📂Lemmas

Generalization of Gaussian Integrals

Formulas¹

For an integer $n \ge 0$, the following expressions are true.

• When multiplied by an even degree polynomial $$ \int_{-\infty}^{\infty} x^{2n} e^{-\alpha x^{2}}dx = \dfrac{(2n)!}{n! 2^{2n}}\sqrt{\dfrac{\pi}{\alpha^{2n+1}}} $$

$$ \int_{0}^{\infty} x^{2n} e^{-\alpha x^{2}}dx = \dfrac{(2n)!}{n! 2^{2n+1}}\sqrt{\dfrac{\pi}{\alpha^{2n+1}}} $$

• When multiplied by an odd degree polynomial

$$ \int_{-\infty}^{\infty} x^{2n+1} e^{-\alpha x^{2}}dx = 0 $$

$$ \int_{0}^{\infty} x^{2n+1} e^{-\alpha x^{2}}dx = \dfrac{n!}{2 \alpha^{n+1}} $$

Explanation

Gaussian Integral

$$ \int_{-\infty}^{\infty} e^{-\alpha x^2} dx= \sqrt{\dfrac{\pi}{\alpha}} $$

This can be seen as a generalization of the Gaussian integral.

If the polynomial being multiplied has an odd degree, it is an odd function, so the integral over the entire range of real numbers is always $0$.

Proof

Even

Differentiate both sides of the Gaussian integral with respect to $\alpha$. Then, by the Leibniz rule, the following is true.

$$ \dfrac{d}{d\alpha}\left( \int_{-\infty}^{\infty} e^{-\alpha x^2} dx \right) = \int_{-\infty}^{\infty} \dfrac{\partial }{\partial \alpha}e^{-\alpha x^2} dx = \dfrac{d}{d\alpha}\sqrt{\dfrac{\pi}{\alpha}} $$

$$ \implies \int_{-\infty}^{\infty} \cancel{-}x^{2}e^{-\alpha x^2} dx = \cancel{-}\dfrac{1}{2}\sqrt{\dfrac{\pi}{\alpha^{3}}} $$

Differentiating again with respect to $\alpha$ gives the following.

$$ \int_{-\infty}^{\infty} \cancel{-}x^{2}x^{2}e^{-\alpha x^2} dx = \cancel{-}\dfrac{1}{2}\dfrac{3}{2}\sqrt{\dfrac{\pi}{\alpha^{5}}} $$

$$ \implies \int_{-\infty}^{\infty} x^{2\cdot2}e^{-\alpha x^2} dx = \dfrac{1 \cdot 3}{2^{2}}\sqrt{\dfrac{\pi}{\alpha^{5}}} $$

Differentiating once more with respect to $\alpha$ gives the following.

$$ \int_{-\infty}^{\infty} \cancel{-}x^{2}x^{2\cdot2}e^{-\alpha x^2} dx = \cancel{-}\dfrac{1 \cdot 3}{2^{2}}\dfrac{5}{2}\sqrt{\dfrac{\pi}{\alpha^{7}}} $$

$$ \implies \int_{-\infty}^{\infty} x^{2\cdot3}e^{-\alpha x^2} dx = \dfrac{1 \cdot 3 \cdot 5}{2^{3}}\sqrt{\dfrac{\pi}{\alpha^{7}}} $$

Product of Consecutive Odd Numbers

For an integer $n \ge 0$, the following is true.

$$ (2n-1) \cdot (2n-3) \cdots 5 \cdot 3 \cdot 1 = \dfrac{(2n)!}{2^{n} (n!)} = (2n-1)!! $$

Hence, by the above formulas, it is generalized as follows.

$$ \int_{-\infty}^{\infty} x^{2n}e^{-\alpha x^2} dx = \dfrac{(2n!)}{n!2^{2n}}\sqrt{\dfrac{\pi}{\alpha^{2n+1}}} $$

$x^{2n}e^{-\alpha x^{2}}$ is an even function, so if the integration range is halved, the value is also halved.

$$ \int_{0}^{\infty} x^{2n}e^{-\alpha x^2} dx = \dfrac{(2n!)}{n!2^{2n+1}}\sqrt{\dfrac{\pi}{\alpha^{2n+1}}} $$

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Odd

First, by substituting as in $\alpha x^{2} \equiv y$, since $2\alpha x dx = dy$, the expression when $n=0$ is as follows.

$$ \int_{0}^{\infty} xe^{-\alpha x^{2}}dx = \dfrac{1}{2\alpha} \int_{0}^{\infty} e^{-y}dy $$

The right side of the above equation can be seen to be $\Gamma (1) = \int_{0}^{\infty} y^{0} e^{-y}dy = 1$ using the Gamma Function. Hence, the following expression is obtained.

$$ \int_{0}^{\infty} x^{1}e^{-\alpha x^{2}}dx = \dfrac{1}{2\alpha} = \dfrac{0!}{2\alpha^{1}} $$

Similar to the even case, by differentiating both sides with respect to $\alpha$, the following is obtained.

$$ \int_{0}^{\infty} \cancel{-}x^{2}xe^{-\alpha x^{2}}dx = \cancel{-}\dfrac{1}{2\alpha^{2}} $$

$$ \implies \int_{0}^{\infty} x^{3}e^{-\alpha x^{2}}dx = \dfrac{1!}{2\alpha^{2}} $$

Differentiating once more with respect to $\alpha$ gives the following.

$$ \int_{0}^{\infty} \cancel{-}x^{2}x^{3}xe^{-\alpha x^{2}}dx = \cancel{-}\dfrac{2 \cdot 1}{2\alpha^{3}} $$

$$ \implies \int_{0}^{\infty} x^{5}xe^{-\alpha x^{2}}dx = \dfrac{2!}{2\alpha^{3}} $$

Differentiating once more with respect to $\alpha$ gives the following.

$$ \int_{0}^{\infty} \cancel{-}x^{2}x^{5}xe^{-\alpha x^{2}}dx = \cancel{-}\dfrac{3 \cdot 2!}{2\alpha^{4}} $$

$$ \implies \int_{0}^{\infty} x^{7}xe^{-\alpha x^{2}}dx = \dfrac{3!}{2\alpha^{4}} $$

Hence, when it is generalized, it is expressed as follows.

$$ \int_{0}^{\infty} x^{2n+1}xe^{-\alpha x^{2}}dx = \dfrac{n!}{2\alpha^{n+1}} $$

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