I am working through a derivation in Nielsen and Chuang's 2016 text book Quantum Computation and Quantum Information, and I have not been able to reproduce a result they get for Equation 5.34. While the text is not a math text, this question is. They have an integral $$ \frac{1}{2}\int_{e-1}^{2^{t-1}-1}dl\frac{1}{l^2}=\frac{1}{2(e-1)} $$ with $ t $ an integer and $ e $ a real number. I have not beeen able to reproduce this result. By hand I get $ \frac{1}{2}\left(\frac{1}{e-1}-\frac{1}{2^{t-1}-1}\right) $. Can someone help me to reduce this to the given result? Or is there some sort of limiting argument? Thank you!
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As you mentioned, perhaps they are taking a limit as $t\to\infty$. Any suggestion of that in the problem or the accompanying text? – paw88789 Oct 29 '21 at 15:53
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It simply says that $ t $ is a $ t $-bit binary expansion. Maybe a large $ t $ is implied here? – Anne Oct 29 '21 at 15:58
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It says that $0 \leq 2^t\delta \leq 1$ but I have no idea what delta implies in this context. – rhkoulen Oct 29 '21 at 16:04
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$ \delta $ is the difference between the measured state $ \phi $ and $ 2^{-t} $. The goal is to find the best $ t $-bit approximation $ b/2^t $ where $ b $ is an integer. I guess that might imply a large $ t $. – Anne Oct 29 '21 at 16:09
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I guess the only thing that makes sense here is a limit for large $ t $. Thanks for the help! – Anne Oct 29 '21 at 16:15