# The integral of e^(-x^2) from –infinity to infinity

Up a level : Integrals
Previous page : The integral of ln(x+1)/(x^2+1) dx from 0 to 1
Next page : The integral of e^(-x^2) from –infinity to infinity a second way The integral $I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx}$

This is definitely one of the classical difficult integrals. This method is, I think, attributed to Gauss. The problem is that you cannot find a primitive function to the integrand, no matter what you do. So what to do? The trick is to realize that an infinite rectangle is the same as an infinite square. So we start by making the integral into a two integrals, one in the x– and the y-direction. ${I^2} = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \int\limits_{ - \infty }^\infty {{e^{ - {y^2}}}dy}$

Since the integral with respect to y can be seen as a constant seen from the respect to x, then we can move the whole integral inside the other. I.e. ${I^2} = \int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {{e^{ - {y^2}}}} {e^{ - {x^2}}}dydx}$

This can be rewritten as ${I^2} = \int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {{e^{ - ({y^2} + {x^2})}}} dydx}$

This is now an integral over a function $z = {e^{ - ({y^2} + {x^2})}}$

as seen in the figure. Seen from above we basically integrate the whole area by adding small areas ΔA=ΔxΔy. We will now change to polar coordinates. We have that ${r^2} = {x^2} + {y^2}$

For our ΔA we will use ΔA=rΔθΔr as seen in the figure. So to integrate over the whole surface instead of integration from minus to plus infinity over both x and y we will integrate for a radius from 0 to infinity and an angle from 0 to 2π.

Putting it all together we get ${I^2} = \int\limits_{r = 0}^\infty {\int\limits_{\theta = 0}^{2\pi } {{e^{ - {r^2}}}} rd\theta dr}$

Now, since the variables are independent we can basically do the opposite we did above when we put the integrals inside each other. We get ${I^2} = \int\limits_{\theta = 0}^{2\pi } {d\theta } \int\limits_{r = 0}^\infty {{e^{ - {r^2}}}rdr}$

For the first part we get $\int\limits_0^{2\pi } {d\theta } = \left[ \theta \right]_o^{2\pi } = 2\pi$

For the second part we have $\int\limits_0^\infty {{e^{ - {r^2}}}rdr}$

To find this integral we can do an ordinal variable substation: $\left| {\begin{array}{*{20}{c}} {u = {r^2},\quad du = 2rdr,\quad \frac{{du}}{2} = rdr} \\ {r = 0 \Rightarrow u = 0,\quad r \to \infty \Rightarrow u \to \infty } \end{array}} \right.$

This gives us $\int\limits_0^\infty {{e^{ - {r^2}}}rdr} = \int\limits_0^\infty {{e^{ - u}}\frac{{du}}{2} = \left[ { - \frac{{{e^{ - u}}}}{2}} \right]_0^\infty } = 0 + \frac{1}{2} = \frac{1}{2}$

Combining these we get ${I^2} = \int\limits_{\theta = 0}^{2\pi } {d\theta } \int\limits_{r = 0}^\infty {{e^{ - {r^2}}}rdr} = 2\pi \cdot \frac{1}{2} = \pi$

and thus finally $I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} = \sqrt \pi$

We yet get another case where π and e are directly connected. Up a level : Integrals
Previous page : The integral of ln(x+1)/(x^2+1) dx from 0 to 1
Next page : The integral of e^(-x^2) from –infinity to infinity a second way Last modified: Jun 17, 2018 @ 16:39