Evaluate $\int_0^{\infty} {\sin(\tan(x)) \over x}dx$

Question

I tried to solve it the Feynman way and defined:

$$I(a):=\int_0^{\infty} {\sin(\tan(a \cdot x)) \over x} \ dx$$

And look what happens when one substitutes $u=ax$ $(a>0)$:

$$I(a)=\int_0^{\infty} {\sin(\tan(u)) \over u} \ du = I(1)$$

Which implies that $I(a)=const$ for $a>0$. More generally $I(a)=c \cdot sign(a)$. I wonder whether this can help.

I recalled that in order to solve $\int_0^{\infty} {\sin(x) \over x} \ dx$ using the Feynman technique one had to define $I(a):=\int_0^{\infty} {\sin(x) \over x} e^{-a \cdot x}\ dx$ and differentiate it. Consequently I suspect we should define $I(a):=\int_0^{\infty} {\sin(\tan x) \over x} e^{-a \cdot x}\ dx$, but differentiation yields:

$$I'(a)=-\int_0^{\infty} {\sin(\tan x)} e^{-a \cdot x}\ dx$$

Which is another difficult integral.

Any help?

(please try to avoid gamma functions)

Answer 1

Notice $\tan x$ is a periodic function with period $\pi$ and recall following expansion:

$$\frac{1}{\tan x} = \lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x + n\pi}$$

The integral we seek $$\mathcal{I} \stackrel{def}{=} \int_0^\infty \frac{\sin\tan x}{x} dx = \frac12 \int_{-\infty}^\infty \frac{\sin\tan x}{x} dx = \frac12 \left(\sum_{n=-\infty}^\infty \int_{(n-\frac12)\pi}^{(n+\frac12)\pi}\right)\frac{\sin\tan x}{x}dx $$ can be rewritten as $$ \mathcal{I} = \frac12 \int_{-\frac12\pi}^{\frac12\pi}\sin\tan x\left(\sum_{n=-\infty}^\infty\frac{1}{x+n\pi}\right) dx = \frac12\int_{-\frac12\pi}^{\frac12\pi}\frac{\sin\tan x}{\tan x} dx $$ Change variable to $t = \tan x$, we get

$$\mathcal{I} = \frac12\int_{-\infty}^{\infty} \frac{\sin t}{t(1+t^2)} dt = \frac12\Im\left[\int_{-\infty}^{\infty}\frac{e^{it}-1}{t(1+t^2)} dt\right]$$

We can evaluate the integral on RHS as a contour integral. By completing the contour in upper half-plane and using the fact the integrand has only two poles at $t = \pm i$, we get:

$$\begin{align} \mathcal{I} &= \frac12\Im\left[ 2\pi i \, \mathop{\text{Res}}_{z = i}\left(\frac{e^{it}-1}{t(1+t^2)}\right)\right] = \pi\Re\left[ \frac{e^{i(i)} - 1}{i(i+i)}\right] = \frac{\pi}{2}\left(1 - \frac1e \right)\\ &\approx 0.9929326518994357602762750999834... \end{align} $$

Update

If one don't want to use contour integral, we can replace the last step by a Feymann trick. Consider the function

$$J(a) = \int_0^\infty \frac{\sin(at)}{t(1+t^2)}dt $$

It is easy to see $\mathcal{I} = J(1)$ and $J(a)$ satisfies following ODE for $a > 0$.

$$\left( -\frac{d^2}{da^2} + 1 \right)J(a) = \int_0^\infty \frac{\sin(at)}{t} dt = \int_0^\infty \frac{\sin t}{t} dt = \frac{\pi}{2}$$

This implies $\displaystyle\;J(a) = \frac{\pi}{2} + A e^a + B e^{-a}\;$ for suitably chosen constants $A, B$. Notice $$\begin{align} J(+\infty) &= \lim_{a\to+\infty} \int_0^\infty \frac{\sin t}{t\left(1 + \left(\frac{t}{a}\right)^2\right)} dt = \int_0^\infty \frac{\sin t}{t} dt = \frac{\pi}{2}\\ J'(0^{+}) &= \lim_{a\to 0^{+}} \int_0^\infty \frac{\cos(at)}{1+t^2} dt = \int_0^\infty \frac{dt}{1+t^2} = \frac{\pi}{2} \end{align}$$

This fixes $\;A = 0$, $\displaystyle\;B = -\frac{\pi}{2}\;$ and hence

$$J(a) = \frac{\pi}{2}\left(1 - e^{-a}\right) \quad\implies\quad \mathcal{I} = J(1) = \frac{\pi}{2}\left( 1 - \frac{1}{e}\right)$$

Answer 2

Real Manipulations $$ \begin{align} \int_0^\infty\frac{\sin(\tan(x))}x\,\mathrm{d}x &=\frac12\int_{-\infty}^\infty\frac{\sin(\tan(\pi x))}x\,\mathrm{d}x\tag{1}\\ &=\frac12\sum_{k\in\mathbb{Z}}\int_0^1\frac{\sin(\tan(\pi x))}{x+k}\,\mathrm{d}x\tag{2}\\ &=\frac\pi2\int_0^1\sin(\tan(\pi x))\cot(\pi x)\,\mathrm{d}x\tag{3}\\ &=\frac\pi2\int_{-1/2}^{1/2}\sin(\tan(\pi x))\cot(\pi x)\,\mathrm{d}x\tag{4}\\ &=\frac12\int_{-\infty}^\infty\frac{\sin(u)}{u(1+u^2)}\,\mathrm{d}u\tag{5}\\ \end{align} $$ Explanation:
$(1)$: $\frac{\sin(\tan(x))}x$ is even, double domain and divide by $2$; substitute $x\mapsto\pi x$
$(2)$: break the domain into unit intervals
$(3)$: $\sum\limits_{k\in\mathbb{Z}}\frac1{x+k}=\pi\cot(\pi x)$ (see this answer)
$(4)$: $\sin(\tan(\pi x))\cot(\pi x)$ has period $\pi$
$(5)$: substitute $u=\tan(\pi x)$

Contour Integration

We will use the counter-clockwise contour $$ \gamma^+=[-R-i/2,R-i/2]\cup Re^{i[0,\pi]}-i/2 $$ and the clockwise contour $$ \gamma^-=[-R-i/2,R-i/2]\cup Re^{-i[0,\pi]}-i/2 $$ Then $$ \begin{align} \frac12\int_{-\infty}^\infty\frac{\sin(u)}{u(1+u^2)}\,\mathrm{d}u &=\frac12\int_{-\infty-\frac i2}^{\infty-\frac i2}\frac{\sin(u)}{u(1+u^2)}\,\mathrm{d}u\tag{6}\\ &=\frac1{4i}\int_{\gamma^+}\frac{e^{iz}}{z(1+z^2)}\,\mathrm{d}z -\frac1{4i}\int_{\gamma^-}\frac{e^{-iz}}{z(1+z^2)}\,\mathrm{d}u\tag{7}\\ &=\frac1{4i}\int_{\gamma^+}e^{iz}\left(\frac1z-\frac{1/2}{z-i}\color{#A0A0A0}{-\frac{1/2}{z+i}}\right)\,\mathrm{d}u\\ &-\frac1{4i}\int_{\gamma^-}e^{-iz}\left(\color{#A0A0A0}{\frac1z-\frac{1/2}{z-i}}-\frac{1/2}{z+i}\right)\,\mathrm{d}u\tag{8}\\ &=\frac{2\pi i}{4i}\left(1-\frac1{2e}\right)-\frac{2\pi i}{4i}\left(\frac1{2e}\right)\tag{9}\\ &=\frac\pi2\left(1-\frac1e\right)\tag{10} \end{align} $$ Explanation:
$(6)$: no singularities in $\small[-R,R]\cup[R,R-i/2]\cup[R-i/2,-R-i/2]\cup[-R-i/2,-R]$
$\hphantom{(6)\text{:}}$ integrand vanishes on vertical segments
$(7)$: $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$, integrals vanish on semi-circular arcs
$(8)$: $\frac1{z(z^2+1)}=\frac1z-\frac{1/2}{z-i}-\frac{1/2}{z+i}$
$(9)$: singularities at $0$ and $i$ are in $\gamma^+$, singularity at $-i$ is in $\gamma^-$
$(10)$: simplify

Answer 3

Lobachevsky Integral Formula: If $f(x)$ meet $f(x+π)=-f(x)$ and $f(−x)=f(x)$.

Then $$\int_0^\infty f(x)\cdot\frac{\sin x}{x}dx=\int_0^\frac{\pi}{2}f(x)\cdot \cos xdx$$

so $$\int_0^\infty\frac{\sin(\tan x)}{x}dx=\int_0^\infty\frac{\sin(\tan x)}{\sin x}\cdot\frac{\sin x}{x}dx$$ where $$f(x)=\frac{\sin(\tan x)}{\sin x}$$

so $$\int_0^\infty\frac{\sin(\tan x)}{x}dx=\int_0^\frac{\pi}{2}\frac{\sin(\tan x)}{\sin x}\cdot \cos xdx=\int_0^\frac{\pi}{2}\frac{\sin(\tan x)}{\tan x}dx$$ $$=\int_0^\infty\frac{\sin \theta}{\theta(1+\theta^2)}d\theta =\int_0^\infty\frac{\sin\theta}{\theta}d\theta-\int_0^\infty\frac{\theta\sin\theta}{1+\theta^2}d\theta=\frac{\pi}{2}-\frac{\pi}{e\pi}=\frac{\pi}{2}\Big(1-\frac{1}{e}\Big)$$

Answer 4

Assume that $N$ is a large positive integer.

The complex function $$f(z) = \frac{e^{i \tan z}}{z} $$ has a simple pole at the origin and essential singularities at the half-integer multiples of $\pi$.

Let's integrate $f(z)$ around a rectangular contour that has vertices at $z= \pm \pi N$, $z=\pm \pi N+i\sqrt{N}$ with small clockwise-oriented semicircles of radius $\epsilon$ about the origin and about the points $z= \pi \left(k+\frac{1}{2} \right)$, $k=-N, -N+1, \ldots -1, 0, 1, \ldots N-2, N-1$.

The contour is similar to the contour used in this answer.

In the upper half-plane, $$|e^{i \tan z}| = |e^{i \tan (x+iy)}| =\exp \left(-\frac{\sinh 2y}{\cos 2x + \cosh 2y} \right) \le 1.$$

Since the height of the contour is $\sqrt{N}$ and the magnitude of $e^{i \tan z}$ is bounded in the upper half-plane, the estimation lemma tells us that integral vanishes on the left and right the sides of the rectangle vanish as $N \to \infty$.

And since the magnitude of $e^{i \tan z}$ is bounded in the upper half-plane, the integral vanishes on the small semicircles about the half-integer multiples of $\pi$ as $\epsilon \to 0$.

So we have

$$ \text{PV} \int_{-\infty}^{\infty} \frac{e^{i \tan x}}{x} \, dx - i \pi \,\text{Res}[f(z),0] -\lim_{N \to \infty} \int_{-N}^{N} \frac{e^{i \tan (t+i\sqrt{N})}}{t+i\sqrt{N}} \, dt =0. $$

But since the magnitude of $e^{i \tan z}- \frac{1}{e}$ tends to zero exponentially fast as $\Im(z) \to +\infty$, we can replace $e^{i \tan(t+ i \sqrt{N})}$ with $\frac{1}{e}$. (Specifically, it's going to zero like $\frac{2}{e} e^{-2 \, \Im(z)}$.)

Therefore,

$$\text{PV} \int_{-\infty}^{\infty} \frac{e^{i \tan x}}{x} \, dx - i \pi - \frac{1}{e} \lim_{N \to \infty} \int_{-N}^{N} \frac{1}{t+i\sqrt{N}} \, dt =0. $$

But if we integrate $g(z) = \frac{1}{z}$ around a similar contour (minus the semicircles about the half-integer multiples of $\pi$), we get

$$ \underbrace{\text{PV} \int_{-\infty}^{\infty} \frac{dx}{x}}_{0} - i \pi \underbrace{\text{Res}[g(z),0]}_{1} - \lim_{N \to \infty} \int_{-N}^{N} \frac{1}{t+i\sqrt{N}} \, dt = 0.$$

Therefore, $$ \lim_{N \to \infty} \int_{-N}^{N} \frac{1}{t+i\sqrt{N}} \, dt = - i \pi, $$ and

$$ \text{PV} \int_{-\infty}^{\infty} \frac{e^{i \tan x}}{x} \, dx - i \pi - \frac{1}{e} (-i \pi) =0.$$

Equating the imaginary parts on both sides of the equation, we have $$\int_{-\infty}^{\infty} \frac{\sin (\tan x)}{x} \, \mathrm dx = \pi \left(1- \frac{1}{e} \right).$$

---

UPDATE:

Integrating the function $\frac{z}{z^{2}+a^{2}} \, e^{i \tan z} $ around almost the same contour shows that $$\int_{-\infty}^{\infty} \frac{x}{x^{2}+a^{2}} \, \sin(\tan x) \, \mathrm dx = \pi \left(e^{- \tanh a}- \frac{1}{e} \right). $$

Answer 5

Ok i will try a contour method approach wihout a reference to the Mittag Leffler expansion of $\tan(x)$.

To begin, write $\sin(\tan[x])=\Im(e^{i\tan[x]})$.

$$ I=\Im\underbrace{\int_{\epsilon}^{\infty}\frac{e^{i\tan[x]}}{x}}_{I_1}+\Im\underbrace{\int_{-\infty}^{-\epsilon}\frac{e^{i\tan[x]}}{x}}_{I_2} $$

where the limit of $\epsilon\rightarrow 0$ is implicit.

Now we come to our crucial step :

We should keep in mind some possible contributions from complex infinity, because Jordan's Lemma is not sufficent here [the oscillations are not regular enough]

We can rewrite $I_1 $ using Cauchy's theorem ( $\tan[\pm ix]=\pm i \tanh[x]$)

$$I_1=\int_{QC^1_0}\frac{e^{-i\tan[z]}}{z}+\int_{QC^1_{\infty}}\frac{e^{-i\tan[z]}}{z}+\int_{\epsilon}^{\infty-\epsilon}\frac{e^{-\tanh[y]}}{iy}$$

Similiar $I_2$

$$I_2=\int_{QC^2_0}\frac{e^{-i\tan[z]}}{z}+\int_{QC^2_{\infty}}\frac{e^{-i\tan[z]}}{z}-\int_{\epsilon}^{\infty-\epsilon}\frac{e^{-\tanh[y]}}{iy}$$

Here $QC^{1,2}_{0,\infty}$ denotes a quarter circle around $0(\infty)$ in the first(second) quadrant and the remaining to straightline integrals are along the imaginary axis.

Adding back $I_1$ and $I_2$ we see that the two integrals along the positve/negative imaginary cancel out so we are left with $$ I_1+I_2=\int_{QC^1_0}\frac{e^{-i\tan[z]}}{z}+\int_{QC^1_{\infty}}\frac{e^{-i\tan[z]}}{z}+\int_{QC^2_0}\frac{e^{-i\tan[z]}}{z}+\int_{QC^2_{\infty}}\frac{e^{-i\tan[z]}}{z} $$

we can achive further simplifiction by exploting the fact that semi circles around the same point

a) enclose singularities in opposite direction

b) have twisted integration regions.

This results in two minus signs which multiply to an overall plus. Therfore:

$$ I_1+I_2=2\int_{QC^1_0}\frac{e^{-i\tan[z]}}{z}+\int_{QC^1_{\infty}}\frac{e^{-i\tan[z]}}{z} $$

Because this are first order poles the integral is finite in a principal value sense. It's now easy to show that the integral is given by

$$ I=2\Im \left[\frac{\pi i}{4}\left(\text{res}(0)+\text{res}(\infty)\right)\right] $$

the residues can be calculated as $\text{res}[0]=1 $ and $\text{res}[\infty]=\frac{-1}{e} $ and therefor

$$ I=\frac{\pi }{2}\left(1-\frac{1}{e}\right) $$

As expected from the other solutions

Answer 6

I give you two kind of representation; both gave me a good result.

1. $$\frac{sin(tan(ax))}{x} = \frac{(tan(a x))}{x} \int_0^1 cos(t~tan(a x)) dt $$

2. $$ \frac{sin(tan(ax))}{x} = \frac{-i}{2x \sqrt\pi} \int_{(-i \infty+\gamma)}^{(i \infty+\gamma)} \frac{(2^{(-1+2 s)} \Gamma(s) \tan^{(1-2 s)}(a x))}{(\Gamma(\frac{3}{2}-s))} ds$$ $$For (0<\gamma<1~and~\tan(a x)>0)$$

You can use the second easily if you are familiar with de $\Gamma$-function.

Good luck!