Integrand with no Indefinite Integral (1)

This note is one of two notes where we will evaluate a definite integral of a function whose indefinite integral does not exist.

We will prove:

\(\displaystyle \int_{-\infty}^{\infty} e^{-x^2} \,dx = \sqrt{\pi}\)

Luckily, this particular integral is easy to evaluate compared to the next one.

Proof:

Firstly, we must take note of the fact that \(f(x) = e^{-x^2}\) is symmetric about \(x=0\), due to the square in the exponent. This means we may write:

\(\displaystyle \int_{-\infty}^{\infty} e^{-x^2} \,dx = 2\displaystyle \int_{0}^{\infty}e^{-x^2} \,dx\)

Now we make the following substitution:

\(x=\sqrt{\phi}\)

\(\Rightarrow\) \(dx=\frac{d\phi}{2\sqrt{\phi}}\)

Now we have:

\(2\displaystyle \int_{0}^{\infty}e^{-x^2} \,dx = 2\displaystyle \int_{0}^{\infty}\frac{1}{2} \frac{e^{-\phi}}{\sqrt{\phi}}\,d\phi = \displaystyle \int_{0}^{\infty} \frac{e^{-\phi}}{\sqrt{\phi}}\,d\phi\)

\( = \displaystyle \int_{0}^{\infty} \phi^{-\frac{1}{2}}{e^{-\phi}}\,d\phi\)

Now, note the definition of the Gamma Function:

\(\Gamma(t) = \displaystyle \int_{0}^{\infty}x^{t-1} e^{-x}\,dx\)

Set:

\(t-1= -\frac{1}{2}\) \(\Rightarrow\) \(t=\frac{1}{2}\)

Then:

\(\displaystyle \int_{0}^{\infty} \phi^{-\frac{1}{2}}{e^{-\phi}}\,d\phi = \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\)

\(\Rightarrow\) \(\displaystyle \int_{-\infty}^{\infty} e^{-x^2} \,dx = \sqrt{\pi}\)

QED

Note: A probability distribution that arises very frequently is the Standard Normal Distribution:

\(F(x) = \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}\)

This is a continuous probability distribution, which requires that:

\(\displaystyle \int_{-\infty}^{\infty} \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}} = 1\)

Hence we have:

\(\displaystyle \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}= \sqrt{2\pi}\)

I personally find it very interesting that there exist functions that are not the derivative of any combination of elementary functions, yet we may still evaluate their integral in a specific region. This fact, to me, suggests that there may exist equivalent non-elementary functions whose derivatives are the functions we wish to indefinitely integrate. This is all speculation, however. I have done no actual work in finding such functions and they may not exist.

Note by Ethan Robinett
3 years, 8 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

There was a really interesting proof of the first indefinite integral in one of MIT's open courseware videos (18.01)

Bogdan Simeonov - 3 years, 8 months ago

Log in to reply

Are you sure it was the indefinite integral? I just looked it up and I saw the definite one, I.e. the one we did above except they changed to polar coordinates.

Just wondering because I was pretty sure that \(e^{-x^2}\) has no indefinite integral

Ethan Robinett - 3 years, 8 months ago

Log in to reply

Oh, sorry!I meant definite integral, the square root of pi one :D.Im sure I saw it in one of the lectures.It was a really cool proof.

Bogdan Simeonov - 3 years, 8 months ago

Log in to reply

@Bogdan Simeonov Oh haha yeah I watched it earlier, it was really good

Ethan Robinett - 3 years, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...