# What is the limit of $\lim\limits_{n\rightarrow\infty}\frac{1}{n^4}\left(\sum_{k=1}^{n}\ k^2\int_{k}^{k+1}x\ln\big((x-k)(k+1-x)\big)dx\right)$

As the topic how to find the limit of $$\lim_{n\rightarrow\infty}\frac{1}{n^4}\left(\sum_{k=1}^{n}\ k^2\int_{k}^{k+1}x\ln\big((x-k)(k+1-x)\big)dx\right)\;.$$

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The title and the body ask two different questions, let me answer the one in the body: decompose the log into the sum of two logs, transform every integral into an integral on $(0,1)$, perform the cancellations which appeared, use the fact that the sum of the $n$ first cubes is approximately $\frac14n^4$ and compute the integral of the log on $(0,1)$. The answer should be $-\frac12$. – Did Jan 12 '12 at 14:58
i dont quite get how to manipulate it. – Mathematics Jan 12 '12 at 15:05
@Mathematics The integral is $$\begin{eqnarray*} &&\int_{k}^{k+1}x\ln \left( \left( x-k\right) \left( k+1-x\right) \right) dx \\ &=&\int_{k}^{k+1}x\ln \left( x-k\right) dx+\int_{k}^{k+1}x\ln \left( k+1-x\right) dx \\ &=&\int_{0}^{1}\left( u+k\right) \ln u\,du+\int_{0}^{1}\left( k+1-u\right) \ln u\,du \\ &=&\ldots \\ &=&-k-\frac{1}{4}. \end{eqnarray*}$$ – Américo Tavares Jan 12 '12 at 15:42
@Américo: Not $=-(2k+1)$? Maybe I am wrong, I did this too quickly and in my head, so... – Did Jan 12 '12 at 16:07
@DidierPiau: Yes, you are right. It is $(-k-1/4)+(-k-3/4)=-(2k+1)$ – Américo Tavares Jan 12 '12 at 16:39

Put $I_k:=\int_k^{k+1}x\ln ((x-k)(k+1-x))dx$; making the substitution $t=x-k$ we have, following @Didier Piau-'s idea $$I_k=\int_0^1(t+k)\ln(t(1-t))dt=\int_0^1t\ln (t(1-t))dt+k\int_0^1\ln (t(1-t))dt.$$ Since $0\leq \frac 1{n^4}\sum_{k=1}^nk^2\leq \frac{n\cdot n^2}{n^4}$, the limit we are looking for is $$l:=\lim_{k\to\infty}\frac 1{n^4}\sum_{k=1}^nk^3\int_0^1(\ln t+\ln(1-t))dt.$$ Since $\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$, we have $$l=\frac 14\cdot 2\int_0^1\ln tdt=\frac 12\left([t\ln t]_0^1-\int_0^1t\frac 1tdt\right)=-\frac 12.$$
If you apply Stolz Cezaro to the original sequence you get exactly $\lim_n \frac{I_n}{n}$. – N. S. Jan 24 '12 at 16:40
Two substitutions, first $x\mapsto x+k$ and then $x\mapsto1-x$ yields \begin{align} &\int_k^{k+1}x\log\left((x-k)(k+1-x)\right)\,\mathrm{d}x\\ &=\int_0^1(x+k)\log\left(x(1-x)\right)\,\mathrm{d}x\\ &=\int_0^1(x+k)\left[\log\left(x\right)+\log\left(1-x\right)\right]\,\mathrm{d}x\\ &=\int_0^1(2k+1)\log(x)\,\mathrm{d}x\\ &=-(2k+1)\tag{1} \end{align} This gives \begin{align} &\lim_{n\to\infty}\frac1{n^4}\sum_{k=1}^nk^2\int_k^{k+1}x\log\left((x-k)(k+1-x)\right)\,\mathrm{d}x\\ &=-\lim_{n\to\infty}\frac1{n^4}\sum_{k=1}^nk^2(2k+1)\\ &=-\lim_{n\to\infty}\sum_{k=1}^n\left(\frac kn\right)^2\left(2\frac kn+\frac1n\right)\frac1n\tag{2} \end{align} and $(2)$ is a Riemann sum for $$-\int_0^12x^3\,\mathrm{d}x=-\frac12\tag{3}$$