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Evaluate the Integral: $$\int_{0}^{\infty}\text{sech}^2(x+\tan(x))dx$$ Source: MIT Integration Bee

My Try:
Applying Glasser's Master Theorem, the value of improper integral doesn't change. Substituting $x$ in place of $x+\tan(x)$ we have $$\int_{0}^{\infty}\text{sech}^2(x)dx=\left[ \tanh (x) \right]_{0}^{\infty}=\lim_{x \to \infty} \tanh(x)=\lim_{x \to \infty} \frac{e^{2x}-1}{e^{2x}+1}= 1$$
I don't know whether the solution is correct; can anyone tell me please? Also Is there any other method of solving it? Any help would be appreciated .

S.Agera1729
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    And shouldn't that be enough to show that the answer must be $1$, calculating the limit? – A. P. Jan 28 '23 at 08:11
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    Your formula is false. If you change the variable $y=x+\tanh(x)$, you have $dy = (2-\tanh^2x)dx$, and $(2-\tanh^2)$ is not a simple function of $x$. Are you sure that a closed formula exists for this integral? – Christophe Leuridan Jan 28 '23 at 08:31
  • I can't see how Glasser's master theorem is related to your integral. – Anne Bauval Jan 28 '23 at 08:36
  • @ChristopheLeuridan $y=x+\tan x,$ not $\tanh$ (but that makes no difference and I agree with you). – Anne Bauval Jan 28 '23 at 08:39
  • Yeah! I have just shared my two approaches of how I tried to solve it . As said by @A.P. the answer is $1$ but I am trying to solve it in that way . – S.Agera1729 Jan 28 '23 at 08:45
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    I don't see both approaches. On the other hand, I mentioned that the answer is $1$ using your approach. Your approach seems correct to me, you just need to better justify the use of the result. – A. P. Jan 28 '23 at 08:48
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    Starting from $$ \pi \cot z=\frac{1}{z} +\sum_{n=1}^\infty\Big(\frac{1}{z-\pi n}+\frac{1}{z-\pi n}\Big)$$ and making the change $ z=\frac{\pi}{2}-x $ , we get $$ \tan x=-\frac{1}{x-\pi/2}-\sum_{n=1}^\infty\Big(\frac{1}{x+\pi n-\pi/2}+\frac{1}{x-\pi n-\pi/2}\Big)$$ Therefore, the conditions of the theorem are met. https://www.ams.org/journals/mcom/1983-40-162/S0025-5718-1983-0689471-1/S0025-5718-1983-0689471-1.pdf $$u=x-\sum_{j=1}^{n-1}\frac{a_j}{x-C_j}$$ $ a_j $- positive and $ C_j $ - real constants. The theorem is applicable even to the infinite series. – Svyatoslav Jan 28 '23 at 09:44
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    Hi @Svyatoslav. Yes, that is correct of course. Although I find it strange that the OP made the part more difficult and did not know how to conclude. That's why I added that it's what the OP needs to justify. – A. P. Jan 28 '23 at 09:53
  • $\int_{-\infty}^{\infty} \text{sech}^2(x-1/x) dx=2$ is truely a Glasser, OP's integral is not Glasser. – Z Ahmed Jan 28 '23 at 09:57
  • @Svyatoslav That is not true. The sum contains two reciprocal functions, not one like in the original definition. – Kamal Saleh Jul 31 '23 at 23:08
  • @Kamal Saleh This is true, because $$u=x+\tan x=x-\sum\frac{a_j}{x-C_j}$$ Please have a look the part between formula (3) and (4) (similar case) in the original paper of M.L. Glasser - following the link that I attached above. – Svyatoslav Aug 01 '23 at 01:58

1 Answers1

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Your method is correct, and your value seems to be correct as well (numerically verified).

We want to find $$I=\int_{0}^{\infty}\text{sech}^2(x+\tan(x))dx$$

Now, consider the integral $$\int_{-\infty}^{\infty}\text{sech}^2(x+\tan(x))dx=2I$$

Next, we consider the theorem stated below (which is basically another variant of the theorem you mentioned in your OP ig?):

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 $f(x) = \operatorname{sech}^2(x)$, we can see that $$\int_{-\infty}^{\infty}\text{sech}^2(x+\tan(x))dx = \int_{-\infty}^{\infty}\text{sech}^2(x)dx = \operatorname{tanh}(x)\Big|^{\infty}_{-\infty} = 2$$

$$2I=2\Longleftrightarrow I=\boxed{1}$$

Max0815
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