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

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 $$\mathcal{L}\left[\frac{\sin t}{t}\right](x)=\cot^{-1}{\frac1x}=\frac{\pi}{2}-\tan^{-1}x$$

• Not saying your question is fruitless, but many integrals of different functions with the same or different bounds are equal. Example $$\int_0^2 2x dx = \int_0^2 2 dx$$, but that doesn't imply $2x = 2$? This statement is especially true when the limits are $0$ and $\infty$
– Masd
Commented Mar 15 at 13:52

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}

• Brilliant! How could you think of the bridge between them! They look so different. Commented Mar 15 at 21:51
• @MathArt - Thanks. I have seen more such related yet quite different integrals, e.g. \begin{align} \int_0^1 \frac{2\sin^{-1} x\cos^{-1} x}{x}dx= \int_0^1 \frac{\ln^2 x}{1-x^2}dx \end{align} Commented Mar 15 at 22:15
• This is a beautiful identity! Thanks for sharing... Commented Mar 16 at 13:50

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}

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