A strange integral: $\int_{-\infty}^{+\infty} {dx \over 1 + \left(x + \tan x\right)^2} = \pi.$

While browsing on Integral and Series, I found a strange integral posted by @Sangchul Lee. His post doesn't have a response for more than a month, so I decide to post it here. I hope he doesn't mind because the integral looks very interesting to me. I hope for you too. :-)

$$\mbox{How does one prove}\quad \int_{-\infty}^{\infty} {{\rm d}x \over 1 + \left[\,x + \tan\left(\, x\,\right)\,\right]^{2}} = \pi\quad {\large ?} $$

Please don't ask me, I really have no idea how to prove it. I hope users here can find the answer to prove the integral. I'm also interested in knowing any references related to this integral. Thanks in advance.

Here is an approach.

We may use the following result, which goes back to G. Boole (1857) :

$$ \int_{-\infty}^{+\infty}f\left(x-\frac{a_1}{x-\lambda_1}-\cdots-\frac{a_n}{x-\lambda_n}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \tag1 $$

with $a_i>0, \lambda_i \in \mathbb{R}$ and $f$ sufficiently 'regular'.

Observe that, for $x\neq n\pi$, $n=0,\pm1,\pm2,\ldots$, we have $$ \cot x = \lim_{N\to +\infty} \left(\frac1x+\frac1{x+\pi}+\frac1{x-\pi}+\cdots+\frac1{x+N\pi}+\frac1{x-N\pi}\right)$$ leading to (see Theorem 10.3 p. 14 here and see achille's answer giving a route to prove it)

$$ \int_{-\infty}^{+\infty}f\left(x-\cot x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \tag2 $$

with $\displaystyle f(x)=\frac{1}{1+\left(\small{\dfrac\pi2 -x }\right)^2}$.

On the one hand, from $(2)$, $$ \begin{align} \int_{-\infty}^{+\infty}f\left(x-\cot x\right)\mathrm{d}x& =\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \\\\ &=\int_{-\infty}^{+\infty}\frac{1}{1+\left(\small{\dfrac\pi2 -x }\right)^2}\: \mathrm{d}x\\\\ &=\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\: \mathrm{d}x\\\\ & =\pi \tag3 \end{align} $$ On the other hand, with the change of variable $x \to \dfrac\pi2 -x$, $$ \begin{align} \int_{-\infty}^{+\infty}\!\!\!f\left(x-\cot x\right)\mathrm{d}x & =\int_{-\infty}^{+\infty} \!\!\!f\left(\dfrac\pi2-x-\tan x\right)\mathrm{d}x \\\\ & =\int_{-\infty}^{+\infty}\frac{1}{1+\left(x+ \tan x \right)^2} \mathrm{d}x \tag4 \end{align} $$ Combining $(3)$ and $(4)$ gives

$$ \int_{-\infty}^{+\infty}\frac{1}{1+\left(x+ \tan x \right)^2} \mathrm{d}x=\pi. $$

Please view this as an supplement of Olivier's answer.

I will derive a sufficient condition on the meromorphic function involved which allow one to apply a result similar to that in Olivier's answer.

Let $\phi(z)$ be any meromorphic function over $\mathbb{C}$ which

1. preserve the extended real line $\mathbb{R}^* = \mathbb{R} \cup \{ \infty \}$ in the sense: $$\begin{cases}\phi(\mathbb{R}) \subset \mathbb{R}^*\\ \phi^{-1}(\mathbb{R}) \subset \mathbb{R}\end{cases} \quad\implies\quad P \stackrel{def}{=} \phi^{-1}(\infty) = \big\{\, p \in \mathbb{C} : p \text{ poles of }\phi(z)\,\big\} \subset \mathbb{R} $$

2. Split $\mathbb{R} \setminus P$ as a countable union of its connected components $\,\bigcup\limits_{n} ( a_n, b_n )\,$. Each connected component is an open interval $(a_n,b_n)$ and on such an interval, $\phi(z)$ increases from $-\infty$ at $a_n^{+} $ to $\infty$ at $b_n^{-}$.

3. There exists an ascending chain of Jordan domains $D_1, D_2, \ldots$ that cover $\mathbb{C}$, $$\{ 0 \} \subset D_1 \subset D_2 \subset \cdots \quad\text{ with }\quad \bigcup_{k=1}^\infty D_k = \mathbb{C} $$ whose boundaries $\partial D_k$ are "well behaved", "diverge" to infinity and $| z - \phi(z)|$ is bounded on the boundaries. More precisely, let $$ \begin{cases} R_k &\stackrel{def}{=}& \inf \big\{\, |z| : z \in \partial D_k \,\big\}\\ L_k &\stackrel{def}{=}& \int_{\partial D_k} |dz| < \infty\\ M_k &\stackrel{def}{=}& \sup \big\{\, |z - \phi(z)| : z \in \partial D_k \,\big\} \end{cases} \quad\text{ and }\quad \begin{cases} \lim\limits_{k\to\infty} R_k = \infty\\ \lim\limits_{k\to\infty} \frac{L_k}{R_k^2} = 0\\ M = \sup_k M_k < \infty \end{cases} $$

Given such a meromorphic function $\phi(z)$ and any Lebesgue integrable function $f(x)$ on $\mathbb{R}$, we have following identity: $$ \int_{-\infty}^\infty f(\phi(x)) dx = \int_{-\infty}^\infty f(x) dx \tag{*1} $$

In order to prove this, we split our integral into a sum over the connected components of $\mathbb{R} \setminus P$. $$\int_\mathbb{R} f(\phi(x)) dx = \int_{\mathbb{R} \setminus P} f(\phi(x)) dx = \sum_n \int_{a_n}^{b_n} f(\phi(x)) dx $$ For any connected component $( a_n, b_n )$ of $\mathbb{R} \setminus P$ and $y \in \mathbb{R}$, consider the roots of the equation $\phi(x) = y$. Using properties $(1)$ and $(2)$ of $\phi(z)$, we find there is a unique root for the equation $y = \phi(x)$ over $( a_n, b_n )$. Let we call this root as $r_n(y)$. Change variable to $y = \phi(x)$, the integral becomes

$$\sum_n \int_{-\infty}^\infty f(y) \frac{d r_n(y)}{dy} dy = \int_{-\infty}^\infty f(y) \left(\sum_n \frac{d r_n(y)}{dy}\right) dy $$ We can use the obvious fact $\frac{d r_n(y)}{dy} \ge 0$ and dominated convergence theorem to justify the switching of order of summation and integral.

This means to prove $(*1)$, one only need to show $$\sum_n \frac{d r_n(y)}{dy} \stackrel{?}{=} 1\tag{*2}$$

For any $y \in \mathbb{R}$, let $R(y) = \phi^{-1}(y) \subset \mathbb{R}$ be the collection of roots of the equation $\phi(z) = y$.

Over any Jordan domain $D_k$, we have following expansion

$$\frac{\phi'(z)}{\phi(z) - y} = \sum_{r \in R(y) \cap D_k} \frac{1}{z - r} - \sum_{p \in P \cap D_k} \frac{1}{z - p} + \text{something analytic}$$

This leads to $$\sum_{r \in R(y)\cap D_k} r - \sum_{ p \in P \cap D_k} p = \frac{1}{2\pi i}\int_{\partial D_k} z \left(\frac{\phi'(z)}{\phi(z) - y}\right) dz$$

As long as $R(y) \cap \partial D_k = \emptyset$, we can differentiate both sides and get

$$\begin{align} \sum_{r_n(y) \in D_k} \frac{dr_n(y)}{dy} &= \frac{1}{2\pi i}\int_{\partial D_k} z \left(\frac{\phi'(z)}{(\phi(z) - y)^2}\right) dz = -\frac{1}{2\pi i}\int_{\partial D_k} z \frac{d}{dz}\left(\frac{1}{\phi(z)-y}\right) dz\\ &= \frac{1}{2\pi i}\int_{\partial D_k}\frac{dz}{\phi(z) - y} \end{align} $$ For those $k$ large enough such that $R_k > 2(M+|y|)$, we can expand the integrand in last line as

$$\frac{1}{\phi(z) - y} = \frac{1}{z - (y + z - \phi(z))} = \frac{1}{z} + \sum_{j=1}^\infty \frac{(y + z - \phi(z))^j}{z^{j+1}}$$ and obtain a bound

$$\left|\left(\sum_{r_n(y) \in D_k} \frac{dr_n(y)}{dy} \right) - 1\right| \le \frac{1}{2\pi}\sum_{j=1}^\infty \int_{\partial D_k} \frac{(|y| + |z-\phi(z)|)^j}{|z|^{j+1}} |dz|\\ \le \frac{(M + |y|)L_k}{2\pi R_k^2}\sum_{j=0}^\infty\left(\frac{M+|y|}{R_k}\right)^j \le \frac{M + |y|}{\pi}\frac{L_k}{R_k^2} $$ Since $\lim\limits_{k\to\infty} \frac{L_k}{R_k^2} = 0$, this leads to

$$\sum_n \frac{dr_n(y)}{dy} = \lim_{k\to\infty} \sum_{r_n(y) \in D_k} \frac{dr_n(y)}{dy} = 1$$

This justifies $(*2)$ and hence $(*1)$ is proved. Notice all the $\frac{dr_n(y)}{dy}$ are positive, there is no issue in rearranging the order of summation in last line.

Back to the original problem of evaluating

$$\int_{-\infty}^\infty \frac{1}{1+(x+\tan x)^2} dx$$

One can take $\phi(z)$ as $z + \tan z$ and $f(x)$ as $\frac{1}{1+x^2}$. It is easy to see $\phi(z)$ satisfies:

• Condition $(1)$ - For any $y \in \mathbb{R}$ and $u + iv \in \mathbb{C} \setminus \mathbb{R}$, we have $$\begin{align} \Im (\phi(u + iv) - y ) &= v + \Im\tan(u+iv) = v + \Im\frac{\tan u + i\tanh v}{1 - i\tan u\tanh v}\\ &= v + \tanh v\frac{1 + \tan^2 u}{1 + \tan^2 u\tanh^2 v} \ne 0 \end{align}$$

• Condition $(2)$ - obvious.

• Condition $(3)$. - Let $D_k$ to be the square $$D_k = \big\{\, u + v i \in \mathbb{C} : |u|, |v| \le k \pi \,\big\}$$ It is not hard to show $|z - \phi(z)| = |\tan z|$ is bounded above by $\frac{1}{\tanh k\pi}$ on $\partial D_k$.

Combine these, we can apply $(*1)$ and deduce

$$ \int_{-\infty}^\infty \frac{1}{1+(x+\tan x)^2} dx = \int_{-\infty}^\infty \frac{1}{1+x^2} dx = \pi $$

The Inverse Function Theorem gives us $$ \int_{-\infty}^{+\infty}f(g(x))\,\mathrm{d}x=\int_{-\infty}^{+\infty}\sum_{g(x)=\alpha}\frac1{\left|g'(x)\right|}\,f(\alpha)\,\mathrm{d}\alpha\tag{1} $$ If we integrate along squares, centered at the origin, whose sides are parallel to the $x$ and $y$ axes with length $2k\pi$, as $k\to\infty$, we get $$ \begin{align} \sum_{x+\tan(x)=\alpha}\frac1{1+\sec^2(x)} &=\frac1{2\pi i}\oint\frac{\mathrm{d}z}{z+\tan(z)-\alpha}\\[6pt] &=1\tag{2} \end{align} $$ Letting $g(x)=x+\tan(x)$, $(1)$ and $(2)$ give $$ \int_{-\infty}^{+\infty}f(x+\tan(x))\,\mathrm{d}x=\int_{-\infty}^{+\infty}f(x)\,\mathrm{d}x\tag{3} $$ Therefore, applying $(3)$ to $f(x)=\frac1{1+x^2}$ yields $$ \begin{align} \int_{-\infty}^{+\infty}\frac{\mathrm{d}x}{1+(x+\tan(x))^2} &=\int_{-\infty}^{+\infty}\frac{\mathrm{d}x}{1+x^2}\\[6pt] &=\pi\tag{4} \end{align} $$