TITLE: Evaluate $\int_0^{\infty} {\sin(\tan(x)) \over x}dx$
QUESTION [18 upvotes]: I tried to solve it the Feynman way and defined:
$$I(a):=\int_0^{\infty} {\sin(\tan(a \cdot x)) \over x} \ dx$$
And look what happens when one substitutes $u=ax$ $(a>0)$:
$$I(a)=\int_0^{\infty} {\sin(\tan(u)) \over u} \ du = I(1)$$
Which implies that $I(a)=const$ for $a>0$. More generally $I(a)=c \cdot sign(a)$. I wonder whether this can help.
I recalled that in order to solve $\int_0^{\infty} {\sin(x) \over x} \ dx$ using the Feynman technique one had to define $I(a):=\int_0^{\infty} {\sin(x) \over x} e^{-a \cdot x}\ dx$ and differentiate it. Consequently I suspect we should define $I(a):=\int_0^{\infty} {\sin(\tan x) \over x} e^{-a \cdot x}\ dx$, but differentiation yields:
$$I'(a)=-\int_0^{\infty} {\sin(\tan x)} e^{-a \cdot x}\ dx$$
Which is another difficult integral.
Any help?
(please try to avoid gamma functions)

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