# Integral from $0$ to Infinity of $e^{-3x^2}$? [duplicate]

How do you calculate the integral from $0$ to Infinity of $e^{-3x^2}$? I am supposed to use a double integral. Can someone please explain? Thanks in advance.

## marked as duplicate by rschwieb, Micah, muzzlator, vonbrand, Davide GiraudoApr 11 '13 at 17:24

• Change of variable then en.wikipedia.org/wiki/Gaussian_integral – Jean-Claude Arbaut Apr 11 '13 at 16:13
• @arbautjc: We never learned Gaussian integral. We are just told to a double integral. – Sue Apr 11 '13 at 16:14
• Nice then, there are proofs of it with double integrals. Did you have a look at the link? By the way, the result is hard enough that the teacher would not ask you to prove it from scratch, so either he assumes you know it, either he should probably give you hints on how to compute it. – Jean-Claude Arbaut Apr 11 '13 at 16:15
• See this post. – David Mitra Apr 11 '13 at 16:15
• Evaluate $\iint e^{-3(x^2+y^2)}\,dx\,dy$ over the plane, switching to polar coordinates. The result is the square of what you want, – André Nicolas Apr 11 '13 at 16:17

There is a neat trick. Set $$I = \int_{0}^\infty e^{-3x^2}dx.$$ Then $$I^2 = \left(\int_0^\infty e^{-3x^2} dx\right) \left(\int_0^\infty e^{-3y^2}dy\right) = \int_0^\infty \int_0^\infty e^{-3(x^2+y^2)} dxdy.$$ Now change to polar coordinates to get $$I^2 = \int_{0}^{\pi/2} \int_0^{\infty} re^{-3r^2} dr d\theta$$ which from here you can solve and then take the square root.

• Thank you, I got the answer. – Sue Apr 11 '13 at 16:28
• No problem! I'm glad it helped. – Suugaku Apr 11 '13 at 16:31
• I just have one question: Why is the upper endpoint for the interval of theta $\frac{\pi}{2}$? I was assuming it would be $2\pi$. – Sue Apr 11 '13 at 16:41
• Note that the original integrals before changing into polar coordinates are both from $0$ to $\infty$. This region is all points $(x,y)$ with $0 < x < \infty$ and $0< y < \infty$. This is just a fancy way of saying the first quadrant. So when we change to $\theta$, we only want to go from the positive $x$-axis to the positive $y$-axis which is a rotation of $\pi/2$. – Suugaku Apr 11 '13 at 16:43
• Oh, I understand. Thanks so much @Suugaku! – Sue Apr 11 '13 at 16:47

Write $y=\sqrt{3}x$, which gives you:

$I=\displaystyle\int^{+\infty}_0 e^{-3x^2}\ dx = \frac{1}{\sqrt{3}}\displaystyle\int^{+\infty}_0 e^{-y^2}\ dy$

This is half the Gaussian integral which is equal to $\frac{\sqrt{\pi}}{2}$. And here you go:

$I=\frac{\sqrt{\pi}}{2\sqrt{3}}$