# Integral of polynomial times exp(-x^2)

• I
• Mr Davis 97
In summary, the conversation discusses strategies for integrating the integral ##\int_{-\infty}^{\infty} x^2 e^{-x^2} ~dx##, with the knowledge that the value of the Gaussian integral is ##\sqrt{\pi}##. Suggestions include using integration by parts and recognizing the even symmetry of the integrand. The conversation also mentions the relevance of this integral in various fields such as chemistry and psychology.
Mr Davis 97
I have the integral ##\int_{-\infty}^{\infty} x^2 e^{-x^2} ~dx##. Is there any simple way to integrate this, given that that I already know that the value of the Gaussian integral is ##\sqrt{\pi}##?

Try integration by parts.

Stephen Tashi said:
Try integration by parts.
What I found was that ##\int_{-\infty}^{\infty}e^{-x^2} ~dx = 2\int_{-\infty}^{\infty}x^2 e^{-x^2} ~dx##, so by original integral is ##\sqrt{\pi} / 2##. To do this though I had to start from the original Gaussian integral and integrate by parts to get the integral that I have. Is that trick of integrating by parts the original Gaussian integral to get what I want what you had in mind?

Mr Davis 97 said:
Is that trick of integrating by parts the original Gaussian integral to get what I want what you had in mind?

No, I was thinking of starting with the original problem and using ##f(x) = x, g'(x) = x e^{-x^2} ##.

for ##t+t^\ast>0\\
\int_{-\infty}^\infty \mathrm{f}(x^2)e ^{-t x^2} dx=\mathrm{f}\left(-\dfrac{d}{dt}\right)\int_{-\infty}^\infty e ^{-t x^2} dx%=\mathrm{f}\left(-\dfrac{d}{dt}\right)\sqrt{\frac{\pi}{t}}##
the case f(x)=x^n imvolves (2n-1)!
that comes up all the time in
chemistry http://www.colby.edu/chemistry/PChem/notes/Integral.pdf
psychology http://www.umich.edu/~chem461/Gaussian Integrals.pdf
http://www.math.tamu.edu/~manshel/papers/Hermite-integrals.pdf
http://mathworld.wolfram.com/GaussianIntegral.html

Last edited:
Another approach would be to realize that the integrand is an even function and write:
$$I(x) =\int_{-\infty}^{\infty}x^2e^{-x^2}dx=2\int_{0}^{\infty}x^2e^{-x^2}dx$$
Let ##u=x^2##, ##du=2xdx## and ##x=\sqrt {u}##
$$I(u)=\int_{0}^{\infty}u^{\frac {1} {2}}e^{-u}du = \Gamma(\frac {3} {2}) = \frac {1} {2} \sqrt {\pi}$$

Peace,
Fred

## 1. What is the general formula for the integral of polynomial times exp(-x^2)?

The general formula for the integral of polynomial times exp(-x^2) is given by:
∫ (P(x) * e^(-x^2)) dx = (C + ∑ (a_n * x^n * e^(-x^2)))
where P(x) is a polynomial, C is a constant, and a_n is a coefficient of the nth term in the polynomial.

## 2. What is the significance of the term exp(-x^2) in the integral of polynomial times exp(-x^2)?

The term exp(-x^2) is significant as it is the Gaussian function, which is commonly used in probability and statistics. It also has important applications in physics, engineering, and other fields.

## 3. How do I solve the integral of polynomial times exp(-x^2)?

To solve the integral of polynomial times exp(-x^2), you can use integration by parts or substitution methods. You can also use the general formula for the integral and simplify the expression by expanding the polynomial and integrating each term separately.

## 4. Can the integral of polynomial times exp(-x^2) be evaluated using numerical methods?

Yes, the integral of polynomial times exp(-x^2) can be evaluated using numerical methods such as the trapezoidal rule or Simpson's rule. These methods are useful when the integral cannot be solved analytically or when the polynomial is of a high degree.

## 5. How is the integral of polynomial times exp(-x^2) related to the error function?

The integral of polynomial times exp(-x^2) is related to the error function, erf(x), by the following equation:
∫ (P(x) * e^(-x^2)) dx = √π/2 * ∫ (P(x) * erf(x)) dx
This relationship is useful in solving integrals involving the Gaussian function and in calculating probabilities and error margins in statistical analysis.

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