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.

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## Comments

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

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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

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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.

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