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I'm stuck on the following problem: given the integral: $$I(N)=\int \prod_{k=1}^N \left(k-\frac{k}{x}\right) \, dx$$ calculate the following limit: $$I_{\infty}(N)=\lim_{x\to\infty}I(N)$$ I know that $$I(N)=x(1-x)^{-N}\frac{(x-1)^N}{x^N}\Gamma(N+1)_{2}\tilde F_1(1-N,-N;2-N;x)\frac{1}{1-N}$$ but how can I calculate this limit for $x\to\infty$? Thanks in advance.

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In the limit of $x \to \infty$ shouldn't you expect asymptote to be $\int \prod_{k=1}^N k \mathrm{d}x = x N!$, and thus the limit is infinity? –  Sasha Sep 19 '12 at 15:51
    
Are the integral limits 1 to infinity and do you mean N approaches infinity instead of of x? –  Apprentice Queue Sep 19 '12 at 15:51
    
@Sasha: Ok. So the $I_\infty(N)=\infty$. Thanks. –  Riccardo.Alestra Sep 19 '12 at 15:57
    
Please supply the limits of the integral. –  joriki Sep 19 '12 at 16:49

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\begin{eqnarray} I_x(N)&=&\int\prod_{k=1}^N\left(k-\frac{k}{x}\right)dx=N!\int\left(1-\frac{1}{x}\right)^Ndx\cr &=&N!\sum_{k=0}^N(-1)^k{N\choose k}\int x^{-k}dx=N!\left[x-N\ln|x|+\sum_{k=1}^{N-1}\frac{(-1)^k}{k}{N\choose k+1}\left(\frac{1}{x}\right)^k\right]+C. \end{eqnarray} It follows that $$ I_\infty(N)=\lim_{x\to \infty}I_x(N)=\lim_{x\to \infty}N!(x-N\ln|x|)=\infty. $$

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