I want to show that $$\frac{1}{N} \left| \int_1^N e^{2 \pi i b \log x }dx \right| \rightarrow \frac{1}{ \sqrt{1 + 4\pi ^2 b^2} } $$ as $N \rightarrow \infty$. This what I have done:

First do a substitution $u = \log x$ to obtain $$ \int_1^N e^{2 \pi i b \log x }dx = \int_0^{ \log N} e^{(1+ 2 \pi i b ) u }du .$$ Then $$\int_0^{ \log N} e^{(1+ 2 \pi i b ) u }du = \frac{ e^{(1+ 2 \pi i b ) \log N } -1 }{ 1+ 2 \pi i b} = \frac{ N e^{2 \pi i b \log N } -1 }{ 1+ 2 \pi i b}. $$

So $$\frac{1}{N} \int_1^N e^{2 \pi i b \log x }dx = \frac{ e^{2 \pi i b \log N } }{ 1+ 2 \pi i b}- \frac{ 1 }{ N(1+ 2 \pi i b)}. $$

The second term goes to zero but I'm not sure how to proceed with the first term.

EDIT: So my question is how does one evaluate the limit $$ \lim_{N \rightarrow \infty} N^{2 \pi i b } $$

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    $\begingroup$ Is it true at all? For example, the calculation gives $-0.4814940580e-1-0.9388404786e-1i$ in the case $N=100$ and $b=1.5.$ Maybe, do you mean $$\frac{1}{N} \left|\int_1^N e^{2 \pi i b \log x }dx\right| \rightarrow \frac{1}{ \sqrt{1 + 4\pi ^2 b^2} } ?$$ $\endgroup$ – user64494 Aug 5 '13 at 19:06
  • $\begingroup$ Yes that is what I mean. Thanks for pointing that out. $\endgroup$ – Henrik Finsberg Aug 5 '13 at 21:06
  • $\begingroup$ Because of your work, it suffices you show $$\left| {\frac{{N{e^{2\pi ib\log N}} - 1}}{N}} \right|\to 1$$ $\endgroup$ – Pedro Tamaroff Aug 5 '13 at 21:12

The solution is straightforward. Notice that $$e^{2\pi i b\log x}= x^{2\pi i b}$$

Then the integral is $$\frac{1}{N}\left|\int_1^N x^{2\pi i b} \rm{d}x\right|=\frac{1}{N}\left|\frac{N^{2\pi i b+1}-1}{2\pi i b +1} \right|=\frac{1}{|1+2\pi i b|}\left|N^{2\pi i b}-\frac{1}{N}\right|=$$ $$=\frac{1}{\sqrt{1+4\pi^2b^2}}\left|N^{2\pi i b}-\frac{1}{N}\right|$$

Taking the limit $$\lim_{N\to\infty}\left|N^{2\pi i b}-\frac{1}{N}\right|=\lim_{N\to\infty}\sqrt{1-\frac{2\cos(2\pi b \log N)}{N}+\frac{1}{N^2}}=1.$$


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