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I'm having trouble evaluating this integral

$$\int_0^\infty {e^{-ax^2}} \,dx $$

My guess is that it would evaluate into something like

$$\int_0^\infty \frac 12e^{-s}s^{\frac 12} \ldots \,dx = \frac {\Gamma\left(\frac 12\right)}{\frac{a^{\frac 12}}{2}}$$

When you do a substitution $ \sqrt{s}= \sqrt{a}x $ so that $ s = ax^2 $. I'm having trouble convincing myself though that $ \frac {d}{ds}\sqrt{s} = \left(\ldots a^\frac 12\right) $ which would satisfy the answer that I provided.

Am I doing something wrong or is my guess wrong?


marked as duplicate by Travis, Narasimham, Community Apr 22 '15 at 13:24

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  • $\begingroup$ See Gaussian integral. $\endgroup$ – Lucian Apr 22 '15 at 12:41
  • $\begingroup$ I guess my question really is a duplicate. What I was really trying to integrate though is $\int_0^\infty {e^(-ax^2+bx+c} \,dx} $ and that was the last part of the problem. Thanks everyone though for helping! $\endgroup$ – Aldon Apr 22 '15 at 13:14

If you want to use the Gamma function the substitution is $a x^2 = t$, so “$dx = \frac{1}{2\sqrt{a}}t^{-1/2}dt$". Then the integral appears as, $$\frac{1}{2\sqrt{a}} \int_0^\infty dt\,t^{-1/2}e^{-t} = \frac{1}{2\sqrt{a}}\Gamma(1/2)\ .$$

That's all.

  • $\begingroup$ So, the result you give in your question has a $2$ in the wrong position in the fraction. It should be in front of the $a^{1/2}$ at the denominator. $\endgroup$ – Oscar Apr 22 '15 at 12:25
  • $\begingroup$ Thanks, this was the step that I was looking for. I guess that I should have differentiated $x$ instead of $s$ so that the Gamma function would show up. $\endgroup$ – Aldon Apr 22 '15 at 13:04

The trick (at least one of them) is to write:

$$I = \int_0^\infty {e^{-ax^2}} dx$$ and since the variable doesn't matter, $$I = \int_0^\infty {e^{-ay^2}} dy$$ So that $$I^2=\int_0^\infty {e^{-ax^2}} dx\int_0^\infty {e^{-ay^2}} dy=\int_0^\infty\int_0^\infty {e^{-a(x^2+y^2)}}dx dy$$. We then switch to circular polar coordinates so that $dxdy=r drd\theta$ and $x^2+y^2=r^2$ and then $I^2$ becomes, with appropriate limits to cover only the first quadrant:

$$I^2 = \int_0^\infty\int_0^{\pi/2} e^{-ar^2} r dr d\theta = {\pi\over 2}\int_0^\infty e^{-ar^2} r dr$$ and then use the substitution $u=r^2$ and $du=2r dr$: $$I^2 = {\pi\over4} \int_0^\infty e^{-au}du = {\pi\over4} \left({1\over a}\right)$$

So finally, $$I = \int_0^\infty {e^{-ax^2}} dx = \sqrt{\pi\over 4a}$$

  • $\begingroup$ you can also get forms like $I = \int_0^\infty {x^{2n} e^{-ax^2}} dx$ by differentiation of both sides with respect to $a$ $\endgroup$ – danimal Apr 22 '15 at 12:33
  • $\begingroup$ I laughed after looking at your answer since I remembered that I could use the trick found in Gaussian integrals which is what you've shown. This is an extremely nice method of solving integrals involving exponentials, but what I'm looking for is a way to use the special functions to solve the integral. Thank you still though for reminding me about this technique, and for that I'm including your answer in the list of answers for my question. Addendum: I guess you can't put the check mark on more than one answer at the same time, so I'm putting the check mark on the Gamma function one sorry. $\endgroup$ – Aldon Apr 22 '15 at 13:08

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