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Glasser's Master Theorem States that if a function $\phi(x)$ can be written as $$|a|x-\sum_{n=1}^N\frac{|\alpha_n|}{x-\beta_n}\tag{1}$$for a constant $a$ and sequences $\alpha_n$ and $\beta_n$ that are defined on the integers between $1$ and $N$, then we have that $$\int_{-\infty}^\infty F(x)dx=\int_{-\infty}^\infty F(\phi(x))dx$$Where the LHS exists and its integrand is integrable. This theorem is used to prove that $$\int_0^\infty \text{sech}^2(x+\tan x)dx=2$$ (given in the MIT integration Bee, see here) which means that $x+\tan x$ can be written in the form of $(1)$. But what would $a,\alpha_n,\beta_n$ and $N$ be? I tried using the taylor series expansion of $\tan x$ but its coefficient is a monster, and the sum is infinite.

This is NOT a duplicate to the second link since I am wondering how $x+\tan x$ can be written in the form of $(1)$.

Kamal Saleh
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    This post has an explanation from user Olivier Oloa – Тyma Gaidash May 16 '23 at 19:23
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    Pls don't take it otherwise, but dude I'm literally speechless that you're just in 8th grade and you know this much calculus +1 – MathStackexchangeIsNotSoBad May 16 '23 at 19:27
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    @MathStackexchangeIsNotSoBad Thanks for the compliment :) To add to the insanity, I only started to learn advanced math less than a year ago. – Kamal Saleh May 16 '23 at 19:46
  • @TymaGaidash Thank you for the link, but it doesn’t quite answer my question. I want to know how to re-write $x+\tan x$ in the form of $(1)$ – Kamal Saleh May 16 '23 at 19:48
  • Here , we have transcendental $\tan$ , hence $N$ has to be infinite. Then $a$ , $\alpha_n$ , $\beta_n$ are not unique. – Prem May 16 '23 at 20:10
  • In https://math.stackexchange.com/questions/4627456/evaluate-int-0-infty-textsech2x-tanxdx (same link as given by OP), a series is introduced for $\tan x$ as$$\tan x=-\frac{1}{x-\pi/2}-\sum_{n=1}^\infty (\frac{1}{x-\pi n-\pi/2}+\frac{1}{x+\pi n-\pi/2}).$$Why does not this help? – Mostafa Ayaz May 16 '23 at 20:11
  • @MostafaAyaz Assalamu alaikum, the two terms in the sum make a denominator with $x^2$ and not $x$ when combined into a single fraction. Maybe it can be simplified but I am not sure since I am on the phone right now. – Kamal Saleh May 16 '23 at 21:11
  • @Prem What do you mean by not unique? I know only so much. – Kamal Saleh May 16 '23 at 22:01
  • Are you sure the equation in $(1)$ is correct? It can't be. Assuming you meant $\sum_{n=1}^{N}$, set every $\alpha_n = 0$ and you can see a contradiction. – Accelerator May 16 '23 at 23:13
  • Both Wikipedia and the Mathworld article use a similar formula to that of the OP – Тyma Gaidash May 16 '23 at 23:26
  • @Accelerator oops, you're right. Thanks – Kamal Saleh May 17 '23 at 00:46
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    And if you don't mind me being nitpicky, the theorem says $\operatorname{PV}\int_{-\infty}^\infty F(x)dx=\operatorname{PV}\int_{-\infty}^\infty F(\phi(x))dx$, which is generally different from $\int_{-\infty}^\infty F(x)dx=\int_{-\infty}^\infty F(\phi(x))dx$. But for $\int_{-\infty}^\infty \text{sech}^2(x+\tan x)dx=2$ it's fine to omit the $\operatorname{PV}$ operator because the integrand has removable discontinuities at $x=(2n+1)\pi/2$ for any $n \in \mathbb{Z}$ and is integrable on $\mathbb{R}$. – Accelerator May 17 '23 at 02:25
  • 'Maybe it can be simplified but I am not sure since I am on the phone right now.' Wow! – A rural reader May 17 '23 at 03:11
  • I will elaborate my Previous Comment : [[1]] When $N$ is finite , we will have rational function which can not give the transcendental function $\tan$ , hence $N$ is infinite here. [[2]] When we make $\alpha_n=1$ , we will get certain $\beta_n$ values. When we make $\alpha_n=n$ , we will get alternate $\beta_n$ values. When we make $\alpha_n=n^2$ , we will get variant $\beta_n$ values. In other words , We can control $\alpha_n$ unknowns to generate the $\beta_n$ unknowns , hence there may be no unique way to get those values. [[3]] Exact Solution is unnecessary , Solution Existence is enough ! – Prem May 17 '23 at 05:11

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