Gaussian integral via power series?

Question

It is well known that the Gaussian integral is $\sqrt \pi$, i.e., $$ \int_0^\infty e^{-x^2} dx = \frac{\sqrt \pi}2.$$ This is something we typically learn in calculus with polar coordinates. We also learn how to integrate $e^{-x^2}$ via power series, e.g., $$ \int_0^t e^{-x^2} dx = \sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}.$$ Hence one has $$ \frac{\sqrt \pi}2 = \lim_{t \to \infty} \left(\sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}\right).$$

Question: Is there a nicer way to rewrite the expression on the right, e.g. with "only one infinity" involved, to get an interesting expression/algorithm for computing $\pi$?

Of course the goal would be "pass the limit through the sum" but this is evidently non-trivial as the tail of the series explodes with $t$. My limited experience with analysis suggests one could try to split up the sum in various ranges ($n \ll t$, $n \approx t$, $n \gg t$) but that middle ranges could get hairy.

Answer 1

In addition to changing the "upper bound" $t$ of integral, you can make the integrand depends on $t$. This allows you to get rid of the tail of the integrand for finite $t$.

For any $n \in \mathbb{Z}_{+}$, let $f_n : [0,\infty) \to \mathbb{R}$ be the function

$$f_n(x) = \begin{cases} (1 - \frac{x^2}{n})^n, & x \le \sqrt{n}\\0, & x \ge \sqrt{n}\end{cases}$$

For any fixed $x \in [0,\infty)$, it is easy to see $f_n(x) \le f_{n+1}(x)$ whenever $x \ge \sqrt{n}$. When $x < \sqrt{n}$, we can apply AM $\ge$ GM to $n$ copies of $1 - \frac{x^2}{n}$ and one copy of $1$ and get

$$1 - \frac{x^2}{n+1} = \frac{n}{n+1}\left(1 - \frac{x^2}{n}\right) + \frac{1}{n+1} \ge \left( 1 - \frac{x^2}{n}\right)^{n/n+1} \implies f_{n+1}(x) \ge f_n(x)$$

This means $f_1(x), f_2(x), \ldots$ is a sequence of pointwise non-decreasing, non-negative functions. Since its pointwise limit equals to $e^{-x^2}$, we can use Monotone convergence theorm to convert the integral on $e^{-x^2}$ to a limit of single variable:

$$\begin{align}\frac{\sqrt{\pi}}{2} = \int_0^\infty e^{-x^2} dx &= \int_0^\infty \lim_{n\to\infty} f_n(x) dx \stackrel{\rm MCT}{=} \lim_{n\to\infty} \int_0^\infty f_n(x) dx\\ &= \lim_{n\to\infty} \int_0^{\sqrt{n}}\left(1 - \frac{x^2}{n}\right)^n dx = \lim_{n\to\infty} \sqrt{n} \int_0^1 \left(1 - t^2\right)^n dt\\ &= \lim_{n\to\infty} \sqrt{n}\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k} \end{align}$$

Answer 2

The method of brackets can be used to calculate this integral, just like other functions that can be expanded into power series. The relevant papers that will help are here: https://arxiv.org/pdf/0812.3356.pdf https://arxiv.org/pdf/1004.2062.pdf http://www.mat.utfsm.cl/scientia/archivos/vol25/articulo8.pdf

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Here is a brief outline of the method

If a function $f(x)$ can be expressed in a power series as $$f(x)=\sum_{n=0}^{+\infty} a_n x^{\alpha n+\beta-1}$$

Then its integral over the half real line is converted into a bracket series:

$$\int_0^{+\infty} f(x)\,dx \mapsto \sum_{n=0}^{+\infty}a_n \langle \alpha n+\beta \rangle $$ Now, let me define the following symbol, called the indicator, as $$\phi_n=\frac{(-1)^n}{\Gamma(n+1)}$$

In this method, the bracket series is given the value (also the value of the integral) according to the following rule: $$\sum_{n=0}^{+\infty} \phi_n \, f(n) \langle \alpha n+\beta \rangle=\frac{1}{|\alpha|} f(-n^*) \Gamma(-n^*)$$ Where $n^*$ is the solution to the equation $$\alpha n+\beta=0$$

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As the function $f(x)=e^{-x^2}$ can be expanded into the following power series $$e^{-x^2}=\sum_{n=0}^{+\infty} \frac{(-1)^n}{n!} x^{2n}=\sum_{n=0}^{+\infty} \phi_n \, x^{2n}$$ The Gaussian integral is converted into the following bracket series: $$I=\int_0^{+\infty} e^{-x^2}\,dx\mapsto \sum_{n=0}^{+\infty} \phi_n \langle 2n+1\rangle $$ Which can be evaluated by the rule I described above, where $f(n)=1$, and $n^*=-\frac{1}{2}$. $$I=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}$$

Answer 3

An approach that emulates the usual integral way of finding the limit might be handy. Let

$$ A = \frac{\sqrt \pi}2 = \lim_{t \to \infty} \left(\sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}\right). $$ Now, we can write that $$ A^2 = \lim_{s\to\infty} \lim_{t \to \infty} \left(\sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}\right)\left(\sum_{m=0}^\infty (-1)^m \frac{s^{2m+1}}{(2m+1)m!}\right) $$ However, the double-limit should hold if $s$ and $t$ are taken to infinity together - that is, if $s=t\to\infty$. So we'll write it as $$\begin{align} A^2 &= \lim_{t \to \infty} \left(\sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}\right)\left(\sum_{m=0}^\infty (-1)^m \frac{t^{2m+1}}{(2m+1)m!}\right)\\ &=\lim_{t \to \infty} \left(\sum_{n=0}^\infty \left[\sum_{k=0}^n\frac{(-1)^k}{(2k+1)k!}\frac{(-1)^{n-k}}{(2(n-k)+1)(n-k)!} \right]t^{n+1}\right) \end{align}$$ where I've found the Cauchy product of the power series and noted that we can replace $t^2$ with $t$ without changing the result.

From here, we can use a Pade approximant (with equal numerator and denominator orders) to get a result that we can take the limit of without running into convergence problems. For example, the first six terms of the power series I have provided for $A^2$ are $$ A^2\approx \lim_{t\to\infty}\left(t-\frac23t^2+\frac{14}{45}t^3-\frac4{35}t^4+\frac{166}{4725}t^5-\frac{292}{31185}t^6+\dots\right) $$ And the pade approximant $[3/3]_A(t)$, expressed with integers, for this power series, is $$ A^2\approx \lim_{t\to\infty}\left(\frac{46193t^3+118860t^2+1742895t}{39266t^3+357819t^2+1280790t+1742895}\right) = \frac{46193}{39266} $$ This isn't the greatest approximation (it's on the wrong side of 1, after all)... but if you increase the number of terms included, the approximation improves (somewhat slowly, though).

Note that you don't have to perform the Cauchy product to get $A^2$ as a power series before finding the Pade approximant, and you really only need the leading coefficients, which can be done relatively efficiently.

If you look at $\lim_{t\to\infty} [x/x]_{A^2}(t)$, the first 20 results look like this:

Answer 4

The Expression Can Be Written As Zeta Function and Zeta Function At infinity is 1 hence using proper methods it may be evaluated