# Evaluate $\int_{0}^{\infty}\text{sech}^2(x+\tan(x))dx$

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 .

• And shouldn't that be enough to show that the answer must be $1$, calculating the limit? Jan 28 at 8:11
• 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? Jan 28 at 8:31
• 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. Jan 28 at 8:48
• 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. ams.org/journals/mcom/1983-40-162/S0025-5718-1983-0689471-1/… $$u=x-\sum_{j=1}^{n-1}\frac{a_j}{x-C_j}$$ $a_j$- positive and $C_j$ - real constants. The theorem is applicable even to the infinite series. Jan 28 at 9:44
• 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. Jan 28 at 9:53

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 $$\int_{-\infty}^{\infty}\text{sech}^2(x+\tan(x))dx = \int_{-\infty}^{\infty}\text{sech}^2(x)dx = \operatorname{tanh}(x)\Big|^{\infty}_{-\infty} = 2$$

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

• So, it is still Master theorem of someone... Jan 28 at 14:08