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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 $-\frac12$.2012-01-12
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    i dont quite get how to manipulate it.2012-01-12
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    @Mathematics The integral is $$\begin{eqnarray*} &&\int_{k}^{k+1}x\ln \left( \left( x-k\right) \left( k+1-x\right) \right) dx \\ &=&\int_{k}^{k+1}x\ln \left( x-k\right) dx+\int_{k}^{k+1}x\ln \left( k+1-x\right) dx \\ &=&\int_{0}^{1}\left( u+k\right) \ln u\,du+\int_{0}^{1}\left( k+1-u\right) \ln u\,du \\ &=&\ldots \\ &=&-k-\frac{1}{4}. \end{eqnarray*}$$2012-01-12
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    @Américo: Not $=-(2k+1)$? Maybe I am wrong, I did this too quickly and in my head, so...2012-01-12
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    @DidierPiau: Yes, you are right. It is $(-k-1/4)+(-k-3/4)=-(2k+1)$2012-01-12
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    Correction: $$\begin{eqnarray*} &&\int_{k}^{k+1}x\ln \left( \left( x-k\right) \left( k+1-x\right) \right) dx \\ &=&\int_{k}^{k+1}x\ln \left( x-k\right) dx+\int_{k}^{k+1}x\ln \left( k+1-x\right) dx \\ &=&\int_{0}^{1}\left( u+k\right) \ln u\,du+\int_{0}^{1}\left( k+1-u\right) \ln u\,du \\ &=&-k-\frac{1}{4}-k-\frac{3}{4} \\ &=&-2k-1 \end{eqnarray*}$$2012-01-12
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    @Mathematics ... and the sum of squares and cubes of the first $n$ natural numbers is ([Wolfram MathWorld](http://mathworld.wolfram.com/PowerSum.html) ) $$\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$$ and $$\sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4}.$$2012-01-12
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    i get it,i thought i could solve it by mean of converting into a form of integration but obviously using the identities is the way to solve2012-01-12

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