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This question already has an answer here:

I know from probability class that the area under the bell curve $e^{-x^2}$ is $\sqrt{\pi}$. I would like to be able to verify this, so in other words, solve this integral: $$\int_{-\infty}^{\infty} e^{-x^2} dx $$ The proof we saw in class used polar coordinates, but I don't like polar coordinates! Is there another way to evalutuate it?

Thanks!

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marked as duplicate by tired, Em., Community Mar 14 '16 at 19:19

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    $\begingroup$ Seriously, seriously, get used to polar coords. You will suffer if you don't. I speak from experience. $\endgroup$ – bob.sacamento Mar 14 '16 at 19:08
  • $\begingroup$ @bob.sacamento I am a math student, I know how to work with polar coordinates but I like the proof in my answer much more. I put it up here just to share because most people that have to learn this don't know about this proof and they just might end up here. $\endgroup$ – Jens Renders Mar 14 '16 at 19:40
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What you could use is the following formula: $$V = \pi \int_a^b f(x)^2 dx $$ Which tells you the volume of the solid of revolution created by rotating $f$ around the $x$-axis. It's easy to see why this is true.

We start of the usual way, by setting your integral equal to $I$, so

$$I^2 = I\cdot I = \int_{-\infty}^{\infty} e^{-x^2} dx \cdot \int_{-\infty}^{\infty} e^{-x^2} dx$$

and rename the variable in the second integral: $$I^2=\int_{-\infty}^{\infty} e^{-x^2} dx\cdot \int_{-\infty}^{\infty} e^{-y^2} dy$$

The first integral is a constant, so we can move it inside the second integral:

$$I^2=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} dx\cdot e^{-y^2} dy$$

And rewrite as

$$I^2=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)}\ dx\ dy$$

This means geometrically that we are looking for the area underneath the graph of $e^{-(x^2+y^2)}$. Notice that this is a function of $x^2+y^2$, which means that it's constant on circles with center $(0,0)$. That means it's a solide of revolution around the $z$-axis!

3D plot

The curve you need to rotate to get this shape can be found by choosing $y=0$, so that we get the image of the function above the $x$-axis. This results in $z=e^{-x^2}$, our original function. Just as a visual refference, it looks like this:

curve

The formula to get the volume only works for solids of revolution around the $x$-axis, so we flip this curve on its side by inverting it (just switch the $z$ and $x$ and solve):

$$x=e^{-z^2} \iff z=\sqrt{-ln(x)}$$

plot of inverse function

Notice that we lost half of the curve because of taking only the positive square root, but when rotating this all the way around the $x$-axis, the same shape is created.

All that is left to do now is use the formula to get the volume of the solid that is created by rotating this function around the $x$-axis.

$$I^2 = \pi \int_0^1 -\ln(x) dx$$

The square cancelled with the square root, and we integrate from 0 to 1 to get the whole solid. The integral is has now become easy and evaluates to 1, so only the $\pi$ is left: $$I^2=\pi \iff I=\sqrt{\pi}$$

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    $\begingroup$ You answered your own question? $\endgroup$ – Ravi Mar 14 '16 at 19:12
  • $\begingroup$ @Jake Yes, see blog.stackoverflow.com/2011/07/… $\endgroup$ – Jens Renders Mar 14 '16 at 19:14
  • $\begingroup$ Oh I had no idea! Learning new things everyday :D $\endgroup$ – Ravi Mar 15 '16 at 3:00
  • $\begingroup$ How did you get $I^2 = \pi \int_0^1 -\ln(x) dx$? I don't see why you can cancel the square root inside the integral, just because there is a square outside the integral. $\endgroup$ – user56834 Jun 4 '17 at 11:52
  • $\begingroup$ @Programmer2134 no, it cancels with the square in te formula of the volume of a solid of revolution (al the way at the top) $\endgroup$ – Jens Renders Jun 4 '17 at 22:12
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Using change of variables, the Gaussian integral is equivalent to $$ \int_{-\infty}^\infty e^{-x^2} dx = \Gamma\left( \frac{1}{2} \right) $$ where $\Gamma$ is the Euler $\Gamma$ function. Since ($t = s^2, dt = 2s\,ds$) $$\begin{align*} \Gamma\left(\frac 1 2\right) &= \int_0^\infty t^{-1/2} e^{-t} dt \\ &= \int_0^\infty \frac{e^{-s^2}}{s} 2s\,ds \\ &= 2\int_0^\infty e^{-s^2} ds = \int_{-\infty}^\infty e^{-x^2} dx \end{align*}$$ $\Gamma$ has a reflection formula $$ \Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin \pi z} $$ Substituting $z = \frac{1}{2}$ yields $\sqrt\pi$. The reflection formula can be proven using contour integration.

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