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## Homework Statement

Show ##\int_{0}^{1}e_n(x)\overline e_k(x) dx = 1## if ##n=k## and ##0## otherwise.

## Homework Equations

##e_n(x) = e^{2\pi inx}##.

## The Attempt at a Solution

Consider 2 cases:

case 1: ##n=k##. Then ##\int_{0}^{1} e_n(x) \bar e_k(x) dx = \int_{0}^{1} e_n(x)e_{-k}(x) dx = \int_{0}^{1} e_0(x) dx = \int_{0}^{1} 1 dx = 1(1-0) = 1.##

case 2: ##n \neq k##. Then ##\int_{0}^{1} e_n(x)\bar e_k(x) dx = \int_{0}^{1} e_{n-k}(x) dx = \int_{0}^{1} e^{2\pi i (n-k)}x = \int_{0}^{1} (e^{2\pi i})^{(n-k)x} dx = \int_{0}^{1} 1^{(n-k)x}dx = 1##? Can someone please point out the mistake..

Also this was from lecture notes today:

case 2: ##n \neq k##. In ##L^2([0,1]) = \lbrace f: [0,1] \rightarrow \mathbb{C} \vert \int_{0}^{1} f^2(x) dx < \infty \rbrace## is an Inner Product space and ##<f, g>^2 = \int_{0}^{1} f(x)\bar g(x) dx##. So ##e_n(x)## and ##e_k(x)## are orthonogal.

--end notes--

So that would imply their dot product is 0..

My questions are: Can I please have help on case 2? and reading my notes I am not sure how we get the conclusion ##e_n(x)## and ##e_k(x)## are orthogonal, how do we do this?