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I am trying to find the limit of the following integral as $n$ grows to infinity.

$$\int_1^{ n+1 }\frac { \ln x }{ x^n } \, dx $$

However, when I am doing an integration by parts, the result looks ugly. It is not in a simple form from which, I think, one can find the limit in a simple way (indeterminate form, I mean).

So there should be a trick in there to find the limit.

Can someone help me on this one please?


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Why does the result of integration by parts look ugly? You get a pretty simple thing. – Daniel Fischer Sep 4 '13 at 22:09
Daniel, just try using Wolfram Alpha, to see the look, it is not a simple thing... – XCoder Sep 4 '13 at 22:14
up vote 3 down vote accepted

$$0<\int_1^{ n+1 }\frac { \ln x }{ x^n } \, dx<\int_1^{\infty}\frac { \ln x }{ x^n } \, dx < \int_1^{\infty }x^{1-n } \, dx=\frac{1}{n-2}$$ but $\lim_{n\to\infty}\frac{1}{n-2}=0$, so by the Squeeze Theorem your integral is $0$.

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Let $u=\ln x$ and $dv=x^{-n}\,dx$. Then $du=\frac{1}{x}\,dx$ and we can take $v=\frac{1}{-n+1}x^{-n+1}$.

The second integration will be easy.

Remark: One can avoid some of the complications by noting the fact that for $x\ge 1$ we have $0\le \ln x\le x-1$.

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That's what I did, André, but the final result is not that simple. I did not reproduce here, because I have trouble typing long equations using MathJax. But looking at the result from Wolfram Alpha, it is not so easy to derive a limit here. Result: (1 - (1 + n)^(1 - n) (1 + (-1 + n) Log[1 + n]))/(-1 + n)^2 – XCoder Sep 4 '13 at 22:15
If you want slightly less ugly, you can use the fact that for $x\ge 1$, we have $0\le \ln x\le x-1$. – André Nicolas Sep 4 '13 at 22:25
For me, this thing always tends to 0 as n grows. Am I correct in saying this ? – XCoder Sep 4 '13 at 22:26
Yes, the limit is $0$. If one has experience, it is obvious. Pick a small $\epsilon$. The integral from $\epsilon$ up goes to $0$, because $\frac{1}{x^n}$ shrinks rapidly. The integral from $1$ to $1+\epsilon$ is also small, because on this interval $\ln x$ is close to $0$. – André Nicolas Sep 4 '13 at 22:29
There are also fancy theorems one could quote, but I am guessing the course is an early one. – André Nicolas Sep 4 '13 at 22:36

$$ \begin{align} & \int_1^{n+1} \underbrace{\Big(\ln x\Big)}_{u} \underbrace{\left(\frac{1}{x^n} \, dx\right)}_{dv} = \int u\,dv = uv - \int v\,du \\[12pt] & = \left.(\ln x)\left(\frac{x^{-n+1}}{-n+1}\right)\right|_1^{n+1} - \int_1^{n+1} \left(\frac{x^{-n+1}}{-n+1}\right) \frac{dx}{x} \\[12pt] & = \frac{\ln(n+1)}{-n+1}\cdot\frac{1}{(n+1)^{n-1}} + \frac{1}{n-1} \int_1^{n+1} x^{-n}\,dx \\[12pt] & = \frac{1}{n-1}\cdot\frac{1}{(n+1)^n} + \frac{1}{n-1}\cdot\left(\frac{1}{(n+1)^n}-1\right) \to 0 \text{ as }n\to\infty. \end{align} $$

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I think you forgot about your $\ln(n+1)$ term while moving from line $3$ to line $4$? – pre-kidney Sep 4 '13 at 23:38

I am seeing a whole lot of zero here. If you do the manual integration then evaluate the result term by term there are -n exponents on all of the terms in the numerators. If you expand your result in mathematica you get something like

$$-\frac{(n+1)^{1-n}}{(n-1)^2}+\frac{1}{(n-1)^2}-\frac{n (n+1)^{1-n} \log (n+1)}{(n-1)^2}+\frac{(n+1)^{1-n} \log (n+1)}{(n-1)^2}.$$

(I just used Expand[] on your result, did TeXForm[%] on that and pasted that result in here between dollar symbols.)

Anyway, as $n \rightarrow \infty$, the numerators (note the -n exponents) approach zero, the denominators diverge, and so the limit of the integral is zero.

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One can say either the sequence approaches zero or the limit is zero, but you shouldn't say the limit approaches zero. – Michael Hardy Sep 4 '13 at 22:53
I will buy that. Thanks, and I must say your derivation is top notch writing. – J. W. Perry Sep 4 '13 at 22:56
Thank you. ${{{{}}}}$ – Michael Hardy Sep 4 '13 at 23:03
Thanks to all for your help. – XCoder Sep 4 '13 at 23:38

\begin{align} \int_{1}^{n + 1}{\ln\left(x\right) \over x^{n}}\,{\rm d}x &= \lim_{m \to 0} {\partial \over \partial m}\int_{1}^{n + 1}{x^{m} \over x^{n}}\,{\rm d}x = \lim_{m \to 0} {\partial \over \partial m} \left\lbrack\left(n + 1\right)^{m - n + 1} - 1 \over m - n + 1\right\rbrack \\[3mm]&= \lim_{m \to 0} {\left(n + 1\right)^{m - n + 1}\ln\left(n + 1\right)\,\left(m - n + 1\right) - \left\lbrack\left(n + 1\right)^{m - n + 1} - 1\right\rbrack \over \left(m - n + 1\right)^{2}} \\[3mm]&= {\left(n + 1\right)^{1 - n}\,\ln\left(n + 1\right)\,\left(1 - n\right) - \left(n + 1\right)^{1 - n} + 1 \over \left(n - 1\right)^{2}} \\[5mm]--------&------------------------------- \end{align}

\begin{align} \lim_{n \to \infty}\int_{1}^{n + 1}{\ln\left(x\right) \over x^{n}}\,{\rm d}x &= \lim_{n \to \infty} {n^{-n}\,\ln\left(n\right)\,\left(-n\right) - n^{-n} + 1 \over n^{2}} \\[3mm]&= -\lim_{n \to \infty}{\ln\left(n\right) \over n^{n + 1}} + \lim_{n \to \infty}{1 \over n^{n + 2}} + \lim_{n \to \infty}{1 \over n^{2}} \\[5mm]& \end{align}

$$ \begin{array}{|c|}\hline \\ \color{#ff0000}{\large\quad% \lim_{n \to \infty}\int_{1}^{n + 1}{\ln\left(x\right) \over x^{n}}\,{\rm d}x = 0\quad} \\ \\ \hline \end{array} $$

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