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Around a couple months ago, I found an interesting integral which I haven't been able to solve yet - it goes as follows:

$$\int_{-\infty}^{\infty} e^{-(x+\tan(x))^2} \mathrm{d}x = \sqrt\pi$$

I've attempted different techniques such as Feynman's technique, Laplace Transform, substitutions, integration by parts, and many more - yet none could crack it.

Further, verifying the result via Wolfram Alpha was unsuccessful.

Gary
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Daddy Myers
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    Glasser's master theorem is a quick way to go. – Sangchul Lee Mar 02 '23 at 12:18
  • To add onto that, see the results in the answers to this famous related question. – KStarGamer Mar 02 '23 at 12:19
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    That integral, as it is, has a fundamental problem: it isn't defined in infinite points on the real line and $;\tan x;$ on these points isn't bounded...Thus I'd say this is not even an improper integral only because of the upper and lower limits, but also the function has some major issues. – DonAntonio Mar 02 '23 at 12:30
  • @DonAntonio The integrand tends to $0$ as $x$ approaches a singularity of $\tan x$. You can define it to be $0$ at those points and you end up with a continuous, bounded integrand. – Gary Mar 03 '23 at 03:11
  • @Gary That's a very good point, Gary. Thanks...though I'm not sure what's the OP allowed to do. That's why I remarked what I did above. – DonAntonio Mar 03 '23 at 07:55

1 Answers1

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We want to find $$\int_{-\infty}^{\infty} e^{-(x+\tan(x))^2} \mathrm{d}x$$

Now, consider the theorem stated below:

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: $$ \int_{-\infty}^\infty f(\phi(x)) dx = \int_{-\infty}^\infty f(x) dx $$

A proof of this theorem can be found here.

Taking $\phi(x) = x+\tan(x)$ and $f(x) = e^{-x^2}$, we can see that $$\int_{-\infty}^{\infty} e^{-(x+\tan(x))^2} \mathrm{d}x = \int_{-\infty}^{\infty} e^{-x^2} \mathrm{d}x = \sqrt\pi$$

The last integral is a standard Gaussian integral.

Max0815
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    Thanks! Although this felt a bit underwhelming after almost 3 months of trying to crack this integral, the fact that such a general fact exists is certainly quite nice. – Daddy Myers Mar 02 '23 at 12:39
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    "such a meromorphic function" - the conditions included in the "such" from the other post should probably be listed, or at least you should mention that the other post does indeed check that $\phi(z)=z+\tan z$ satisfy them. Either way it's a neat solution, +1 – Lorago Mar 02 '23 at 12:47
  • @Lorago thanks! also idk I think it should be pretty intuitive to see that $\phi(x)=z+\tan(z)$ is meromorphic since only the tangent contributes to poles, and $\tan(x)$ itself is meromorphic. – Max0815 Mar 02 '23 at 13:01
  • Did you not fully read the post you linked then? The condition is not only that $\phi$ is meromorphic, but rather that it is meromorphic and satisfies three conditions listed in the post you linked – Lorago Mar 02 '23 at 13:04
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    The function being meromorphic is very clear, but the rest perhaps not – Lorago Mar 02 '23 at 13:05
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    @Lorago I did read the rest of the post, but what i reproduced here was only the statement of the theorem from it, since i felt like a link to the post itself would be enough of a reference to the rest. Now that you mention it though on second thought I will edit in the rest then :P – Max0815 Mar 02 '23 at 13:20