Cannot find the limit of this integral 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?
Regards,
 A: 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 $0\le \ln x\le x-1$.  
A: $$0<\int_1^{ n+1 }\frac { \ln x }{ x^n } \, dx<\int_1^{\infty}\frac { \ln x }{ x^n } \, dx < \int_1^{\infty }x^{1-n } \, dx=\frac{1}{n-2}$$
but $\lim_{n\to\infty}\frac{1}{n-2}=0$, so by the Squeeze Theorem your integral is $0$.
A: $$
\begin{align}
& \int_1^{n+1} \underbrace{\Big(\ln x\Big)}_{u} \underbrace{\left(\frac{1}{x^n} \, dx\right)}_{dv} = \int u\,dv = uv - \int v\,du \\[12pt]
& = \left.(\ln x)\left(\frac{x^{-n+1}}{-n+1}\right)\right|_1^{n+1} - \int_1^{n+1} \left(\frac{x^{-n+1}}{-n+1}\right) \frac{dx}{x} \\[12pt]
& = \frac{\ln(n+1)}{-n+1}\cdot\frac{1}{(n+1)^{n-1}} + \frac{1}{n-1} \int_1^{n+1} x^{-n}\,dx \\[12pt]
& = \frac{1}{n-1}\cdot\frac{1}{(n+1)^n} + \frac{1}{n-1}\cdot\left(\frac{1}{(n+1)^n}-1\right) \to 0 \text{ as }n\to\infty.
\end{align}
$$
A: 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.
A: \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}
$$
\begin{array}{|c|}\hline
\\
\color{#ff0000}{\large\quad%
\lim_{n \to \infty}\int_{1}^{n + 1}{\ln\left(x\right) \over x^{n}}\,{\rm d}x
=
0\quad}
\\ \\
\hline
\end{array}
$$
