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In a recent question, it was stated in a comment, without proof, that

$$ PV \int_{-\infty}^{\infty} \frac{\tan x}{x}dx = \pi$$

What is the easiest way to prove this? I was able to show that

$$ PV \int_{-\infty}^{\infty} \frac{\tan x}{x}dx = -PV\int_{-\infty}^{\infty} \frac{1}{x \tan x}dx \\ PV \int_{-\infty}^{\infty} \frac{1}{x \sin x}dx = 0 \\ \int_{-\infty}^{\infty} \frac{\sin x}{x}dx = \pi $$

but failed to compute the original integral from this.

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The question is : What would be a right way to define the principal value of this integral, knowing that it has infinitely many singularities at the points $\frac{\pi}{2}+\pi\Bbb{Z}$ ? I will propose the following $$ PV\int_0^\infty\frac{\tan x}{x}dx~\buildrel{\rm def}\over{=}~ \lim_{\lambda\to0}\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx $$ Next I will show that the limit in this definition does exist and that its value is $\frac{\pi}{2}$.

First, note that the convergence of the integral $\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx$ is easy to prove using integration by parts. Now $$\eqalign{ \int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx &=\frac{1}{2}\lim_{n\to\infty}\int_{-\pi n}^{\pi (n+1)}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx\cr &=\frac{1}{2}\lim_{n\to\infty}\sum_{k=-n}^{n}\int_{\pi k}^{\pi(k+1)}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx\cr &=\frac{1}{2}\lim_{n\to\infty}\sum_{k=-n}^{n}\int_{0}^{\pi}\frac{\sin x\cos x}{(x+ \pi k)(\cos^2 x+\lambda^2)}dx\cr &=\frac{1}{2}\lim_{n\to\infty}\int_{0}^{ \pi}\left(\sum_{k=-n}^{n}\frac{1}{x+ \pi k}\right) \frac{\sin x\cos x}{ \cos^2x+\lambda^2}dx\cr &=\frac{1}{2}\lim_{n\to\infty}\int_{0}^{ \pi}U_n(x) \frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx\cr } $$ where $$ U_n(x)=\tan(x)\left(\sum_{k=-n}^{n}\frac{1}{x+ \pi k}\right) $$ But using the well-known expansion of the cotangent function, it is easy to see that $\{U_n \}_n$ converges point-wise to $1$, and that this sequence is bounded uniformely on the interval $[0,\pi]$. Thus, we can interchange the signs of integral and limit in the above formula to get $$ \int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx =\frac{1}{2} \int_{0}^{ \pi} \frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx = \int_{0}^{ \pi/2} \frac{ \cos^2 x}{ \cos^2x+\lambda^2}dx $$ Finally, taking the limit as $\lambda\to0$ we get $$ PV\int_0^\infty\frac{\tan x}{x}dx~=~ \lim_{\lambda\to0}\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx=\frac{\pi}{2}. $$

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  • $\begingroup$ Thank you. Upvoted, accepted, and bounty awarded. Could you comment, though, on why you chose this particular definition for the PV? Is there no more 'canonical' choice? $\endgroup$ – user111187 May 6 '14 at 16:37
  • $\begingroup$ This definition allows me to treat the infinitely many singularities at the same time. The other more conventional way is to define $PV\int^{π(k+1)}_{πk}\tan x dx$ as $\lim\limits_{\epsilon\to0}\left(\int_{\pi k}^{\pi(k+1/2-\epsilon)}\frac{\tan x}{x}dx+\int_{\pi (k+1/2+\epsilon)}^{\pi(k+1)}\frac{\tan x}{x}dx\right)$, and then to add(with justification) all these principal values. I thought my procedure is easier. $\endgroup$ – Omran Kouba May 6 '14 at 16:53
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\pp\int_{-\infty}^{\infty}{\tan\pars{x} \over x}\,\dd x = \pi:\ {\large ?}}$

$$ \mbox{Note that}\quad \pp\int_{-\infty}^{\infty}{\tan\pars{x} \over x}\,\dd x =2\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}{\tan\pars{x} \over x}\,\dd x =2\int_{0}^{\infty}{\tan\pars{x} \over x}\,\dd x\tag{1} $$

With $\ds{N \in {\mathbb N}}$, let's consider the following integral: \begin{align} &\color{#c00000}{\int_{0}^{N\pi}{\tan\pars{x} \over x}\,\dd x} =\int_{0}^{\pi}{\tan\pars{x} \over x}\,\dd x +\int_{\pi}^{2\pi}{\tan\pars{x} \over x}\,\dd x + \cdots +\int_{\pars{N - 1}\pi}^{N\pi}{\tan\pars{x} \over x}\,\dd x \\[3mm]&=\int_{0}^{\pi}\tan\pars{x}\sum_{n = 0}^{N - 1}{1 \over x + n\pi}\,\dd x ={1 \over \pi}\int_{0}^{\pi}\tan\pars{x} \color{#00f}{\sum_{n = 0}^{N - 1}{1 \over n + x/\pi}}\,\dd x\tag{2} \end{align}

\begin{align} &\color{#00f}{\sum_{n = 0}^{N - 1}{1 \over x + n\pi}} =\sum_{n = 0}^{\infty}\pars{{1 \over n + x/\pi} - {1 \over n + N + x/\pi}} =N\sum_{n = 0}^{\infty}{1 \over \pars{n + x/\pi}\pars{n + N + x/\pi}} \\[3mm]&=N\,{\Psi\pars{x/\pi} - \Psi\pars{N + x/\pi} \over \pars{x/\pi} - \pars{N + x/\pi}} =\color{#00f}{\Psi\pars{N + {x \over \pi}} - \Psi\pars{x \over \pi}}\tag{3} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function and we used A&S table formula ${\bf\mbox{6.3.16}}$.

We replace $\pars{3}$ in $\pars{2}$: \begin{align} &\color{#c00000}{\int_{0}^{N\pi}{\tan\pars{x} \over x}\,\dd x} ={1 \over \pi}\int_{0}^{\pi}\tan\pars{x}\bracks{% \Psi\pars{N + {x \over \pi}} - \Psi\pars{x \over \pi}}\,\dd x \\[3mm]&=\int_{0}^{1}\tan\pars{\pi x}\bracks{\Psi\pars{N + x} - \Psi\pars{x}}\,\dd x \\[3mm]&=\int_{-1/2}^{1/2}\bracks{-\cot\pars{\pi x}}\bracks{% \Psi\pars{N + x + \half} - \Psi\pars{x + \half}}\,\dd x \\[3mm]&=-\int_{0}^{1/2}\!\!\!\!\!\!\!\cot\pars{\pi x}\bracks{% \Psi\pars{N + x + \half} - \Psi\pars{x + \half} - \Psi\pars{N - x + \half} + \Psi\pars{-x + \half}}\,\dd x \end{align} Hoever ( see A&S table identity ${\bf\mbox{6.3.7}}$ ): $$ -\Psi\pars{x + \half} + \Psi\pars{-x + \half} =\pi\cot\pars{\pi\bracks{x + \half}} = -\pi\tan\pars{\pi x} $$

such that \begin{align} &\color{#c00000}{\int_{0}^{N\pi}{\tan\pars{x} \over x}\,\dd x} ={\pi \over 2}-\ \overbrace{\int_{0}^{1/2}\cot\pars{\pi x}\bracks{% \Psi\pars{N + x + \half} - \Psi\pars{N - x + \half}}\,\dd x} ^{\ds{\to 0\quad\mbox{when}\quad N \to \infty}} \end{align} Since ( see A&S table asymptotic expansion ${\bf\mbox{6.3.18}}$ ) $$ \Psi\pars{z} \sim \ln\pars{z} - {1 \over 2z} - {1 \over 12z^{2}} + \cdots\,, \qquad \verts{z} \gg 1\,,\quad \verts{{\rm arg}\pars{z}} < \pi $$ we'll have $\ds{\int_{0}^{\infty}{\tan\pars{x} \over x}\,\dd x = {\pi \over 2}}$ and from $\pars{1}$:

$$\color{#00f}{\large% \pp\int_{-\infty}^{\infty}{\tan\pars{x} \over x}\,\dd x = \pi} $$

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  • $\begingroup$ How do you interpret $\int_0^\pi\frac{\tan x}{x}dx$ around $x=\pi/2$ ? $\endgroup$ – Omran Kouba May 13 '14 at 18:12
  • $\begingroup$ @OmranKouba We can consider infinite principal values at intervals $\large\left[N\pi,\left(N + 1\right)\pi\right]$ with the 'singularity' at $\large\left(N + 1/2\right)\pi$. However, it was a too long answer. At the end, we switch from $\large\left(0,1\right)$ to $\large\left(0,1/2\right)$ which I believe it 'repairs' the 'neglecting' at the beginning. It could be written in a more elaborated way but it will be a too long answer. In this way, the main idea is shown. I saw how you avoid that problems and that's is quite fine. Maybe, later I will make some refinement. Thanks a lot. $\endgroup$ – Felix Marin May 13 '14 at 20:53
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I found a solution, but it is not a very satisfying one, because I am not really able to justify applying the below theorem to the non-convergent integral.

Theorem If $f$ is an odd function with period $a$, then $$\int_0^{\infty} \frac{f(x)}{x}dx= \frac{\pi}{a} \int_0^{a/2} \frac {f(x)} {\tan(\pi x / a)} dx $$ This theorem follows from Frullani's theorem and the power series expansion of $\tan$.

Because $\tan$ is odd and periodic with period $\pi$, the above theorem immediately yields $$\int_0^{\infty} \frac{\tan x}{x}dx = \frac{\pi}{2}$$

If someone knows with a better (i.e. more rigorous) or otherwise interesting solution, I would love to hear it.

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