I tried to solve this integral and find its original value, unfortunately I did not succeed. I made some variable changes, but I think it has become more complicated.
Numerical Solution: 1.15745
Other: 0.99 [This solution](https://math.stackexchange.com/questions/1271585/evaluate-int-0-infty-sin-tanx-over-xdx)
My solution is as follows:

$$I=\int\_0^{\infty}\frac{\sin(\tan(x))}{x}dx$$
$$I(a)=\int\_0^{\infty}\frac{\sin(a\tan(x))}{x}dx$$
$$I'(a)=\int\_0^{\infty}\frac{\tan(x)\cos(a\tan(x))}{x}dx$$
$$u=tan(x) \iff \arctan{u}=x \Rightarrow \frac{du}{1+u^2}=dx$$
$$I'(a)=\int\_0^{\infty}\frac{u\cos(a\*u)}{\arctan{u}}\frac{dx}{1+u^2}$$

And the other hand:
$$I(a,b)=\int\_0^{\infty}\frac{\sin(\tan(bx))}{x}e^{-ax}dx$$