$$I =\int_0^\infty e^{-x^2}\mathrm{d}x$$

such that $$ \int_0^\infty \int _0^\infty e^{-(x^2+y^2)} \mathrm{d}x \ \mathrm{d}y = I^2 $$

Change the variable to polar coordinates in order to evaluate $$ \int_0^\infty \int _0^\infty e^{-(x^2+y^2)} \mathrm{d}x \mathrm{d}y = I^2 $$


Here is where my question lies in these steps $$\int _0^\frac{\pi}{2} \int_0^\infty\ e^{-r^2} r \mathrm{d}r\mathrm{d}\theta$$ Why $\frac{\pi}{2}$ not any other $\theta$ value in the integral? That's a solution I found in a textbook, regardless the following steps, (conducting a non proper integral and computing the definte integral's value), this is the point where I don't understand.

  • And why do we consider the $e^{-a^2} =0$ in:

$$\frac{-1}{2} \left( \lim\limits_{a \to \infty}\ e^{-a^2}-1 \right)\int_0^\infty \mathrm{d}\theta=\frac{\pi}{4}$$


1 Answer 1


You're only interested in the first quadrant of $\Bbb R^2$ in the double integral. That is, the double integral is integrating over the region $(0,\infty)\times(0,\infty)$, which is precisely the first quadrant. If you think about the unit circle values for $\theta$ and determine which angles lie in the first quadrant, you'll see that the first quadrant corresponds to $\theta\in(0,\pi/2)$.

For the values of $r$, well, you want to cover the entire first quadrant. This means that along any ray with an angle of $\theta$ (as measured from the positive $x$-axis), you'll need to take $r$ from $0$ to $\infty$. Since you do this for every value of $\theta$, you end up with a double integral: one that integrates with respect to $\theta$ and another that integrates with respect to $r$.

  • $\begingroup$ and why am I concerned with just the first integeral ? $\endgroup$
    – M12567
    Commented Oct 18, 2018 at 22:10
  • $\begingroup$ sorry I mean why am I concerned with just the first quadrat and not the rest ? and does this apply to for the triple integeral too? $\endgroup$
    – M12567
    Commented Oct 18, 2018 at 22:12
  • $\begingroup$ @Mayar: I've edited my answer. Does it make sense now why you're only concerned with the first quadrant? $\endgroup$
    – Clayton
    Commented Oct 18, 2018 at 22:14
  • $\begingroup$ The triple integral is going to have a similar trick, too. There you'll convert to spherical coordinates, if I recall correctly, though, so some of the constants will look a little different. $\endgroup$
    – Clayton
    Commented Oct 18, 2018 at 22:17
  • 2
    $\begingroup$ @Mayar: You'd have to explain further what you mean. Here, the region is $(0,\infty)\times(0,\infty)$ which corresponds to the first quadrant; in other words, it's the same region as $(0,\pi/2)\times(0,\infty)$ in polar coordinates. If our region were $(-\infty,\infty)\times(0,\infty)\subset\Bbb R^2$, then $\theta$ would range between $0$ and $\pi$ since this would encompass the upper half plane. $\endgroup$
    – Clayton
    Commented Oct 18, 2018 at 22:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .