0
$\begingroup$

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.

  • 0
    Please supply the limits of the integral.2012-09-19

1 Answers 1

0

\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. $