Show $\int_{0}^{2\pi}e^{int}\mathrm dt=0$ for all $n\neq0$

Question

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 $0$.

Answer 1

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 $n$.

Answer 2

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.

$$\therefore~\int_0^{2\pi}e^{int}\mathrm dt~=~0~~~,\forall n\in\mathbb Z\setminus\{0\}$$

Answer 3

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.