Gaussian integral

The Gaussian integral (also the Euler – Poisson integral or the Poisson integral ^[1] ) is the integral of the Gaussian function :

\int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}.

{\ displaystyle \ int _ {- \ infty} ^ {\ infty} e ^ {- x ^ {2}} \, dx = {\ sqrt {\ pi}}.}

\ int _ {- \ infty} ^ {\ infty} e ^ {- x ^ {2}} \, dx = {\ sqrt {\ pi}}.

Evidence

Consider the function

(1+t)e^{-t}

{\ displaystyle (1 + t) e ^ {- t}}

(1+t)e^{-t}

. It is bounded above by a unit in the interval

(-\infty ;+\infty )

{\ displaystyle (- \ infty; + \ infty)}

(-\infty ;+\infty )

, and from below by zero on the interval

[-1;+\infty )

{\ displaystyle [-1; + \ infty)}

[-1;+\infty )

. In particular, setting

t=\pm x^{2}

{\ displaystyle t = \ pm x ^ {2}}

t=\pm x^{2}

we get when

x\not =0

{\ displaystyle x \ not = 0}

x\not =0

:

\left\{{\begin{matrix}(1-x^{2})e^{x^{2}}<1\\(1+x^{2})e^{-x^{2}}<1\end{matrix}}\right.

{\ displaystyle \ left \ {{\ begin {matrix} (1-x ^ {2}) e ​​^ {x ^ {2}} <1 \\ (1 + x ^ {2}) e ​​^ {- x ^ {2}} <1 \ end {matrix}} \ right.}

\left\{{\begin{matrix}(1-x^{2})e^{x^{2}}<1\\(1+x^{2})e^{-x^{2}}<1\end{matrix}}\right.

We restrict the change in the first inequality $x$ ${\ displaystyle x}$ $x$ gap $(0,1)$ ${\ displaystyle (0,1)}$ $(0,1)$ , and in the second - the interval $(0;\infty )$ ${\ displaystyle (0; \ infty)}$ $(0;\infty )$ , raise both inequalities to the power $n(n\in N)$ ${\ displaystyle n (n \ in N)}$ $n(n\in N)$ , since inequalities with positive terms can be raised to any positive degree. We get:

\left\{{\begin{matrix}(1-x^{2})^{n}<e^{-nx^{2}}\\0<x<1\end{matrix}}\right.

{\ displaystyle \ left \ {{\ begin {matrix} (1-x ^ {2}) ^ {n} <e ^ {- nx ^ {2}} \\ 0 <x <1 \ end {matrix}} \ right.}

\left\{{\begin{matrix}(1-x^{2})^{n}<e^{-nx^{2}}\\0<x<1\end{matrix}}\right.

and

\left\{{\begin{matrix}e^{-nx^{2}}<{\frac {1}{(1+x^{2})^{n}}}\\x>0\end{matrix}}\right.

{\ displaystyle \ left \ {{\ begin {matrix} e ^ {- nx ^ {2}} <{\ frac {1} {(1 + x ^ {2}) ^ {n}}} \\ x> 0 \ end {matrix}} \ right.}

\left\{{\begin{matrix}e^{-nx^{2}}<{\frac {1}{(1+x^{2})^{n}}}\\x>0\end{matrix}}\right.

Integrating inequalities within the indicated limits and reducing them into one, we obtain

\int \limits _{0}^{1}(1-x^{2})^{n}\,dx<\int \limits _{0}^{1}e^{-nx^{2}}\,dx<\int \limits _{0}^{\infty }e^{-nx^{2}}\,dx<\int \limits _{0}^{\infty }{\frac {1}{(1+x^{2})^{n}}}\,dx

{\ displaystyle \ int \ limits _ {0} ^ {1} (1-x ^ {2}) ^ {n} \, dx <\ int \ limits _ {0} ^ {1} e ^ {- nx ^ {2}} \, dx <\ int \ limits _ {0} ^ {\ infty} e ^ {- nx ^ {2}} \, dx <\ int \ limits _ {0} ^ {\ infty} {\ frac {1} {(1 + x ^ {2}) ^ {n}}} \, dx}

\int \limits _{0}^{1}(1-x^{2})^{n}\,dx<\int \limits _{0}^{1}e^{-nx^{2}}\,dx<\int \limits _{0}^{\infty }e^{-nx^{2}}\,dx<\int \limits _{0}^{\infty }{\frac {1}{(1+x^{2})^{n}}}\,dx

When replacing $u=x{\sqrt {n}}$ ${\ displaystyle u = x {\ sqrt {n}}}$ $u=x{\sqrt {n}}$ we get

\int \limits _{0}^{\infty }e^{-nx^{2}}\,dx={\frac {1}{\sqrt {n}}}\cdot K,\;K=\int \limits _{0}^{\infty }e^{-x^{2}}\,dx.

{\ displaystyle \ int \ limits _ {0} ^ {\ infty} e ^ {- nx ^ {2}} \, dx = {\ frac {1} {\ sqrt {n}}} \ cdot K, \; K = \ int \ limits _ {0} ^ {\ infty} e ^ {- x ^ {2}} \, dx.}

\int \limits _{0}^{\infty }e^{-nx^{2}}\,dx={\frac {1}{\sqrt {n}}}\cdot K,\;K=\int \limits _{0}^{\infty }e^{-x^{2}}\,dx.

Assuming $x=\cos t$ ${\ displaystyle x = \ cos t}$ $x=\cos t$ we obtain, respectively,

\int \limits _{0}^{1}(1-x^{2})^{n}\,dx=\int \limits _{0}^{\pi /2}\sin ^{2n+1}t\,dt={\frac {2n!!}{(2n+1)!!}}

{\ displaystyle \ int \ limits _ {0} ^ {1} (1-x ^ {2}) ^ {n} \, dx = \ int \ limits _ {0} ^ {\ pi / 2} \ sin ^ {2n + 1} t \, dt = {\ frac {2n !!} {(2n + 1) !!}}}

\int \limits _{0}^{1}(1-x^{2})^{n}\,dx=\int \limits _{0}^{\pi /2}\sin ^{2n+1}t\,dt={\frac {2n!!}{(2n+1)!!}}

Replacing the integration limits is due to the fact that when the variable changes $t$ ${\ displaystyle t}$ from 0 to $\pi /2$ ${\ displaystyle \ pi / 2}$ value $\cos t$ ${\ displaystyle \ cos t}$ varies from 0 to 1.

And replacing $x=\mathrm {ctg} \,t$ ${\ displaystyle x = \ mathrm {ctg} \, t}$ we get

\int \limits _{0}^{\infty }{\frac {1}{(1+x^{2})^{n}}}\,dx=\int \limits _{0}^{\pi /2}\sin ^{2n-2}t\,dt={\frac {\pi (2n-3)!!}{2(2n-2)!!}}

{\ displaystyle \ int \ limits _ {0} ^ {\ infty} {\ frac {1} {(1 + x ^ {2}) ^ {n}}} \, dx = \ int \ limits _ {0} ^ {\ pi / 2} \ sin ^ {2n-2} t \, dt = {\ frac {\ pi (2n-3) !!} {2 (2n-2) !!}}}

Here with integration limits is similar: $\mathrm {ctg} \,t$ ${\ displaystyle \ mathrm {ctg} \, t}$ changes from infinity to zero as the variable changes $t$ ${\ displaystyle t}$ from 0 to $\pi /2$ ${\ displaystyle \ pi / 2}$ .

The last two integrals can be found as follows: twice integrating them in parts, we obtain recurrence relations, resolving which we arrive at the results on the right-hand side.

Thus, the desired K can be included in the interval

{\sqrt {n}}\cdot {\frac {2n!!}{(2n+1)!!}}<K<{\sqrt {n}}\cdot {\frac {\pi (2n-3)!!}{2(2n-2)!!}}

{\ displaystyle {\ sqrt {n}} \ cdot {\ frac {2n !!} {(2n + 1) !!}} <K <{\ sqrt {n}} \ cdot {\ frac {\ pi (2n -3) !!} {2 (2n-2) !!}}}

To find K, we square the entire inequality and transform it. As a result, everything is greatly simplified to

{\frac {n}{2n+1}}\cdot {\frac {(2n!!)^{2}}{(2n+1)((2n-1)!!)^{2}}}<K^{2}<{\frac {n}{2n-1}}\cdot {\frac {(2n-1)((2n-3)!!)^{2}}{((2n-2)!!)^{2}}}\cdot {\frac {\pi ^{2}}{4}}

{\ displaystyle {\ frac {n} {2n + 1}} \ cdot {\ frac {(2n !!) ^ {2}} {(2n + 1) ((2n-1) !!) ^ {2} }} <K ^ {2} <{\ frac {n} {2n-1}} \ cdot {\ frac {(2n-1) ((2n-3) !!) ^ {2}} {((2n -2) !!) ^ {2}}} \ cdot {\ frac {\ pi ^ {2}} {4}}}

It follows from the Wallis formula that both left and right expressions tend to $\pi /4$ ${\ displaystyle \ pi / 4}$ at $n\rightarrow \infty$ ${\ displaystyle n \ rightarrow \ infty}$

Consequently, $K^{2}={\frac {\pi }{4}}\Longleftrightarrow K={\frac {\sqrt {\pi }}{2}}$ ${\ displaystyle K ^ {2} = {\ frac {\ pi} {4}} \ Longleftrightarrow K = {\ frac {\ sqrt {\ pi}} {2}}}$

Due to the parity of the function $e^{-x^{2}}$ ${\ displaystyle e ^ {- x ^ {2}}}$ we get that

\int \limits _{-\infty }^{\infty }e^{-x^{2}}\,dx=2\int \limits _{0}^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}.

{\ displaystyle \ int \ limits _ {- \ infty} ^ {\ infty} e ^ {- x ^ {2}} \, dx = 2 \ int \ limits _ {0} ^ {\ infty} e ^ {- x ^ {2}} \, dx = {\ sqrt {\ pi}}.}

Proof 2

Gaussian integral can be represented as

I=\int \limits _{-\infty }^{\infty }e^{-x^{2}}\,{\rm {d}}x=\int \limits _{-\infty }^{\infty }e^{-y^{2}}\,{\rm {d}}y

{\ displaystyle I = \ int \ limits _ {- \ infty} ^ {\ infty} e ^ {- x ^ {2}} \, {\ rm {d}} x = \ int \ limits _ {- \ infty } ^ {\ infty} e ^ {- y ^ {2}} \, {\ rm {d}} y}

. Consider the square of this integral

I^{2}

{\ displaystyle I ^ {2}}

. Introducing two-dimensional Cartesian coordinates , passing from them to polar coordinates

(x=r\cos \phi

{\ displaystyle (x = r \ cos \ phi}

,

y=r\sin \phi

{\ displaystyle y = r \ sin \ phi}

,

r^{2}=x^{2}+y^{2})

{\ displaystyle r ^ {2} = x ^ {2} + y ^ {2})}

and integrating over

\phi

{\ displaystyle \ phi}

(from 0 to

2\pi

{\ displaystyle 2 \ pi}

), we get:

I^{2}=\int \limits _{-\infty }^{\infty }\int \limits _{-\infty }^{\infty }\,e^{-x^{2}-y^{2}}{\rm {d}}x\,{\rm {d}}y=\int \limits _{0}^{2\pi }d\varphi \int \limits _{0}^{\infty }e^{-r^{2}}r\;{\rm {d}}r=2\pi \int \limits _{0}^{\infty }e^{-r^{2}}r\;{\rm {d}}r=\pi \int \limits _{0}^{\infty }e^{-r^{2}}\;{\rm {d}}r^{2}=\pi .

{\ displaystyle I ^ {2} = \ int \ limits _ {- \ infty} ^ {\ infty} \ int \ limits _ {- \ infty} ^ {\ infty} \, e ^ {- x ^ {2} -y ^ {2}} {\ rm {d}} x \, {\ rm {d}} y = \ int \ limits _ {0} ^ {2 \ pi} d \ varphi \ int \ limits _ {0 } ^ {\ infty} e ^ {- r ^ {2}} r \; {\ rm {d}} r = 2 \ pi \ int \ limits _ {0} ^ {\ infty} e ^ {- r ^ {2}} r \; {\ rm {d}} r = \ pi \ int \ limits _ {0} ^ {\ infty} e ^ {- r ^ {2}} \; {\ rm {d}} r ^ {2} = \ pi.}

Consequently, $I=\int \limits _{-\infty }^{\infty }e^{-x^{2}}\,{\rm {d}}x={\sqrt {\pi }}$ ${\ displaystyle I = \ int \ limits _ {- \ infty} ^ {\ infty} e ^ {- x ^ {2}} \, {\ rm {d}} x = {\ sqrt {\ pi}}}$ .

Proof 3

Gaussian integral can be represented as

I=\int \limits _{-\infty }^{\infty }{{e^{-{x^{2}}}}dx}=\int \limits _{-\infty }^{\infty }{{e^{-{y^{2}}}}dy}=\int \limits _{-\infty }^{\infty }{{e^{-{z^{2}}}}dz}

{\ displaystyle I = \ int \ limits _ {- \ infty} ^ {\ infty} {{e ^ {- {x ^ {2}}}} dx} = \ int \ limits _ {- \ infty} ^ { \ infty} {{e ^ {- {y ^ {2}}}} dy} = \ int \ limits _ {- \ infty} ^ {\ infty} {{e ^ {- {z ^ {2}}} } dz}}

. Consider the cube of this integral

I^{3}

{\ displaystyle I ^ {3}}

. Introducing three-dimensional Cartesian coordinates , passing from them to spherical coordinates :

$\left\{{\begin{array}{l}x=r\sin \theta \cos \varphi \\y=r\sin \theta \sin \varphi \\z=r\cos \theta \\{r^{2}}={x^{2}}+{y^{2}}+{z^{2}}\end{array}}\right.$ ${\ displaystyle \ left \ {{\ begin {array} {l} x = r \ sin \ theta \ cos \ varphi \\ y = r \ sin \ theta \ sin \ varphi \\ z = r \ cos \ theta \ \ {r ^ {2}} = {x ^ {2}} + {y ^ {2}} + {z ^ {2}} \ end {array}} \ right.}$ , the Jacobian transform is equal to $J={r^{2}}\sin \theta$ ${\ displaystyle J = {r ^ {2}} \ sin \ theta}$ ,

and integrating over $\theta$ ${\ displaystyle \ theta}$ (from $0$ ${\ displaystyle 0}$ before $\pi$ ${\ displaystyle \ pi}$ ), by $\varphi$ ${\ displaystyle \ varphi}$ (from $0$ ${\ displaystyle 0}$ before $2\pi$ ${\ displaystyle 2 \ pi}$ ), by $r$ ${\ displaystyle r}$ (from $0$ ${\ displaystyle 0}$ before $\infty$ ${\ displaystyle \ infty}$ ), we get:

${\begin{array}{l}{I^{3}}=\int \limits _{-\infty }^{\infty }{\int \limits _{-\infty }^{\infty }{\int \limits _{-\infty }^{\infty }{{e^{-{x^{2}}-{y^{2}}-{z^{2}}}}dx}dy}dz}=\int \limits _{0}^{2\pi }{\int \limits _{0}^{\pi }{\int \limits _{0}^{\infty }{{e^{-{r^{2}}}}Jdr}d\theta }d\varphi }=\int \limits _{0}^{2\pi }{d\varphi \int \limits _{0}^{\pi }{\sin \theta d\theta \int \limits _{0}^{\infty }{{e^{-{r^{2}}}}{r^{2}}dr}}}=\\=2\pi \left({(-\cos \pi )-(-\cos 0)}\right)\left({-{\frac {1}{2}}\int \limits _{0}^{\infty }{rd{e^{-{r^{2}}}}}}\right)=-2\pi \left({\left.{\left({r{e^{-{r^{2}}}}}\right)}\right|_{0}^{\infty }-\int \limits _{0}^{\infty }{{e^{-{r^{2}}}}dr}}\right)=-2\pi (0-{\frac {I}{2}})=\pi I\end{array}}$ ${\ displaystyle {\ begin {array} {l} {I ^ {3}} = \ int \ limits _ {- \ infty} ^ {\ infty} {\ int \ limits _ {- \ infty} ^ {\ infty } {\ int \ limits _ {- \ infty} ^ {\ infty} {{e ^ {- {x ^ {2}} - {y ^ {2}} - {z ^ {2}}}} dx} dy} dz} = \ int \ limits _ {0} ^ {2 \ pi} {\ int \ limits _ {0} ^ {\ pi} {\ int \ limits _ {0} ^ {\ infty} {{e ^ {- {r ^ {2}}}} Jdr} d \ theta} d \ varphi} = \ int \ limits _ {0} ^ {2 \ pi} {d \ varphi \ int \ limits _ {0} ^ {\ pi} {\ sin \ theta d \ theta \ int \ limits _ {0} ^ {\ infty} {{e ^ {- {r ^ {2}}}} {r ^ {2}} dr}} } = \\ = 2 \ pi \ left ({(- \ cos \ pi) - (- \ cos 0)} \ right) \ left ({- {\ frac {1} {2}} \ int \ limits _ {0} ^ {\ infty} {rd {e ^ {- {r ^ {2}}}}}} right) = - 2 \ pi \ left ({\ left. {\ Left ({r {e ^ {- {r ^ {2}}}} \ right)} \ right | _ {0} ^ {\ infty} - \ int \ limits _ {0} ^ {\ infty} {{e ^ {- {r ^ {2}}}} dr}} \ right) = - 2 \ pi (0 - {\ frac {I} {2}}) = \ pi I \ end {array}}}$

Consequently, $I=\int \limits _{-\infty }^{\infty }{{e^{-{x^{2}}}}dx}={\sqrt {\pi }}$ ${\ displaystyle I = \ int \ limits _ {- \ infty} ^ {\ infty} {{e ^ {- {x ^ {2}}}} dx} = {\ sqrt {\ pi}}}$ .

Variations

Gaussian integrals of a scaled Gaussian function

\int \limits _{-\infty }^{\infty }\alpha e^{-x^{2}/\beta ^{2}}\,dx=\alpha \beta {\sqrt {\pi }}

{\ displaystyle \ int \ limits _ {- \ infty} ^ {\ infty} \ alpha e ^ {- x ^ {2} / \ beta ^ {2}} \, dx = \ alpha \ beta {\ sqrt {\ pi}}}

and multidimensional Gaussian integrals

\int \alpha e^{-(x^{2}/\beta _{1}^{2}+y^{2}/\beta _{2}^{2}+z^{2}/\beta _{3}^{2}+\dots )}\,dxdydz\dots =\alpha \beta _{1}\beta _{2}\beta _{3}\dots {\sqrt {\pi ^{n}}}

{\ displaystyle \ int \ alpha e ^ {- (x ^ {2} / \ beta _ {1} ^ {2} + y ^ {2} / \ beta _ {2} ^ {2} + z ^ {2 } / \ beta _ {3} ^ {2} + \ dots)} \, dxdydz \ dots = \ alpha \ beta _ {1} \ beta _ {2} \ beta _ {3} \ dots {\ sqrt {\ pi ^ {n}}}}

elementarily reduced to the usual one-dimensional, described first (here and below, integration throughout the whole space is implied).

The same applies to multidimensional integrals of the form

\int e^{xMx}\,dx_{1}dx_{2}dx_{3}\dots dx_{n}={\sqrt {\frac {\pi ^{n}}{|\det(M)|}}}

{\ displaystyle \ int e ^ {xMx} \, dx_ {1} dx_ {2} dx_ {3} \ dots dx_ {n} = {\ sqrt {\ frac {\ pi ^ {n}} {| \ det ( M) |}}}}

where x is a vector, and M is a symmetric matrix with negative eigenvalues, since such integrals are reduced to the previous one if we make a coordinate transformation diagonalizing the matrix M.

Practical application (for example, to calculate the Fourier transform of a Gaussian function) often finds the following relation

\int \limits _{-\infty }^{\infty }e^{-ax^{2}+bx+c}\,dx={\sqrt {\frac {\pi }{a}}}\,e^{{\frac {b^{2}}{4a}}+c},

{\ displaystyle \ int \ limits _ {- \ infty} ^ {\ infty} e ^ {- ax ^ {2} + bx + c} \, dx = {\ sqrt {\ frac {\ pi} {a}} } \, e ^ {{\ frac {b ^ {2}} {4a}} + c},}

History

For the first time, the one-dimensional Gaussian integral was calculated in 1729 by Euler , then Poisson found a simple way to calculate it ^[2] . In this regard, he received the name of the Euler – Poisson integral.

Notes

↑ Poisson Integral - an article from the Great Soviet Encyclopedia .
↑ See ibid.

[1] Poisson Integral - an article from the Great Soviet Encyclopedia .

[2] See ibid.

Gaussian integral

Evidence

Variations

History

See also

Notes

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