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This question already has an answer here:

I'm trying to solve this

$$\int_{0}^{\infty} e^{-4x^2}dx$$

I saw how $\int_{\infty}^{\infty} e^{-x^2}dx$ it's done and how it covers the whole plane, then I assume that from $0 \ to \ \infty$ it's the right part of the plane, My problem comes when I try to cover the first and the fourth quadrant of the plane, I know that in polar coordinates $0\le r \le \infty$, but

What about $\theta$ for this case? How $\theta$ moves in order to cover the right half of the plane?

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marked as duplicate by zahbaz, Community Jun 1 '16 at 1:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Note that your integrand is an even function. That might help. $\endgroup$ – EuYu Jun 1 '16 at 0:15
  • $\begingroup$ $\left(\int_0^\infty e^{-4x^2}dx\right)^2 = \int_0^\infty e^{-4y^2}dy\int_0^\infty e^{-4x^2}dx = \int_0^\frac{\pi}{2}d\theta \int_0^\infty re^{-4r^2}dr$. It's not the right part of the plane--it's only the 1st quadrant. $\endgroup$ – Jared Jun 1 '16 at 0:18
  • $\begingroup$ If you want to carry out the idea without borrowing the answer, let $\theta$ go from $-\pi/2$ to $\pi/2$. $\endgroup$ – André Nicolas Jun 1 '16 at 0:19
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Using the fact that $$\int_{-\infty}^\infty e^{-x^2}\ dx=\sqrt \pi$$ and the even-ness of $x^2$, we have $$\int_0^\infty e^{-x^2}\ dx=\frac{\sqrt\pi}{2}.$$ Let $u=2x$. Then $du=2dx$. The limits won't change and we have $$\int_0^\infty e^{-4x^2}\ dx=\frac{1}{2}\int_0^\infty e^{-u^2}\ du=\frac{\sqrt\pi}{4}.$$

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I'm not totally sure what "proof" you're talking about--I'm assuming it's the trick that the square of the integral can be written as a double integral in the familiar 2D $xy$-plane then converted to an easy integral using polar coordinates:

$$ \left(\int\limits_0^\infty e^{-4x^2}dx\right)^2 = \int_0^\infty e^{-4x^2}dx\int\limits_0^\infty e^{-4y^2}dy = \int\limits_0^\infty\int\limits_0^\infty e^{-4\left(x^2+y^2\right)}dxdy $$

Before you convert to polar coordinates, you must understand that this region is only the first quadrant--which means the angle runs from $0$ to $\frac{\pi}{2}$ (as opposed to $[0,2\pi)$--as is the case when it's $-\infty$ to $\infty$).

Converting to polar coordinates gives:

$$ \int\limits_0^\infty\int\limits_0^\infty e^{-4\left(x^2+y^2\right)}dxdy = \int\limits_0^\frac{\pi}{2}d\theta\int\limits_0^\infty re^{-4r^2}dr $$

The $r$-integral is easy through $u$-substitution and is $\frac{1}{8}$. The $\theta$-integral is independent of the $r$-integral so is just $\frac{\pi}{2}$. Giving:

$$ \left(\int\limits_0^\infty e^{-4x^2}dx\right)^2 = \frac{\pi}{16} $$

And thus:

$$ \int\limits_0^\infty e^{-4x^2}dx = \frac{\sqrt{\pi}}{4} $$

(just as can be retrieved from a u-substitution used from the result that $\int\limits_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}$).

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For a non polar coordinates approach see my answer to a related question: Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \dfrac{\sqrt \pi}{2}$

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