I am struggling to see how this result holds for non-integer $n$ because $$\int_{0}^{2\pi}e^{int}\mathrm dt=\int_{0}^{2\pi}[\cos(nt)+i\sin(nt)]\mathrm dt$$ and this works out to be $$\frac{\sin(2\pi n)}{n}+i\left(-\frac{\cos(2\pi n)}{n}+\frac{1}{n}\right)$$ and I only see this equaling $0$ when $n$ is an integer not equal to $0$.
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It's fairly common for $n$ to denote an integer. Preumably what was meant was to show this for all integers $n>0$. As you've observed, it's not true for arbitrary real $n$.
Daniel McLaury
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but cant we use Cauchy's theorem to say it is true for all n not 0? – math111 Mar 25 '19 at 12:35
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No, because it isn't true for non-integer $n$. – Daniel McLaury Mar 25 '19 at 12:36
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Cauchy's theorem is about integrating the value of an analytic function around a closed path. For non-integer $n$ there's no way to view this integral as the path integral of any analytic function around a closed path. – Daniel McLaury Mar 25 '19 at 12:39
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If $n$ is a non-zero real number then $\int_0^{2\pi}e^{int}\, dt \equiv \frac {e^{i2n\pi}-1} {in}=0$ if and only if $n$ is an integer.
Kavi Rama Murthy
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@DLB $z^{n}$ is multi-valued in general. It is single valued and analytic when $n$ is an integer. So Cauchy's Theorem does not apply when $n$ is not an integer. – Kavi Rama Murthy Mar 25 '19 at 12:36
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You are perfectly right! For arbitrarely $n$ the given equality does not hold. One may note that your final expression can also be written as
$$\frac{\sin(2\pi n)}{n}+i\left(-\frac{\cos(2\pi n)}{n}+\frac{1}{n}\right)=\frac1{in}(e^{2\pi in}-1)$$
Since $e^{2\pi in}=1$ for all integer $n$ $($beside $n=0$$)$ we can conclude that the integral equals indeed zero.
$$\therefore~\int_0^{2\pi}e^{int}\mathrm dt~=~0~~~,\forall n\in\mathbb Z\setminus\{0\}$$
mrtaurho
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