TITLE: Computing $I=\int_0^{\infty}\frac{\sin(\tan(x))}{x}dx$
QUESTION [5 upvotes]: 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:
$$I(a,b)=\int_0^{\infty}\frac{\sin(\tan(bx))}{x}e^{-ax}dx$$

REPLY [3 votes]: 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
$$\int_0^\infty\frac{\sin(\tan x)}{x}dx=\int_0^\frac{\pi}{2}\frac{\sin(\tan x)}{\sin x}\cdot \cos xdx=\int_0^\frac{\pi}{2}\frac{\sin(\tan x)}{\tan x}dx$$
$$=\int_0^\infty\frac{\sin \theta}{\theta(1+\theta^2)}d\theta
=\int_0^\infty\frac{\sin\theta}{\theta}d\theta-\int_0^\infty\frac{\theta\sin\theta}{1+\theta^2}d\theta=\frac{\pi}{2}-\frac{\pi}{e\pi}=\frac{\pi}{2}\Big(1-\frac{1}{e}\Big)$$