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 .
Best Answer
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?):
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 $$\int_{-\infty}^{\infty}\text{sech}^2(x+\tan(x))dx = \int_{-\infty}^{\infty}\text{sech}^2(x)dx = \operatorname{tanh}(x)\Big|^{\infty}_{-\infty} = 2$$
$$2I=2\Longleftrightarrow I=\boxed{1}$$