Integral of the day XXIV

A miserious integral related to Lobachevsky

Today we show the proof of this strange integral posted by @integralsbot \[ \int_{0}^\infty \frac{\sin(\tan(x))}{x} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right) \] The trick to solve this integral is a variation to the Lobachevsky integral formula for odd functions and the application of the residue theorem.

Proof:

Recall the Lobachevsky integral formula for odd functions:

Theorem. Let $f(x)$ an odd, complex-valued, $\pi$-periodic function such that $\displaystyle \frac{f(x)}{\tan x}$ is absolutely integrable over a single period.

Therefore \[ \int_{0}^{\infty} \frac{f(x) }{x}dx = \int_{0}^{\frac{\pi}{2}} \frac{f(x)}{\tan\left(x \right)} dx \tag{*} \] Back to the integral:

Note that the function \[ f(x) = \sin(\tan(x)) \] is an odd function that satisfies all the hypothesis of the theorem.

Hence \begin{align*} I =& \int_{0}^\infty \frac{\sin(\tan(x))}{x} dx \\ = & \int_{0}^{\frac{\pi}{2}} \frac{\sin(\tan(x))}{\tan(x)} dx \quad \textrm{ from (*) } \\ =& \int_{0}^{\frac{\pi}{2}} \frac{\sin(w)}{w(1+w^2)} dw \quad \left(w \mapsto \tan x\right)\\ =& \int_{0}^\infty \frac{\sin (w)}{w} dw -\int_{0}^{\infty} \frac{w\sin(w)}{1+w^2} dw \\ =& \int_{0}^\infty \frac{\sin (w)}{w} dw -\frac{1}{2}\int_{-\infty}^{\infty} \frac{w\sin(w)}{1+w^2} dw \quad (\textrm{the integrand is even}) \\ =& \underbrace{\int_{0}^\infty \frac{\sin (w)}{w} dw}_{\textrm{Dirichlet integral}} - \Im \left(\underbrace{\int_{-\infty}^{\infty} \frac{we^{iw}}{1+w^2} dw }_{J}\right) \\ =& \frac{\pi}{2}- \Im \left(\underbrace{\int_{-\infty}^{\infty} \frac{we^{iw}}{1+w^2} dw }_{J}\right) \\ \end{align*} The first integral is the classic Dirichlet integral, the second is a little bit more difficult: Let $R>1$ and consider the following complex function \[ f(z)= \frac{ze^{iz}}{1+z^2} \] This function has two poles at $z= \pm i$.

Now consider the following contour (half upper circle): \[ C_{1} = t \quad t\in (-R,R)\] \[ C_{2} = Re^{i\theta} \quad \theta \in \left[0,\pi\right]\] Hence \[ \oint_{C=C_{1}\cup C_{2}} \frac{ze^{iz}}{1+z^2} dz = \int_{C_{1}} f(z) dz + \oint_{C_{2}} f(dz)dz = \int_{-R}^{R} \frac{te^{it}}{1+t^2} dt + \int_{0}^{\pi} \frac{iR^2e^{2i\theta}e^{iRe^{i\theta}}}{1+Re^{i\theta}} d\theta \] We will show that the integral in the right hand side will vanish as $R\to \infty$. For that end we will make use of the Jordan's Lemma:

Jordan's lemma. Consider a complex-valued, continuous function $f$, defined on a semicircular contour \[C_R = \{R e^{i \theta} \mid \theta \in [0, \pi]\}\] of positive radius $R$ lying in the upper half-plane, centered at the origin. If the function $f$ is of the form \[f(z) = e^{i a z} g(z) , \quad z \in C_R ,\] with a positive parameter $a$, then Jordan's lemma states the following upper bound for the contour integral: \[\left| \int_{C_R} f(z) \, dz \right| \le \frac{\pi}{a} M_R \quad \text{where} \quad M_R := \max_{\theta \in [0,\pi]} \left| g \left(R e^{i \theta}\right) \right| \] with equality when $g$ vanishes everywhere, in which case both sides are identically zero.

Applying the lemma to $f(z)$: \[ f(z)= \frac{ze^{iz}}{1+z^2} = e^{iz} g(z) \quad \textrm{ where } g(z) = \frac{z}{1+z^2} \] Hence \[\left| \int_{C_2}\frac{ze^{iz}}{1+z^2} \, dz \right| \le \pi M_{R} \] where \begin{align*} M_R = \max_{\theta \in [0,\pi]} \left| g \left(R e^{i \theta}\right) \right| = &\max_{\theta \in [0,\pi]} \left| \frac{Re^{i\theta}}{1+R^2e^{i2\theta}} \right|\\ =& \max_{\theta \in [0,\pi]} \frac{R}{\left| 1+R^2e^{i2\theta}\right|} \\ =& \max_{\theta \in [0,\pi]} \frac{R}{\left| 1-(-R^2e^{i2\theta})\right|} \\ \leq & \frac{R}{\left| 1-R^2\right|}\\ =& \frac{R}{R^2-1} \quad (R>1) \end{align*} Therefore \[\left| \int_{C_2}\frac{ze^{iz}}{1+z^2} \, dz \right| \le \pi M_{R} \leq \frac{\pi R}{R^2-1} \] Taking the limit as $R \to \infty$ \[ \int_{C_2}\frac{ze^{iz}}{1+z^2} \, dz =\int_{0}^{\pi} \frac{iR^2e^{2i\theta}e^{iRe^{i\theta}}}{1+Re^{i\theta}} d\theta= 0 \] Hence, since the only pole in the upper half circle is $z=i$ we have by the residue theorem \[ J = \int_{-\infty}^{\infty} \frac{te^{it}}{1+t^2} dt = \oint_{C} \frac{ze^{iz}}{1+z^2} dz = 2\pi i \operatorname{Res}(f,i) = \lim_{z \to i } 2\pi i \frac{ze^{iz}}{(z+i)} = \frac{\pi i}{e} \] Therefore \[ I = \frac{\pi}{2} - \Im(J) = \frac{\pi}{2} -\frac{1}{2}\Im \left(\frac{\pi i}{e}\right) = \frac{\pi}{2} - \frac{\pi}{2e} = \frac{\pi}{2}\left(1-\frac{1}{e}\right) \] Hence, we can conclude \[\boxed{ \int_{0}^\infty \frac{\sin(\tan(x))}{x} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right) } \]