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Prove the following

$$\int^\infty_0 \frac{\tan(x)}{x}=\frac{\pi}{2} $$

This question was posted on some forum, but i think it should be rewritten as $$PV\int^\infty_0 \frac{\tan(x)}{x}=\frac{\pi}{2} $$ Because if the discontinuoities of the zeros of $\cos(x)$.

My attempt

Consider the following function

$$f(x) = \frac{\tan(x)}{x}$$

On the interval $\left(-\frac{\pi}{2},\frac{\pi}{2} \right)$, clearly the function is symmetric and positive around the origin.

Let us consider $x \in \left(0,\frac{\pi}{2} \right)$

$$f'(x) = \frac{\sec^2(x) (2x - \sin(2 x))}{2x^2} > 0$$

Note that $\lim_{x\to 0} f(x) = 1$, we deduce that the function is increasing on the interval $\left(0,\frac{\pi}{2} \right)$. Hence

$$\int^{\pi/2}_{-\pi/2}\frac{\tan(x)}{x}\,dx = 2 \int^{\pi/2}_0 \frac{\tan(x)}{x}\,dx >2 \int^{\pi/2}_0\,dx = \pi$$ Also note that

Near $\pi/2$ the integral acts like $\frac{1}{\pi/2-x}$ which diverges to inifnity. The visual of $f$ is on that interval

enter image description here

From the graph of $f$ on the real line it seems the integrals on left and right are also divergent to -infinity and contribute to the infinity at the middle to cause a convergent value.

enter image description here

Question

I need a proof if the principal value exists or not?

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    It is worth it noticing that $$ \cot x = \lim_{N\to\infty} \sum_{n=-N}^{N} \frac{1}{x - \pi n}. $$ – 2017-02-12

2 Answers 2

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Let $Z = \{k\pi +\frac{\pi}{2} : k \in \Bbb{Z}$ be the set of poles of $\tan x$ and define

$$ I(N,\epsilon) = \int\limits_{\substack{ \text{dist}(x,Z) > \epsilon \\ 0 < x < N\pi}} \frac{\tan x}{x} \, dx. $$

Then I will prove that

$$ \lim_{\substack{\epsilon &\to 0^+ \\ N&\to\infty}} I(N,\epsilon) = \frac{\pi}{2}. $$

Indeed, let $\epsilon \in (0,\pi/2)$. If we denote by $D(\epsilon) = \{x \in [0,\pi] : |x - \frac{\pi}{2}| > \epsilon \}$, then

\begin{align*} I(N,\epsilon) &= \frac{1}{2} \sum_{k=0}^{N-1} \int_{D(\epsilon)} \left( \frac{1}{x + k\pi} + \frac{1}{x - (k+1)\pi} \right) \tan x \, dx \\ &= \frac{1}{2} \int_{D(\epsilon)} \left( \sum_{k=0}^{N-1} \frac{2x-\pi}{(x + k\pi)(x - (k+1)\pi)} \right) \tan x \, dx \tag{*} \end{align*}

Here are several observations:

  • The function $f(x) = \dfrac{2x-\pi}{x(x-\pi)} \tan x$ extends to a bounded continuous function on $[0,\pi]$.

  • On the interval $(0,\pi)$, we have the following bound $$ \left|\frac{x(x-\pi)}{(x + k\pi)(x - (k+1)\pi)}\right| \leq \begin{cases} 1, & k = 0, \\ \frac{1}{k^2}, & k \geq 1 \end{cases} $$

This shows that the integrand of $\text{(*)}$ converges uniformly on $(0,\pi)$ to

$$ \lim_{N\to\infty} \sum_{k=0}^{N-1} \left( \frac{1}{x+k\pi} + \frac{1}{x-(k+1)\pi} \right) \tan x = \cot x \cdot \tan x = 1. $$

Therefore it follows that

$$ \lim_{\substack{\epsilon &\to 0^+ \\ N&\to\infty}} I(N,\epsilon) = \frac{1}{2}\int_{0}^{\pi} dx = \frac{\pi}{2}. $$

Addendum. In fact, the integrand of $\text{(*)}$ is always non-negative and monotone-increases to $1$ as $N \to \infty$:

enter image description here

Thus the same conclusion can be obtained by the monotone convergence theorem.

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{n \in \mathbb{N}_{\ \geq\ 0}}$:

\begin{align} \mrm{P.V.}\int_{n\pi}^{n\pi + \pi}{\tan\pars{x} \over x}\,\dd x & = \mrm{P.V.}\int_{0}^{\pi}{\tan\pars{x} \over x + n\pi}\,\dd x = -\,\mrm{P.V.}\int_{-\pi/2}^{\pi/2}{\cot\pars{x} \over x + n\pi + \pi/2}\,\dd x \\[5mm] & = -\int_{0}^{\pi/2}\cot\pars{x}\pars{% {1 \over x + n\pi + \pi/2} - {1 \over -x + n\pi + \pi/2}}\,\dd x \\[5mm] & = -\,{1 \over \pi}\int_{0}^{\pi/2}\cot\pars{x}\pars{% {1 \over n + 1/2 + x/\pi} - {1 \over n + 1/2 - x/\pi}}\,\dd x \end{align}

Then, \begin{align} \mrm{P.V.}\int_{0}^{\infty}{\tan\pars{x} \over x}\,\dd x & = \sum_{n = 0}^{\infty}\mrm{P.V.}\int_{n\pi}^{n\pi + \pi}{\tan\pars{x} \over x} \,\dd x \\[5mm] & = -\,{1 \over \pi}\int_{0}^{\pi/2}\cot\pars{x} \bracks{\Psi\pars{{1 \over 2} - {x \over \pi}} - \Psi\pars{{1 \over 2} + {x \over \pi}}}\,\dd x \\[5mm] & = -\,{1 \over \pi}\int_{0}^{\pi/2} \cot\pars{x}\braces{\pi\cot\pars{\pi\bracks{{1 \over 2} + {x \over \pi}}}} \,\dd x \\[5mm] & = -\,{1 \over \pi}\int_{0}^{\pi/2}\cot\pars{x}\bracks{-\pi\tan\pars{x}}\,\dd x =\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{\pi \over 2}} \end{align}