4
$\begingroup$

I would like to show that

$ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx \sim_{n\rightarrow \infty} \frac{1}{n}$

Using the change of variable $u=x^n$:

$ I_{n}=\frac{1}{n^2} \int_0^1 \frac{u^{1/n} \ln u}{u^{1/n}-1} \mathrm du=\frac{1}{n^2}\left(\int_0^1 \ln x \mathrm dx+\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \right)=\frac{-1}{n^2}+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx=o(1/n)+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx$

So I have to show that

$ \int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \sim_{n \rightarrow \infty} n$

Could you help me?

  • 0
    Oddly by my slightly non-rigorous methods, I am getting your original integral ($I_n$) to be roughly $\zeta(2) - H_{n,2}$. So, that would suggest as $n \rightarrow \infty$, the value to go to zero so I am quite baffled by your question at the moment. (Or it could be that I am horribly wrong in my method :-) )2012-06-10

3 Answers 3

5

$\frac{x^n}{x-1} = \frac{x^n-1}{x-1} + \frac{1}{x-1}= \left(1+ x+ \cdots + x^{n-1}\right) +\frac{1}{x-1}.$

So $I_n = \int^1_0 (1+x+\cdots + x^{n-1}) \log x + \int^1_0 \frac{\log x}{x-1} dx.$

Integration by parts shows $ \int^1_0 x^k \log x = -1/(k+1)^2$ and expanding $\log$ by Taylor series will show $\displaystyle \int^1_0 \frac{\log x}{x-1} dx = \frac{\pi^2}{6}$ so $I_n = \sum_{k=n+1}^{\infty} \frac{1}{k^2}.$

Thus $nI_n = \frac{1}{n} \sum_{k=n+1}^{\infty} \frac{1}{(k/n)^2} \to \int_1^{\infty} \frac{1}{x^2} dx=1$

so $I_n \sim 1/n.$


You can skip showing $\displaystyle \int^1_0 \frac{\log x}{x-1} dx = \frac{\pi^2}{6}$ if you expand $x^n/(x-1)$ as a geometric series from the start.

Instead of using Riemann sums we could also have noted that $1/x^2$ is monotonically decreasing and use the well known theorem that if $f$ is monotone then $\displaystyle \int^n_1 f(x) dx \sim \sum_{k=1}^n f(n).$

  • 0
    @RagibZaman In fact we don't even need to full force of integration of that estimate. It's easy to show that \sum_{k>n}1/k(k+1) < \sum_{k>n}k^{-2} < \sum_{k>n}1/k(k-1), so 1/(n+1) < I_n < 1/n. Hence $I_n = \Theta(n^{-1})$. I use Euler-Maclaurin formula to show that the more precise asymptotics can be obtained through a systematic way.2012-06-11
2

I propose different approach. First, consider integral: $J_n = \int_0^1 \frac{x^n -1}{x-1} \, dx$ than we have $I_n = \tfrac{dJ_n}{dn}$. Observe that: $J_n = \int_0^1 \frac{x^n - 1 }{x-1} \, dx = H_n \sim \ln n + \ldots$ So it's now easy to note that: $I_n \sim \frac{1}{n}$ for large $n$.

  • 0
    @TenaliRaman you're right, I haven't noticed this divergence before. Thanks!2012-06-10
0

Actually, a closed form solution may be given:

$I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}dx=-\int_0^1[x^n(1+x^2+x^3+...)\ln x]dx=$

$=-\sum_{k=0}^\infty \int_0^1 (x^{k+n}\ln x)dx=\sum_{k=0}^\infty\frac{1}{(k+n+1)^2}=$

$=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...=\zeta(2,n+1) $ where $\zeta(s,a)=\sum_{k=0}^\infty\frac{1}{(k+a)^s}$ is so called Hurwitz Zeta Function