I was presented to the following integral:

$(1)$ $\int_{-\infty}^{\infty}e^{-x^{2}}dx$

Let's call the value of this integral $A$ for "Answer".

I was told that it would help to think of a tridimensional analog to the curve in $(1)$ by thinking about this other integral:

$(2)$ $\iint_R e^{-(x^2+y^2)}dA$, where $R$ is the whole $xy$ plane

So I made the following substitution to polar coordinates:

$(3)$ $\int_0^{2\pi} \int_0^\infty e^{-r^2} rdrd\theta$

Which indeed got me the correct result for $(2)$, which is $\pi$.

Then this is how my thinking went:

Well, the area that $(1)$ is computing (i.e, the area under the curve $y = e^{-x^{2}}$) is just an infinitesimally thin slice of $(3)$ (more precisely, it's the area under the curve that lies on the intersection of the plane $y=0$ with the surface $z=e^{-(x^2+y^2)}$).

Because of rotational symmetry, it must be the case that if I were to generate a solid by rotating the area described by $(1)$ around the $y$-axis, I will get a solid whose volume is $\pi$, because that's what $(3)$ was computing.

That is, rotating the whole area of $(1)$ around the $y$-axis is the same as computing $\int_0^{2\pi}\int_0^{\infty}e^{-r^2} rdrd\theta$

But the inner integral $\int_0^{\infty}e^{-r^2} rdr$ is computing the area under the surface from the origin to infinity (if I understand polar coordinates well), which is half the area of $(1)$, since in $(1)$ the area goes from negative infinity to positive infinity.

So it must be the case that $\int_0^\infty e^{-r^2}rdr$ is half the value of $A$.

So it must be true that:

$(4)$ $\int_0^\infty e^{-r^2}rdr$ = $\frac{A}{2}$

Substituing this into $(3)$ we get:

$(5)$ $\int_0^{2\pi} \frac{A}{2} d\theta= \pi$

Solving for $A$, we get $A = 1$, which is wrong. The correct answer is $A = \sqrt{\pi}$

I was then presented to the correct way of solving the problem, but I still couldn't see where my flaw was. Could someone clarify?

Thanks in advance!

  • 1
    $\begingroup$ You have $\int \mathrm{e}^{-r^2}dr$ and $\int \mathrm{e}^{-r^2}rdr$ being equivalent. $\endgroup$
    – Chinny84
    Apr 25 '21 at 21:03
  • $\begingroup$ Fixed that typo. It was not part of the assumptions I made. $\endgroup$ Apr 25 '21 at 21:13
  • $\begingroup$ It doesn’t matter if $r$ is radius and $x$ is Cartesian - the fact you are still integrating along $r$ is the same way so $\int_0^\infty f(x) dx = \int_0^\infty f(r)dr$ if you have the same function. Further more $\int r\mathrm{e}^{-r^2}dr =-\frac{1}{2} \int \frac{d}{dr} \mathrm{e}^{-r^2}dr$ which you don’t need me to tell you what that is. $\endgroup$
    – Chinny84
    Apr 25 '21 at 21:35

(2) is just the square of (1).

$$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)/2} \, dx \, dy = \left(\int_{-\infty}^\infty e^{-x^2/2} \, dx\right) \left(\int_{-\infty}^\infty e^{-y^2/2} \, dy\right)$$

  • $\begingroup$ I know that (2) is just the square of (1). But why is my line of reasoning incorrect? What assumptions have I made that shouldn't have been made? $\endgroup$ Apr 25 '21 at 21:15
  • 1
    $\begingroup$ @DiegoOliveira As Chinny84 points out, the inner integral $\int_0^\infty e^{-r^2/2} r \, dr$ is not the area under the surface of $e^{-(x^2+y^2)}$ along some ray to infinity; it is the area under $e^{-(x^2+y^2)/2} \sqrt{x^2+y^2}$ along the ray, which is altogether different. The extra $r$ only enters as a scaling factor to compute the area of a little wedge at radius $r$ and angle $d\theta$ when making the rotation argument, and has no interpretation when thinking about the original integrand $e^{-(x^2+y^2)/2}$ along a ray. $\endgroup$
    – angryavian
    Apr 25 '21 at 22:05
  • $\begingroup$ I see! That is true, indeed! How can I close the question, then? The solutions you guys gave were in the comments, is there a way I can accept an answer in the comment section? $\endgroup$ Apr 28 '21 at 0:15

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