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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)

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    $\begingroup$ would you mind to show us the solutions, maybe helpful to find a way $\endgroup$ – tired May 7 '15 at 16:21
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    $\begingroup$ Shouldn't it be $I'(1) $, not $I(1)$, which is of course zero? Also it seems like this integral tends to $1$. $\endgroup$ – Cameron Williams May 7 '15 at 21:27
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    $\begingroup$ Observation: We have an even integrand, so we can change limits of integration to $-\infty, \infty$. Therefore we could possibly use contour integration $\endgroup$ – tired May 7 '15 at 21:35
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    $\begingroup$ I did some numerical tests and it seems to converge to $1$. $\endgroup$ – Cameron Williams May 8 '15 at 1:55
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    $\begingroup$ Hint: $\sum_{n=-\infty}^\infty \frac{1}{x-n\pi} = \frac{1}{\tan x}$ and the integral is $\frac{\pi}{2}(1-\frac{1}{e}) \approx 0.992932651899435760276275$ $\endgroup$ – achille hui May 8 '15 at 10:20
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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

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

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  • $\begingroup$ Simply brilliant ! $\endgroup$ – Lucian May 8 '15 at 12:26
  • $\begingroup$ (+999) Great work $\endgroup$ – tired May 8 '15 at 12:44
  • $\begingroup$ Thanks for this great proof :) $\endgroup$ – Chris May 8 '15 at 13:50
  • $\begingroup$ After a cursory glance at your answer, I thought you were doing something different. Now that I've finished my answer, I see that the first part of your answer pretty much outlines what I did. I hope you don't mind me filling in some details. If you do, I will delete my answer. $\endgroup$ – robjohn May 8 '15 at 18:10
  • $\begingroup$ @robjohn I don't mind. the second part of you answer shows another way to perform the contour integral which others may find useful. $\endgroup$ – achille hui May 8 '15 at 18:31
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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

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  • $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn May 9 '15 at 4:29
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EDIT: I modified the contour slightly.


Similar to the answer HERE, we can integrate the complex function $$f(z) = \frac{e^{i \tan z}}{z} $$ around a rectangular contour with vertices at $z= N$, $z=N+i\sqrt{N}$, $z= -N + i \sqrt{N}$, and $z=-N$, where $N$ is some positive integer.

Due to the presence of a simple pole at the origin and essential singularities at the half-integers, the contour needs to be indented along the real axis.

But since $$|e^{i \tan z}| = |e^{i \tan (x+iy)}| =\exp \left(-\frac{\sinh 2y}{\cos 2x + \cosh 2y} \right) \le 1$$ in the upper half-plane, the total contribution from the indentations around the essential singularities is vanishingly small.

Also, 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 integrals along the sides of the rectangle vanish as $N \to \infty$.

So letting the radius of the indentation around the origin go to zero and then letting $N \to \infty$, we get

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

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 $\frac{2}{e} e^{-2 \, \Im(z)}$.)

So we have

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

But if we integrate $g(z) = \frac{1}{z}$ around a similar contour, 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,

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

The result then follows if we equate the imaginary parts on both sides of the equation.

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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

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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!

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  • $\begingroup$ Okay, after intensive thinking I still don't know how to use number 1. Can you give more details? $\endgroup$ – Chris May 7 '15 at 20:36
  • $\begingroup$ I found $\frac{\pi}{2}(1-\frac{1}{e})$; It seems to be the answer for the different answers provided from my last visit. $\endgroup$ – user230283 May 8 '15 at 16:27
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This is just a guess. Use $$\int_0^{\infty} x^{a-1} \cdot {{\sin(\tan(x))} \over x}$$ After differentiation, you'll get $$\int_0^{\infty} x^{a-2} \cdot ln(x) \cdot \sin(\tan(x))$$ This can be recognized as a Mellin Transform, and might be evaluable (by hand) using Ramunajan's Master Theorem (You'll need knowledge of the Gamma Function). After that, I'd assume you just follow the rest of the steps...

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