Prove that $\int_0^1 \frac{1}{1+\ln^2 x}\,dx = \int_1^\infty \frac{\sin(x-1)}{x}\,dx $ I've found the following identity.

$$\int_0^1 \frac{1}{1+\ln^2 x}\,dx = \int_1^\infty \frac{\sin(x-1)}{x}\,dx $$ 

I could verify it by using CAS, and calculate the integrals in term of exponential and trigonometric integrals, then using identities between them. However, I think there is a more elegent way to prove it.

How could we prove this identity?

Also would be nice to see some references.
 A: Hint. One may observe that

$$
\frac1{1+\ln^2 x}=-\Im \frac1{i-\ln x}=-\Im \int_0^{\infty}e^{-(i-\ln x)t}dt,\quad x \in (0,1),
$$ 

gives

$$
\begin{align}
\int_0^1 \frac{1}{1+\ln^2 x}\,dx&=-\Im \int_0^1\!\!\int_0^{\infty}e^{-(i-\ln x)t}dt\:dx\\\\
&=-\Im \int_0^{\infty}\!\!\left(\int_0^1x^t dx\right)e^{-it}dt\\\\
&=-\Im \int_0^{\infty}\!\!\frac1{t+1} e^{-it}dt\\\\
&=\int_0^{\infty}\!\! \frac{\sin t}{t+1}  dt\\\\
&= \int_1^\infty \frac{\sin(x-1)}{x}\,dx
\end{align}
$$ 

as announced.
A: A variant:
$$
\int_1^{\infty}\frac{\sin(x-1)}{x}dx=\Im\int_1^{\infty}\frac{e^{i(x-1)}}{x}dx= \quad (1)\\ \Im\int_1^{\infty}\int_0^{\infty}e^{i(x-1)-xt}dtdx=\Im\int_0^{\infty}\int_1^{\infty}e^{i(x-1)-xt}dxdt=\quad (2) \\
\Im\int_0^{\infty}\frac{e^{-t}}{t-i}dt=-\int_0^{\infty}\frac{e^{-t}}{1+t^2}dt=\quad (3) \\
\int_0^1\frac{1}{1+\log^2(x)}dx\quad (4)
$$
Explanation 
(1)   $\quad\Im (e^{i x})=\sin(x)$
(2) $\quad$ $\frac{1}{x}=\int_{0}^{\infty} e^{-xt}dt$ for $x>0$
(3) $\quad$ straightforward integration and $\Im\left(\frac{1}{z}\right)=\frac{\Im(\bar{z})}{|z|^2}$
(4) $\quad    e^{-t}=x,dt=\frac{-1}{x}$
A: Define $I(a)=\int_0^{\infty} e^{-(x+1)a}\frac{\sin(x)}{x+1}dx$.
Then $\displaystyle I'(a)=-\int_0^{\infty} e^{-(x+1)a}\sin(x)dx=-\frac{e^{-a}}{a^2+1}$, and since $\lim_{a\to \infty}  I(a)=0$, we have  $\ \displaystyle I(0)=\int_0^{\infty} \frac{\sin(x)}{x+1}=\int_0^{\infty}\frac{e^{-a}}{a^2+1}da=\int_0^1\frac{dx}{1+\ln^2 x}.$
