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

Question

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.

Answer 1

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.

Answer 2

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 $\ \displaystyle I(0)=\int_0^{\infty} \frac{\sin(x)}{x+1}=\int_0^{\infty}\frac{e^{-a}}{a^2+1}da=\int_0^1\frac{dx}{1+\ln^2 x}.$

Answer 3

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) $\quad e^{-t}=x,dt=\frac{-1}{x}$