\pi}{2}$.
And $ \displaystyle \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx$ converges since $\displaystyle \frac{\sin (\tan x)}{x}$ is bounded near $\pi(n-\frac{1}{2})$ and $\pi(n+\frac{1}{2})$.
But does $ \displaystyle \sum_{n=1}^{\infty} \int_{\pi(n-\frac{1}{2})}^{\pi(n+\frac{1}{2})} \frac{\sin (\tan x)}{x} \ dx$ converge?

Yes.
Note that 
$$\begin{align} I_n:=\int_{\pi(n-\frac12)}^{\pi(n+\frac12)}\frac{\sin(\tan(x))}{x}\,\mathrm dx&=\int_0^{\frac\pi2}\left(\frac1{n\pi+x}-\frac1{n\pi-x}\right)\sin(\tan(x))\,\mathrm dx\\&=\int_0^{\frac\pi2}\frac{-2x}{n^2\pi^2-x^2}\sin(\tan(x))\,\mathrm dx\\
\end{align}$$
With $A:=\int_0^{\frac\pi2}\max\{-2x\sin(\tan (x)),0\}\,\mathrm dx$, $B:=\int_0^{\frac\pi2}\min\{-2x\sin(\tan (x)),0\}\,\mathrm dx$ (which both converge), we can thus estimate
$$  \frac{1}{n^2\pi^2-\pi^2/4}B+\frac{1}{n^2\pi^2-0}A\le I_n\le \frac{1}{n^2\pi^2-\pi^2/4}A+\frac{1}{n^2\pi^2-0}B,$$
The difference between these bounds and $\frac1{n^2\pi^2}(A+B)$ is governed by 
$$\frac{1}{n^2\pi^2-\frac{\pi^2}{4}}-\frac1{n^2\pi^2}=\frac1{4\pi^2 n^4-\pi^2 n^2},$$
hence 
$$ I_n=\frac1{n^2\pi^2}(A+B)+O(n^{-4}).$$
We conclude that $\sum I_n$ converges at least as good as $\sum\frac1{n^2}$.
