Is there any relationship between these two equal integrals $\int_0^\infty\frac{\sin x}{x + x^2}dx=\int_0^\infty\frac{\pi-2\tan^{-1}x}{2 e^x}dx$

Question

With some intermediate derivation results I found the two integrals are exactly the same. But why? \begin{align} I=\int_0^\infty\frac{\sin x}{x + x^2}\mathrm dx=\int_0^\infty\frac{\pi-2\tan^{-1}x}{2 e^x}\mathrm dx \end{align}

Both are equal to \begin{align} I&=\operatorname{Si}(1)\cos(1)-\operatorname{Ci}(1)\sin(1)+\frac{\pi}{2}\left(1-\cos(1)\right)\\ &\approx0.9493467025590832615920\ldots \end{align} where $\operatorname{Ci}$ and $\operatorname{Si}$ are cosine and sine integrals, respectively.

Note: The nominator in integrand is the double of Laplace transform of $\operatorname{sinc}$ function $$\mathcal{L}\left[\frac{\sin t}{t}\right](x)=\cot^{-1}{\frac1x}=\frac{\pi}{2}-\tan^{-1}x$$

Answer 1

Their equivalency is established below\begin{align} &\int_0^\infty\frac{\pi-2\tan^{-1}x}{2 e^x}dx\\ =& \int_0^\infty \left(\frac\pi2-\tan^{-1}x\right)d(1-e^{-x}) \overset{ibp} = \int_0^\infty \frac{1-e^{-x}}{1+x^2}dx\\ = &\int_0^\infty (1-e^{-x})\left( \int_0^\infty \sin y\ e^{-xy}\ dy \right)dx\\ = & \int_0^\infty \sin y \left(\int_0^\infty e^{-xy}(1-e^{-x})dx\right)dy\\ =& \int_0^\infty\frac{\sin y}{y + y^2}dy \end{align}

Answer 2

Integrate the function $$f(z) = \frac{e^{iz}}{z(1+z)} $$ around the contour $$[r, R] \cup Re^{i[0,\pi/2]} \cup [iR, ir] \cup re^{i[\pi/2,0]}. $$

$Re^{i[0,\pi/2]}$ is large quarter-circle in the first quadrant of the complex plane of radius $R$, and $re^{i[\pi/2,0]} $ is a small quarter-circle in the first quadrant of the complex plane of radius $r$.

Let's call the small quarter-circle $C_{r}$.

Letting $R \to \infty$, the integral vanishes on the large quarter-circle by Jordan's lemma.

So we have $$\int_{r}^{\infty} \frac{e^{ix}}{x(1+x)} \, \mathrm dx - \int_{r}^{\infty} \frac{e^{-x}}{x(1+ix)} \, \mathrm dx + \int_{C_{r}} f(z) \, \mathrm dz= 0.$$

Since $f(z)$ is a simple pole and $C_{r}$ is a clockwise-oriented quarter-circle, $\int_{C_{r}} f(z) \, \mathrm dz$ goes to $-\frac{i \pi}{2} \operatorname*{Res}_{z = 0}f(z) = -\frac{i \pi}{2}$ as $r \to 0$.

Then equating the imaginary parts on both sides of equation, we have $$ \begin{align} \int_{0}^{\infty} \frac{\sin (x)}{x(1+x)} \, \mathrm dx &= - \int_{0}^{\infty} \frac{e^{-x}}{1+x^{2}} \, \mathrm dx + \frac{\pi}{2} \\ &\overset{ibp}{=} - \int_{0}^{\infty} \arctan(x) e^{-x} \, \mathrm dx + \frac{\pi}{2} \\ &= - \int_{0}^{\infty} \arctan(x) e^{-x} \, \mathrm dx + \frac{\pi}{2} \int_{0}^{\infty} e^{-x} \, \mathrm dx \\ &= \int_{0}^{\infty} \left(\frac{\pi}{2} - \arctan(x) \right)e^{-x} \, \mathrm dx \end{align}$$

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If we equate the real parts on both sides of the equation, we get $$\int_{0}^{\infty} \left( \frac{\cos(x)}{1+x} - \frac{e^{-x}}{1+x^{2}} \right) \frac{\mathrm dx}{x} = 0. $$