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.


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 $$ PV\int_0^\infty\frac{\tan x}{x}dx~=~ \lim_{\lambda\to0}\int_0^{\infty}\frac{\sin x\cos x}{x(\cos^2 x+\lambda^2)}dx=\frac{\pi}{2}. $$

  • $\begingroup$ 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

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\pp\int_{-\infty}^{\infty}{\tan\pars{x} \over x}\,\dd x = \pi:\ {\large ?}}$

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

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 $\pars{1}$:

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

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.