# 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?

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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.