Computing $I=\int_0^{\infty}\frac{\sin(\tan(x))}{x}dx$ 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
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:
$$I(a,b)=\int_0^{\infty}\frac{\sin(\tan(bx))}{x}e^{-ax}dx$$
 A: The integrand function is even, hence
$$ I = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{\sin(\tan x)}{x}\,dx $$
which due to the periodicity of the tangent function equals
$$ I = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\sin(\tan x)\sum_{k\in\mathbb{Z}}\frac{dx}{x+k\pi}.$$
The sum over the integers has to be intended in the symmetric sense, i.e. as
$$ \frac{1}{x}+\sum_{n\in\mathbb{N}}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right)=\cot(x) $$
where the last equality follows from considering the logarithmic derivative of the Weierstrass product for the sine function (or directly from Herglotz' trick). As a consequence,
$$ I = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{\sin(\tan x)}{\tan(x)}\,dx = \int_{0}^{\pi/2}\frac{\sin(\tan x)}{\tan x}\,dx = \int_{0}^{+\infty}\frac{\sin(u)}{u(1+u^2)}\,du = \frac{\pi}{2}-\int_{0}^{+\infty}\frac{u\sin u}{1+u^2}\,du $$
where the last equality follows from a partial fraction decomposition and Dirichlet's integral $\int_{0}^{+\infty}\frac{\sin(u)}{u}\,du=\frac{\pi}{2}$.
By the residue theorem, or by the self-adjointness of the Laplace transform, the last integral equals $\frac{\pi}{2e}$. The conclusion is that
$$ \boxed{\int_{0}^{+\infty}\frac{\sin(\tan x)}{x}\,dx = \color{red}{\frac{\pi}{2}\left(1-\frac{1}{e}\right)}\approx 0.992933} $$
and I wonder if there are slick ways to prove $I<1$ without resorting to explicit computations (or, at least, using only a very limited amount of them). Update: I found one. The Archimedean approximation $\pi < \frac{22}{7}$ can be deduced from the fact that the integral $\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx$ is positive and it equals $\frac{22}{7}-\pi$. The inequality $e<\frac{11}{4}$ can be deduced from the fact that
$$ e=2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots < 2+\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{3^2}+\ldots\right)=\frac{11}{4}.$$
These inequalities lead to $\frac{\pi}{2}<\frac{11}{7}$ and $1-\frac{1}{e} < \frac{7}{11}$, so $\frac{\pi}{2}\left(1-\frac{1}{e}\right)<1$.
A: Lobachevsky Integral Formula: If $f(x)$ meet $f(x+π)=-f(x)$ and $f(−x)=f(x)$.
Then
$$\int_0^\infty f(x)\cdot\frac{\sin x}{x}dx=\int_0^\frac{\pi}{2}f(x)\cdot \cos xdx$$
so
$$\int_0^\infty\frac{\sin(\tan x)}{x}dx=\int_0^\infty\frac{\sin(\tan x)}{\sin x}\cdot\frac{\sin x}{x}dx$$
where
$$f(x)=\frac{\sin(\tan x)}{\sin x}$$
so
$$\int_0^\infty\frac{\sin(\tan x)}{x}dx=\int_0^\frac{\pi}{2}\frac{\sin(\tan x)}{\sin x}\cdot \cos xdx=\int_0^\frac{\pi}{2}\frac{\sin(\tan x)}{\tan x}dx$$
$$=\int_0^\infty\frac{\sin \theta}{\theta(1+\theta^2)}d\theta
=\int_0^\infty\frac{\sin\theta}{\theta}d\theta-\int_0^\infty\frac{\theta\sin\theta}{1+\theta^2}d\theta=\frac{\pi}{2}-\frac{\pi}{e\pi}=\frac{\pi}{2}\Big(1-\frac{1}{e}\Big)$$
