Solving a Gaussian integral using a power series?

In summary: 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}##.
  • #1
patric44
300
39
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 ?
 
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  • #2
There is the standard trick using polar coordinates in 2D and no power series. Is this of interest?

Okay, I'm guessing no.
 
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  • #3
Paul Colby said:
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 :smile: , i need a formal way to evaluate the Gaussian series .
 
  • #4
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?
 
  • #5
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.
 
  • #6
Fred Wright said:
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.
 
  • #7
Paul Colby said:
None of the term integrals will converge.
I think the idea of the exercise was to prove,
$$
\lim_{x \rightarrow +\infty} \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)n!}=\frac{\pi}{2}
$$
 
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  • #8
Fred Wright said:
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.
 
  • #9
Fred Wright said:
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.
Fred Wright said:
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.
 
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  • #10
Fred Wright said:
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}## .
 

Related to Solving a Gaussian integral using a power series?

1. What is a Gaussian integral?

A Gaussian integral, also known as a normal integral, is a type of integral that involves the Gaussian function, which is a bell-shaped curve. It is commonly used in statistics and probability to calculate the area under the curve.

2. Why would I need to solve a Gaussian integral using a power series?

Sometimes, a Gaussian integral cannot be solved using traditional integration techniques. In these cases, a power series can be used as an alternative method to approximate the solution.

3. How does a power series help in solving a Gaussian integral?

A power series is an infinite sum of terms that can be used to represent a function. By using a power series, we can approximate the value of a Gaussian integral by adding up a finite number of terms in the series.

4. What are the steps for solving a Gaussian integral using a power series?

The steps for solving a Gaussian integral using a power series are as follows: 1) Rewrite the Gaussian function as a power series using the Maclaurin series expansion, 2) Substitute the power series into the integral, 3) Integrate each term in the power series, 4) Evaluate the integral by plugging in the limits of integration, and 5) Add up the terms to get an approximate solution.

5. Are there any limitations to using a power series to solve a Gaussian integral?

Yes, there are limitations to using a power series to solve a Gaussian integral. The power series may not always converge, meaning that the approximation may not be accurate. Additionally, the more terms that are added in the series, the more computationally intensive the calculation becomes.

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