Does the definite integral of (1 - tanh t) from 0 to x diverge as x goes to infinity?

Question

Decidedly in the category of things I used to know how to prove but have forgotten: Does

$$ \int_0^x (1 - \tanh t) \,dt $$

converge or diverge as $x \to \infty$? (I know that the indefinite integral is $x - \ln \cosh x + C$, which suggests to me that it does diverge, but that's hardly a proof.)

Answer 1

A direct approach, in terms of your indefinite integral:

We want to show that $\lim_{x\to\infty}x-\log\cosh x=c\in\Bbb R$ in order to prove that the given integral converges. In fact, we will show that $c=\log 2$, and thus also prove that your definite integral has value $\log 2$. (Note that your $C=0$, seen by evaluating at $x=0$.)

\begin{align}\cosh x=\frac{e^{-x}+e^x}2\implies x-\log\cosh x&=\log e^x-\log(e^{-x}+e^x)+\log2\\ &=\log2-\log(e^{-2x}+1) \end{align}

For large $x$, $e^{-2x}\to0$, so the expression approaches $\log 2$ as desired.

Answer 2

We are integrating $$1-\frac{e^t-e^{-t}}{e^t+e^{-t}},$$ that is, $$\frac{2e^{-t}}{e^t+e^{-t}}.$$ This is positive and less than $2e^{-2t}$. The integral $\int_0^\infty 2e^{-2t}\,dt$ converges, so by Comparison so does our integral.

Remark: The expression $x-\ln(\cosh x)$ strongly suggests convergence, since $\cosh x$, for large $x$, is essentially indistinguishable from $\frac{e^x}{2}$. The integral to $\infty$ is in fact $\ln 2$.