Find an upper bound to $\int_0^1 \frac{x^n}{\ln (1+x)} dx$.

Question

I've conjectured that

$$\int_0^1 \frac{x^n}{\ln (1+x)} dx $$

where $n$ is a positive integer, goes to $0$ as $n$ grows larger. Knowing this, I wanted to find a non-constant function $f(n)$ such that

$$\int_0^1 \frac{x^n}{\ln (1+x)}dx \leq f(n)$$

where $f(n)$ goes to $0$ as $n$ goes to infinity. Notice that we can write the integral as

$$\lim_{\epsilon \to 0^+}\int_\epsilon^{1+\epsilon} x^n/\ln(1+x)dx $$

since it's improper.

My attempt was

$$\int_0^1\frac{x^n}{\ln (1+x)}dx \leq [\max_{x\in(0,1)} (x)]^n \int_0^1 1/\ln(1+x)dx.$$

However the max is $1^n=1$, so it does not depend on $n$. Moreover, the last integral does not converge.

Any help is appreciated.

Answer 1

Your approach also won't work because $\int_{0}^1\frac{dx}{\log(1+x)}$ doesn't converge. $\log(1)=0$ makes the $x=0$ problematic.

Show for $x\in[0,1]$ that:

$$e^{x/2}\leq 1+x\tag1$$

The inequality follows from the power series for $e^{x/2},$ but it takes a few steps to prove.

Taking the log of $(1)$ you get:

$$\frac x2\leq \log(1+x)$$

From this, deduce a suitable $f.$

Answer 2

$\lim_{x\to0}\frac{\ln(1+x)}x=1$ hence the continuous function $x\mapsto\frac x{\ln(1+x)}$ is bounded on $(0,1]$, say by some $M>0,$ so $$\int_0^1\frac{x^n}{\ln(1+x)}\,\mathrm dx\le M\int_0^1x^{n-1}\,\mathrm dx=\frac Mn.$$

Answer 3

You have $${\ln(1+ x) \over x} = {\ln(1 + x) - \ln(1 + 0) \over x - 0}$$ By the mean value theorem, for each $x$ there's a $y \in [0,x]$ such that $${\ln(1 + x) - \ln(1 + 0) \over x - 0} = {1 \over 1 + y}$$ Since $0 \leq y \leq 1$ we therefore have $${1 \over 2} \leq {\ln(1 + x) - \ln(1 + 0) \over x - 0} \leq 1$$ In other words, we have $${1 \over 2} \leq {\ln(1+ x) \over x} \leq 1$$ Equivalently, $${x \over 2} \leq \ln(1 + x) \leq x$$ Inserting this into your integral should give you what you want.