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Generalization of Gaussian Integrals 📂Lemmas

Generalization of Gaussian Integrals

Formulas1

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

  • When multiplied by an even degree polynomial ∫−∞∞x2ne−αx2dx=(2n)!n!22nπα2n+1 \int_{-\infty}^{\infty} x^{2n} e^{-\alpha x^{2}}dx = \dfrac{(2n)!}{n! 2^{2n}}\sqrt{\dfrac{\pi}{\alpha^{2n+1}}}

∫0∞x2ne−αx2dx=(2n)!n!22n+1πα2n+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

∫−∞∞x2n+1e−αx2dx=0 \int_{-\infty}^{\infty} x^{2n+1} e^{-\alpha x^{2}}dx = 0

∫0∞x2n+1e−αx2dx=n!2αn+1 \int_{0}^{\infty} x^{2n+1} e^{-\alpha x^{2}}dx = \dfrac{n!}{2 \alpha^{n+1}}

Explanation

Gaussian Integral

∫−∞∞e−αx2dx=πα \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 00.

Proof

Even

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

ddα(∫−∞∞e−αx2dx)=∫−∞∞∂∂αe−αx2dx=ddαπα \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}}

  ⟹  ∫−∞∞−x2e−αx2dx=−12πα3 \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 α\alpha gives the following.

∫−∞∞−x2x2e−αx2dx=−1232πα5 \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}}}

  ⟹  ∫−∞∞x2⋅2e−αx2dx=1⋅322πα5 \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 α\alpha gives the following.

∫−∞∞−x2x2⋅2e−αx2dx=−1⋅32252πα7 \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}}}

  ⟹  ∫−∞∞x2⋅3e−αx2dx=1⋅3⋅523πα7 \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≥0n \ge 0, the following is true.

(2n−1)⋅(2n−3)⋯5⋅3⋅1=(2n)!2n(n!)=(2n−1)!! (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.

∫−∞∞x2ne−αx2dx=(2n!)n!22nπα2n+1 \int_{-\infty}^{\infty} x^{2n}e^{-\alpha x^2} dx = \dfrac{(2n!)}{n!2^{2n}}\sqrt{\dfrac{\pi}{\alpha^{2n+1}}}

x2ne−αx2x^{2n}e^{-\alpha x^{2}} is an even function, so if the integration range is halved, the value is also halved.

∫0∞x2ne−αx2dx=(2n!)n!22n+1πα2n+1 \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 αx2≡y\alpha x^{2} \equiv y, since 2αxdx=dy2\alpha x dx = dy, the expression when n=0n=0 is as follows.

∫0∞xe−αx2dx=12α∫0∞e−ydy \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 Γ(1)=∫0∞y0e−ydy=1\Gamma (1) = \int_{0}^{\infty} y^{0} e^{-y}dy = 1 using the Gamma Function. Hence, the following expression is obtained.

∫0∞x1e−αx2dx=12α=0!2α1 \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 α\alpha, the following is obtained.

∫0∞−x2xe−αx2dx=−12α2 \int_{0}^{\infty} \cancel{-}x^{2}xe^{-\alpha x^{2}}dx = \cancel{-}\dfrac{1}{2\alpha^{2}}

  ⟹  ∫0∞x3e−αx2dx=1!2α2 \implies \int_{0}^{\infty} x^{3}e^{-\alpha x^{2}}dx = \dfrac{1!}{2\alpha^{2}}

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

∫0∞−x2x3xe−αx2dx=−2⋅12α3 \int_{0}^{\infty} \cancel{-}x^{2}x^{3}xe^{-\alpha x^{2}}dx = \cancel{-}\dfrac{2 \cdot 1}{2\alpha^{3}}

  ⟹  ∫0∞x5xe−αx2dx=2!2α3 \implies \int_{0}^{\infty} x^{5}xe^{-\alpha x^{2}}dx = \dfrac{2!}{2\alpha^{3}}

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

∫0∞−x2x5xe−αx2dx=−3⋅2!2α4 \int_{0}^{\infty} \cancel{-}x^{2}x^{5}xe^{-\alpha x^{2}}dx = \cancel{-}\dfrac{3 \cdot 2!}{2\alpha^{4}}

  ⟹  ∫0∞x7xe−αx2dx=3!2α4 \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.

∫0∞x2n+1xe−αx2dx=n!2αn+1 \int_{0}^{\infty} x^{2n+1}xe^{-\alpha x^{2}}dx = \dfrac{n!}{2\alpha^{n+1}}

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  1. Stephen J. Blundell and Katherine M. Blundell, 열 물리학(Concepts in Thermal Physics, 이재우 역) (2nd Edition, 2014), p590-591 ↩︎