$\frac{1}{N} | \int_1^N e^{2 \pi i b \log x }dx |\rightarrow \frac{1}{ \sqrt{1 + 4\pi ^2 b^2} } $ as $N \rightarrow \infty$ 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
$$ \lim_{N \rightarrow \infty} N^{2 \pi i b } $$

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