-1
$\begingroup$

For a test prep question:

Let $g(x)=\tanh (x)$, and let $\mu$ be the measure generated by $g$. Which subsets of the reals are $\mu$-measurable? Are polynomials integrable? And finally, is $|\sinh|$ integrable?

  • 0
    Since $\int f(x) d \tanh(x) = \int f(x) \mathbb{sech}^2(x) dx$ and $\int |f(x) |\mathbb{sech}^2(x) dx \leq 2\int |f(x)| e^{-2x}dx$, it follows that all polynomials are integrable. It also follows from this that $\sinh$ is integrable.2012-09-13

1 Answers 1

1

Measurable sets are the same as the usual case because $\tanh$ is strictly increasing. Integration with function $f$ becomes $\int f(x)\tanh'(x) dx = \int \frac{f(x)}{\cosh^2(x)} dx$. Substituting a polynomial or $|\sinh|$ for $f$ will make the integral finite, so polynomials and $|\sinh|$ are integrable. (The domain is $(-\infty, \infty)$.)

  • 0
    $\int_0^\infty \frac{\sinh}{\cosh^2(x)} dx = \int_0^{\infty} \operatorname{sech}(x)\tanh(x)dx = -\operatorname{sech}(x)|_0^{\infty} = 1$.2012-09-13