# sum to integral inequality step in a proof of Kolmogorov

If I have $N$ numbers $x_j$ very very close to $N$th roots of unity. How could I show $$\frac{1}{N} \sum_{j=1}^N \left|\sin(\tfrac{1}{2}(t-x_j))\right|^{-1} > \int_{1/N}^\pi \left|\sin(\tfrac{1}{2}s)\right|^{-1}ds$$ by very very close I mean $$\left|x_j - \frac{2 \pi j}{N}\right| < \frac{1}{N^2}$$ the reason I ask this is because it's the last step in a proof which got skipped over and I haven't been able to link these myself. Thanks very much for any suggestions.

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this is on page 61 of Katznelson intro to harmonic analysis. –  sperners lemma Nov 21 '12 at 20:46
The left hand side is a function of $t$, whereas on the right hand side $t$ is integrated over? Also, there is an $i$ vs. $j$ mismatch on the left. –  Hagen von Eitzen Nov 21 '12 at 21:26
@HagenvonEitzen, thank you I've fixed it the best I can I think that the RHS is just a constant independent of t. –  sperners lemma Nov 21 '12 at 21:48
So, the sum is a Riemann sum for the integral. Maybe it's just a question of looking at the graph, and comparing areas of some rectangles to area under the curve. –  Gerry Myerson Nov 22 '12 at 0:12