## Friday, July 2, 2010

### Gaussian Integral

We will prove that $$\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$$. The common trick that is shown is the use of polar coordinates over the plane. My problem with this method is that it needs a lot of justification. We need to know the justification behind Jabcobi's theorem and how it applies to this integral, we need to know the justification behind Fubini's theorem and how it applies to this integral, these justfications take place over a finite integral that in the limit it taken over the whole line. So there is a lot of mathematical justifications that need to be done to prove such an integral. The polar trick is easy but it has a lot of hidden work behind it. Here is a method that does not have hidden work behind it, all the work will be presented, but the actual derivation takes more work. This is the method I prefer because it is complete unlike the classic polar coordinate trick.

Consider the two curves $$te^{\pi i/4}\pm \tfrac{1}{2}$$ for $$-R\leq t \leq R$$, one with a positive sign for 1/2 and one with a negative sign for 1/2, R will be taken to be a really large positive real number. These are clearly lines in the complex plane. The slope of these lines is $$\tan \tfrac{\pi}{4} = 1$$, so these lines are parallel with $$y=x$$. The first line has x-intercept at -1/2 and the second line has x-intercept at 1/2. So these are a pair of parallel lines. Now connect $$Re^{\pi i/4}-\tfrac{1}{2}$$ and $$Re^{\pi/4} + \tfrac{1}{2}$$ with a horizontal line. Likewise do the same with $$-Re^{\pi i/4} - \tfrac{1}{2}$$ and $$-Re^{\pi i/4} + \tfrac{1}{2}$$. We have constructed a parallelogram. The horizontal sides are always fixed with length 1 and the sloped parallel sides are very long when R is taken to be large. So our parallelogram is a very thin parallelogram. Now let $$\Gamma_R$$ represent this contour with parameter R.

Let $$f(z) = \frac{e^{\pi i z^2}}{\sin \pi z}$$. This is a meromorphic function on the complex plane with simple poles at the integers. The only pole which is inside $$\Gamma_R$$ is at $$z=0$$. This is a simple pole, so it is simple (pun) to compute the residue by $$\lim_{z\to 0}zf(z) = \tfrac{1}{\pi}$$. Therefore, by the residue theorem we have that:

$$\oint_{\Gamma_R} \frac{e^{\pi iz^2}}{\sin \pi z} dz = 2i$$

We will compute this integral by definition, that is, by actually integrating it out. First, we will focus on the sloped sides which are parametrized at $$te^{\pi i/4}\pm \tfrac{1}{2}$$ for $$-R\leq t \leq R$$. Before we write down the integral out let us make some observations:

i) $$\exp \left[\p i (te^{\pi i/4} \pm \tfrac{1}{2})^2 \right] = \exp \left[ -\pi t^2 \pm \pi i e^{\pi i/4} t + \tfrac{\pi i}{4} \right]$$. Which can be further written as $$e^{-\pi t^2} e^{\pm \pi i e^{\pi i /4}t} e^{\pi i/4}$$.

ii) $$\sin \left[ \pi (te^{\pi i/4} \pm \tfrac{1}{2}) \right ] = \pm\cos \pi e^{\pi i/4} t$$.

Therefore, it follows from (i) and (ii) that we have:

$$-f(te^{\pi i/4} - \tfrac{1}{2}) + f(te^{\pi i/4} + \tfrac{1}{2})$$ =
$$\frac{ e^{-\pi t^2} e^{\pi i/4} \left( e^{\pi i e^{\pi i/4} t} + e^{-\pi i e^{\pi i/4} t}\right) }{ \cos \pi e^{\pi i /4} t}$$

By Euler's identity $$2\cos \mu = e^{i\mu} + e^{-i\mu }$$ and so we get that:

$$-f(te^{\pi i/4} - \tfrac{1}{2}) + f(te^{\pi i/4} + \tfrac{1}{2}) = \frac{2e^{-\pi t^2} e^{\pi i/4} \cos \pi e^{\pi i /4} t}{\cos \pi e^{\pi i /4} t}$$

So all that mess reduces to just:

$$-f(te^{\pi i/4} - \tfrac{1}{2}) + f(te^{\pi i/4} + \tfrac{1}{2}) =2 e^{-\pi t^2} e^{\pi i/4}$$.

Notice that when we take the contour integral over $$\Gamma_R$$ we go counterclockwise. So the contribution of the line segment $$\left[-Re^{\pi i/4} + \tfrac{1}{2}, Re^{\pi i/4} + \tfrac{1}{2}]$$ is positive, while the contribution of the line segment $$\left[ -Re^{-\pi i/4} - \tfrac{1}{2},-Re^{\pi i/4} -\tfrac{1}{2}\right]$$ is negative. Therefore, the two integrals when combined together around this counter give:

$$e^{\pi i /4} \int_{-R}^R f(te^{\pi i/4} + \tfrac{1}{2}) - f(te^{\pi i/4} - \tfrac{1}{2}) dt$$

Which, by our above simplifications, reduces to:

$$2e^{\pi i/4} \int_{-R}^R e^{\pi i/4} e^{-\pi t^2} dt = 2i \int_{-R}^R e^{-\pi t^2} dt$$

If we can show that as $$R\to \infty$$ the horizontal integrals go to zero then we have established that:

$$2i \cdot \mbox{PV} \int_{-\infty}^{\infty} e^{-\pi x^2} dx = 2i$$

This would complete the proof because we can drop the Cauchy principal value condition here, it is equivalent to integral convergence.

Of course, the proof is not complete without showing why the horizontal integrals go to zero. It is boring, but I guess I have to do this. Let us just do it for the upper horizontal integral. The line is parametrized by $$Re^{\pi i /4} + t$$ where $$-\tfrac{1}{2} \leq t \leq \tfrac{1}{2}$$.

Notice that, $$|\exp \left[ \pi i (Re^{\pi i/4} + t )\right]| = e^{-\pi R^2 -\pi \sqrt{2} R t } \leq e^{-\pi R^2 + \pi R /\sqrt{2}}$$.

By the basic inequality, $$|\sin (a+bi)| \geq \sinh b$$ we have that:

$$|\sin \pi(Re^{\pi i/4} + t ) | \geq \sinh (\pi R/\sqrt{2})$$

Therefore,

$$\left| \int_{-1/2}^{1/2} f(Re^{\pi i/4} + t ) dt \right| \leq \frac{e^{-\pi R^2 + \pi R/\sqrt{2}}}{\sinh (\pi R/\sqrt{2})}\to 0$$

So this is the proof. It might be a lot messier than the polar trick but the polar trick has a lot of stuff to do before it is a complete proof. This one is not so bad. Just some easy but messy estimates which are bound (pun) to come up in analytic proofs.