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 2020-03-17, 12:45 #1 wildrabbitt   Jul 2014 3·149 Posts something about a sum Hi, can anyone explain this? $S=\sum_{v=-\infty}^{\infty}\int_0^Ne^{2\pi ivx+2\pi i\frac{x^2}{N}} \mathrm{d}x$ $=N\sum_{v=-\infty}^{\infty}\int_0^1e^{2\pi iN(x^2+vx)} \mathrm{d}x$ What I'm hoping for is some intermediate steps which get from the first, step by step to the second that make sense. Please help. Last fiddled with by wildrabbitt on 2020-03-17 at 12:53
2020-03-17, 13:19   #2
Chris Card

Aug 2004

100000102 Posts

Quote:
 Originally Posted by wildrabbitt Hi, can anyone explain this? $S=\sum_{v=-\infty}^{\infty}\int_0^Ne^{2\pi ivx+2\pi i\frac{x^2}{N}} \mathrm{d}x$ $=N\sum_{v=-\infty}^{\infty}\int_0^1e^{2\pi iN(x^2+vx)} \mathrm{d}x$ What I'm hoping for is some intermediate steps which get from the first, step by step to the second that make sense. Please help.
Do you know how to do integration by substitution?
If so, try setting x = Ny and rewrite the integral in terms of y instead of x.

Chris

2020-03-17, 13:19   #3
Dr Sardonicus

Feb 2017
Nowhere

32·641 Posts

Quote:
 Originally Posted by wildrabbitt Hi, can anyone explain this? $S=\sum_{v=-\infty}^{\infty}\int_0^Ne^{2\pi ivx+2\pi i\frac{x^2}{N}} \mathrm{d}x$ $=N\sum_{v=-\infty}^{\infty}\int_0^1e^{2\pi iN(x^2+vx)} \mathrm{d}x$ What I'm hoping for is some intermediate steps which get from the first, step by step to the second that make sense. Please help.
Obvious substitution. You said you knew how to make substitutions in integrals.

It is perhaps unfortunate that the variables in the integrals on both sides have the same name.

 2020-03-17, 14:36 #4 wildrabbitt   Jul 2014 3·149 Posts Thanks to both of you. I do understand integration by substitution but I didn't know what the substitution required was. I should be able to do it now.

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