# Integrating an exponential function

## 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?

fresh_42
Mentor

## 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..
You used arithmetic rules for real numbers and applied them to complex numbers without checking if they were still valid. See https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/
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?
You get the correct results, if you use ##e^{2\pi i m x}=\cos(2\pi m x)+i \sin(2\pi m x)##.

You used arithmetic rules for real numbers and applied them to complex numbers without checking if they were still valid. See https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

You get the correct results, if you use ##e^{2\pi i m x}=\cos(2\pi m x)+i \sin(2\pi m x)##.

So from the article, i can't do ##\int_0^1 e^{2\pi i (n-k)x} = \int_0^1 (e^{2\pi i})^{(n-k)x} ## because we'd need ##2\pi i## and ##(n-k)x## to be positive and real, and ##2\pi i## is not real.

So for case 2: ##n \neq k##. Let ##l = n-k##. Observe, ##I = \int_0^1 e_n(x)e_k(x) dx = \int_0^1 e_{n-k}(x)dx = \int_0^1 e^{2\pi ilx}dx = \int_0^1 \cos(2\pi ilx) + i\sin(2\pi ilx) dx##. Let ##u = 2\pi ilx##. Then ##du = 2\pi il dx## i.e., ##dx = \frac{du}{2\pi il}##. So, ##\int_{0}^{2\pi il} \frac{\cos(u) + isin(u)}{2\pi il} du = \int_0^{2\pi il}\frac{\cos(u)}{2\pi il} du + \int_0^{2\pi il}\frac{\sin(u)}{2\pi il} du =\frac{\sin(2\pi il) - \sin(0)}{2\pi il} + i\frac{-\cos(2\pi il) - -\cos(0)}{2\pi il} = \frac{0 - 0}{2\pi il} + i \frac{-1 - -1}{2\pi il} = 0## which is the result we wanted.

fresh_42
Ray Vickson
Homework Helper
Dearly Missed

## 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?

I know you have already done the question, but for future reference you should realize that for ##n \neq k## you could use the result
$$\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C,$$ with ##a = 2\pi i (n-k)##,

I know you have already done the question, but for future reference you should realize that for ##n \neq k## you could use the result
$$\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C,$$ with ##a = 2\pi i (n-k)##,
Thank you for letting me know, ill keep this in mind for the future.