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So the Gaussian integral basically states that:

$$ I = \int_{-\infty}^{\infty} e^{-x^2} \ dx =\sqrt{\pi}$$

So the way to solve this is by converting to polar co-ordinates and doing a double integration.

Since I haven’t learnt double integration, I have searched a lot to solve this kind of integral using single integration. But to no avail. I have even tried it myself a couple of times but have been unsuccessful.

So here’s my question, is it possible to integrate the above using only single integration and if so how? If this is not possible to integrate using single integration then what is the reason behind it.

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    $\begingroup$ There is a way to evaluate this integral using contour integration in the complex plane, but I am guessing that if you are unfamiliar with double integrals then you are at least as unfamiliar with complex integration. $\endgroup$ – Ron Gordon Mar 22 '18 at 18:44
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    $\begingroup$ @RonGordon yes, you’re right. Haven’t heard about it though! $\endgroup$ – physics2000 Mar 22 '18 at 18:46
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    $\begingroup$ Check this out if you want. Again, it is a pretty sophisticated use of contour integration, but this is a good way to go. math.stackexchange.com/questions/1266856/… $\endgroup$ – Ron Gordon Mar 22 '18 at 18:48
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A famous idea is to approximate the integral $\int_{0}^{+\infty}e^{-x^2}\,dx$ with $\int_{0}^{+\infty}\frac{dt}{(1+t^2/n)^n}$ and $\int_{0}^{\sqrt{n}}(1-s^2/n)^n\,ds$, with $n\in\mathbb{N}$ tending to $+\infty$. These integrals are elementary: they can be computed through the substitutions $t=\sqrt{n}\tan\theta$, $s=\sqrt{n}\sin\varphi$ and repeated integration by parts.

The outcome is the double bound $$ L(n)=\frac{\sqrt{n}4^n}{\binom{2n}{n}(2n+1)}\leq \int_{0}^{+\infty}e^{-x^2}\,dx \leq \frac{\pi n \sqrt{n}\binom{2n}{n}}{(2n-1) 4^n}=R(n)\tag{1}$$ holding for any $n\geq 1$. The statement $$ \lim_{n\to +\infty}\frac{R(n)}{L(n)}=1\tag{2} $$ is equivalent to Wallis' product and the statement $$ \lim_{n\to +\infty} R(n)L(n)=\frac{\pi}{4}\tag{3} $$ is trivial. By squeezing it follows that $\int_{0}^{+\infty}e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}$ and $\int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}$.


Truth to be told, this is a bit of a fraud, too. We are not really avoiding the $\Gamma$ function, we are just relying on the reflection/duplication formulas for $\Gamma$ without making an explicit mention of $\Gamma$. Wallis' product itself is an instance of $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$.

On the other hand, why to avoid the $\Gamma$ function? To say the least, it plays a major role in many relevant probability distributions, and as one of my menthors (C.Viola) likes to say, "a good mathematician or phycisist should not be afraid to manipulate the $\Gamma$ function or the sine function, also because they are not that different. The sooner one gets introduced to $\Gamma$, the better".


A (very!) shortened approach is to notice that $\Gamma\left(\frac{1}{2}\right)$ is the value we are looking for and $\Gamma\left(\frac{1}{2}\right)^2$ is the area of the unit circle.

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  • $\begingroup$ This is fantastic, but perhaps you could demonstrate to the OP where the inequality comes from. $\endgroup$ – Ron Gordon Mar 22 '18 at 18:53
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    $\begingroup$ Easily done: $\exp$ is convex and $\exp(x)=\frac{1}{\exp(-x)}$, hence $e^{-x^2}$ is bounded between $\max(0,(1-x^2))$ and $\frac{1}{1+x^2}$. It is enough to write $e^{-x^2}$ as $\left(e^{-x^2/n}\right)^n$ to let the fun begin. $\endgroup$ – Jack D'Aurizio Mar 22 '18 at 18:57
  • $\begingroup$ Nice proof ! thanks. I have read some articles of C. Viola about irrationality of some constants ($\zeta(3)$,...) $\endgroup$ – FDP Mar 22 '18 at 20:35
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Here's a proof using the gamma function.

The function $$e^{-x^2}=e^{-(-x)^2}$$ is even, so $$\int_{-\infty}^\infty e^{-x^2}\,dx=2\int_0^\infty e^{-x^2}\,dx$$ Let $t=x^2\implies\dfrac{dx}{dt}=\dfrac12t^{-1/2}$. Hence $$\int_{-\infty}^\infty e^{-x^2}\,dx=2\int_0^\infty e^{-t}\cdot\frac12t^{-1/2}\,dt=\int_0^\infty e^{-t}t^{-1/2}\,dt=\Gamma\left(\frac12\right)=\sqrt\pi$$

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  • $\begingroup$ Hey thanks, but I do know how to solve it by the gamma function. I’m not looking for that, I want to just directly integrate it. Hope you understand. Thanks $\endgroup$ – physics2000 Mar 22 '18 at 18:21
  • $\begingroup$ I only know of using the double integral if you want to integrate the expression directly, unfortunately... $\endgroup$ – TheSimpliFire Mar 22 '18 at 18:22
  • $\begingroup$ yeah, you see that’s the problem I’m facing $\endgroup$ – physics2000 Mar 22 '18 at 18:22
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    $\begingroup$ This "method" is simple circular logic, i.e., how do we evaluate $\Gamma \left ( \frac12 \right )$? By using the Gaussian integral we are trying to evaluate! Accordingly, this is not really an answer to the OP's question. $\endgroup$ – Ron Gordon Mar 22 '18 at 18:50
  • $\begingroup$ @RonGordon I was taught this proof and $\Gamma\left(\frac12\right)=\sqrt\pi$ was just an identity. $\endgroup$ – TheSimpliFire Mar 22 '18 at 18:54

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