# 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

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

> 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} $$\boxed{123}$$

Sort by:

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

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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.

- 3 years, 8 months ago

Oh haha yeah I watched it earlier, it was really good

- 3 years, 8 months ago