# Solving a Gaussian integral using a power series?

• patric44
I see, thanks. If one replaces the upper integration limit with ##x## finite there is no problem. Each term is 0 for the lower limit so that takes care of itself. It seems one still requires the usual 2D integration trick to evaluate the integral.Eachterm is 0 for the lower limit so that takes care of itself.However, the series does converge to the Gaussian value ##\frac{\sqrt{\pi}}{2}##.f

#### patric44

Homework Statement
try to solve the Gaussian using power series ?
Relevant Equations
∫e^-(x^2)dx
hi guys

i am trying to solve the Gaussian integral using the power series , and i am suck at some point : the idea was to use the following series :
$$\lim_{x→∞}\sum_{n=0}^∞ \frac{(-1)^{n}}{2n+1}\;x^{2n+1} = \frac{\pi}{2}$$
to evaluate the Gaussian integral as its series some how slimier :
$$\int_{0}^{∞} e^{-x^{2}}dx = \lim_{x→∞}\sum_{n=0}^∞ \frac{(-1)^{n}}{(2n+1)n!}\;x^{2n+1}$$
so i want to evalue the series of the gaussian in terms of the other series or any other way to get ##\frac{\pi^{0.5}}{2}## , is it possible ?

Delta2
There is the standard trick using polar coordinates in 2D and no power series. Is this of interest?

Okay, I'm guessing no.

patric44
There is the standard trick using polar coordinates in 2D and no power series. Is this of interest?
i know the polar trick but solving it using power series seems like a good challenge , i need a formal way to evaluate the Gaussian series .

It doesn't seem obviously easy to me. There might be a clever trick though. Do you have a particular reason to think you can do this?

Expand ##e^{-x^2}##,
$$e^{-x^2}=(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+...)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{n!}$$
Now integrate term by term and take the limit.

Expand ##e^{-x^2}##,
$$e^{-x^2}=(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+...)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{n!}$$
Now integrate term by term and take the limit.
None of the term integrals will converge.

None of the term integrals will converge.
I think the idea of the exercise was to prove,
$$\lim_{x \rightarrow +\infty} \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)n!}=\frac{\pi}{2}$$

Paul Colby
I think the idea of the exercise was to prove,
I see, thanks. If one replaces the upper integration limit with ##x## finite there is no problem. Each term is 0 for the lower limit so that takes care of itself. It seems one still requires the usual 2D integration trick to evaluate the integral.

Expand ##e^{-x^2}##,
$$e^{-x^2}=(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+...)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{n!}$$
Now integrate term by term and take the limit.
I think the idea of the exercise was to prove,
$$\lim_{x \rightarrow +\infty} \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)n!}=\frac{\pi}{2}$$
@patric44 wants to show that ##\int_0^\infty e^{-x^2}\,dx = \sqrt{\pi}/2## using series methods, as an alternative to the usual technique of calculating the integral. He already has the series. He wants to figure out how to show it converges to the right answer.

patric44
I think the idea of the exercise was to prove,
$$\lim_{x \rightarrow +\infty} \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)n!}=\frac{\pi}{2}$$

but shouldn't that series converge to the Gaussian value ##\frac{\sqrt{\pi}}{2}## . i want to know how to show the convergence to ##\frac{\sqrt{\pi}}{2}## .