# Convergence of $\int_0^\infty \frac1{\sqrt x \log(1 + e^x)}$

I am trying to prove the convergence of the following Newton's integral:

$$\int_0^\infty f(x) = \int_0^\infty \frac1{\sqrt x \ln(1 + e^x)}.$$

If $f \in \mathcal{N}(0, \infty)$, then $f \in \mathcal{N}(0, 1)$ and $f \in \mathcal{N}(1, \infty)$. (Where $\mathcal{N}(a, b)$ is the set of functions that have a convergent Newton integral on $(a, b)$ where $a, b \in \mathbb{R}$ and $a < b$.)

We have that $$\forall x \in (0, \infty): 0 < \frac1{\sqrt x \ln(1 + e^x)} < \frac1{\sqrt x \ln(e^x)} = \frac1{\sqrt x x} = x^{-\frac32}.$$

From the comparison test, we get that $\int_1^\infty \frac1{\sqrt x \ln(1 + e^x)}$ is convergent.

Using the limit comparison test, we obtain that

$$\lim_{x\to1^-} \frac{\frac1{\sqrt x \ln(1 + e^x)}}{x^{-\frac32}} = 1 \in (0, \infty)$$

and therefore the integral $\int_0^1 \frac1{\sqrt x \ln(1 + e^x)}$ converges if and only if the integral $\int_0^1 x^{-\frac32}$ converges. Because $\int_0^1 x^{-\frac32}$ is divergent, we get that the given integral $\int_0^\infty \frac1{\sqrt x \ln(1 + e^x)}$ diverges.

Is my proof correct?

Added later: I went wrong where I use the limit comparison test: I would need that $f(x)$ is continuous on $[0, 1)$, which I have forgotten to verify. As pointed out in the answers and comments, the function behaves on $(0, 1)$ as $x^{-\frac12}/\ln2$, and thus converges.

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No, there is a mistake. Note that if $x>0$ then $\log(1+e^x)>\log(2)$. Then $\int_0^1 f\le C \int_0^1 x^{-1/2}$. – Pocho la pantera Sep 6 '13 at 15:22
I don't think a limit comparison test at $1$ gets you anything useful. – dfeuer Sep 6 '13 at 15:23
This improper integral has two potential badnesses: the function blows up near $0$, and the region of integration is infinite. In such situations, it is often useful to separate into two integrals, $\int_0^a$ and $\int_a^\infty$, where $a$ is any positive number you like, say $1$. – André Nicolas Sep 6 '13 at 15:56
@dfeuer I think it would have - if I used it properly. – David Čepelík Sep 6 '13 at 16:02
The integrand is well-behaved around $1$—there's nothing to see there. – dfeuer Sep 6 '13 at 17:49

No. At $x=0$, the integrand behaves as $x^{-1/2}/\log{2}$, which is an integrable singularity. As $x\to\infty$, the integrand approaches $x^{-3/2}$, which leads to a convergent result there. As there are no poles in between, the integral converges.