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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?

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I don't understand the last equality sign in the first line of your computation. –  Phira Jun 10 '12 at 11:11
    
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 :-) ) –  TenaliRaman Jun 10 '12 at 11:36

3 Answers 3

up vote 5 down vote accepted

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

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 $\displaystyle \int^n_1 f(x) dx \sim \sum_{k=1}^n f(n).$

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+1 Typo $I_n \sim 1/n$. Nice, I reached $I_n = \zeta(2) - H_{n,2}$ but somehow thought since $H_{n,2} \rightarrow \zeta(2)$, $I_n$ must go to zero, but your final step was so simply done. –  TenaliRaman Jun 10 '12 at 12:01
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@TenaliRaman Thanks for pointing that out. All your thoughts are correct, so perhaps you just misunderstood what the question was asking. It is indeed true that $I_n = \zeta(2) - H_{n,2}$ and that $H_{n,2}\to \zeta(2) $ so $I_n \to 0$, but the question was a more precise statement: How quickly does it go to zero? We showed that $\lim_{n\to\infty} nI_n=1$ so for very large values of $n$ we will have $I_n \approx 1/n.$ –  Ragib Zaman Jun 10 '12 at 12:09
    
Ah, it was indeed the case of me completely misunderstanding the question. Thanks for that clarification :-) –  TenaliRaman Jun 10 '12 at 12:15
    
@RagibZaman The $\sum_{k\ge{}n}k^{-2}$ can also be estimated by Euler-Maclaurin formula, which gets $n^{-1}+\frac12n^{-2}+O(n^{-3})$, so $\sum_{k>n}k^{-2} = n^{-1}-\frac12n^{-2}+O(n^{-3})$. –  Frank Science Jun 10 '12 at 12:41
    
@Frank Science In fact we don't even need to full force of Euler-Maclaurin for that estimate, the Trapezoidal rule is enough to tell us that. –  Ragib Zaman Jun 10 '12 at 13:16

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 $n$.

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1  
Differentiating both sides of an asymptotic formula is troublesome, for example consider $n+\sqrt{n}\sin n \sim n.$ –  Ragib Zaman Jun 10 '12 at 11:54
    
@RagibZaman sure, but here $I_n$ behaves fine enough. –  qoqosz Jun 10 '12 at 12:03
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@qoqosz This was precisely how I approached to solve and find the value of the integral, though I considered $J_n$ to be $\int_0^1 \frac{x^n - 1}{x - 1}$ since the $J_n$ as currently defined will have a diverging term. $J_n$ then by my definition is simply $H_n$ whose derivative comes out to be $\zeta(2) - H_{n,2}$ as given here. –  TenaliRaman Jun 10 '12 at 12:24
    
@TenaliRaman you're right, I haven't noticed this divergence before. Thanks! –  qoqosz Jun 10 '12 at 12:31

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

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