The Gaussian integral, also known as the Euler–Poisson integral, is the integral of the Gaussian function f ( x )
Integral Gauss, juga dikenal dengan nama integral Euler–Poisson, merupakan integral dari fungsi Gauss f ( x )e − x 2 }} di sepanjang garis riil.
^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}.} Abraham de Moivre originally discovered this type of integral in 1733, while Gauss published the precise integral in 1809, attributing its discovery to Laplace.
Integral ini dinamai dari matematikawan Jerman Carl Friedrich Gauss, dan dituliskan secara matematis sebagai ∫ − ∞ ∞ e − x 2 d xπ ^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}} Jenis integral ini awalnya ditemukan oleh Abraham de Moivre pada tahun 1733, tetapi Gauss menerbitkan bentuk integral yang rinci pada tahun 1809.
The integral has a wide range of applications.
Integral ini dapat diaplikasikan untuk berbagai macam hal.
For example, with a slight change of variables it is used to compute the normalizing constant of the normal distribution.
Contohnya, dengan sedikit perubahan dalam variabel, integral ini digunakan untuk menghitung konstanta normalisasi dari distribusi normal.
That is, there is no elementary indefinite integral for ∫ e − x 2 d x , }\,dx,} but the definite integral ∫ − ∞ ∞ e − x 2 d x ^{\infty }e^{-x^{2}}\,dx} can be evaluated.
Ini mengartikan integral ∫ − ∞ ∞ e − x 2 d x ^{\infty }e^{-x^{2}}\,dx} dapat dihitung cukup dengan menghitung limit dari lim a → ∞ I ( a ) .
^{\infty }e^{-a(x+b)^{2}}\,dx={\sqrt {\frac {\pi }{a}}}.}
^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}.}
A standard way to compute the Gaussian integral, the idea of which goes back to Poisson, is to make use of the property that: ( ∫ − ∞ ∞ e − x 2 d x ) 2∫ − ∞ ∞ e − x 2 d x ∫ − ∞ ∞ e − y 2 d y∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 )
Salah satu cara umum menghitung integral Gauss, berdasar pada ide yang dapat ditelusuri kembali ke Poisson, adalah menggunakan sifat: ( ∫ − ∞ ∞ e − x 2 d x ) 2∫ − ∞ ∞ e − x 2 d x ∫ − ∞ ∞ e − y 2 d y∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 )
^{\infty }e^{-x^{2}}\,dx\right)^{2}=\int _{-\infty }^{\infty }e^{-x^{2}}\,dx\int _{-\infty }^{\infty }e^{-y^{2}}\,dy=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-\left(x^{2}+y^{2}\right)}\,dx\,dy.} Consider the function e − ( x 2 + y 2 )
^{\infty }e^{-x^{2}}\,dx\right)^{2}=\int _{-\infty }^{\infty }e^{-x^{2}}\,dx\int _{-\infty }^{\infty }e^{-y^{2}}\,dy=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-\left(x^{2}+y^{2}\right)}\,dx\,dy.} Selanjutnya, pertimbangkan fungsi e − ( x 2 + y 2 )
e − r 2 +y^{2}\right)}=e^{-r^{2}}} on the plane R 2 ^{2}} , and compute its integral two ways: on the one hand, by double integration in the Cartesian coordinate system, its integral is a square: ( ∫ e − x 2 d x ) 2 ; }\,dx\right)^{2};} on the other hand, by shell integration (a case of double integration in polar coordinates), its integral is computed to be π Comparing these two computations yields the integral, though one should take care about the improper integrals involved. ∬ R 2 e − ( x 2 + y 2 )
e − r 2 +y^{2}\right)}=e^{-r^{2}}} pada bidang R 2 ^{2}} , yang integralnya dapat dihitung menggunakan dua cara: Di ruas kiri, menggunakan integral lipat dua di sistem koordinat Kartesius, yang hasilnya adalah sebuah kuadrat: ( ∫ e − x 2 d x ) 2 ; {\textstyle \left(\int e^{-x^{2}}\,dx\right)^{2};} Di ruas kanan, menggunakan metode kulit (sautu bentuk integral lipat dua dalam koordinat polar), hasil integralnya sama dengan π Integral Gauss dapat dihitung dengan menggabungkan kedua cara tersebut, walau perhatian khusus diperlukan saat berurusan dengan bentuk integral takwajar: ∬ R 2 e − ( x 2 + y 2 )
If the integral ∫ − ∞ ∞ e − x 2 d x ^{\infty }e^{-x^{2}}\,dx} were absolutely convergent we would have that its Cauchy principal value, that is, the limit lim a → ∞ I ( a ) I(a)} would coincide with ∫ − ∞ ∞ e − x 2 d x .
Jika hasil dari integral ∫ − ∞ ∞ e − x 2 d x ^{\infty }e^{-x^{2}}\,dx} konvergen absolut, maka integral tersebut memiliki nilai utama Cauchy; dengan kata lain limit dari lim a → ∞ I ( a ) {\textstyle \lim _{a\to \infty }I(a)} akan sama dengan hasil integral tersebut.
To see that this is the case, consider that ∫ − ∞ ∞ | e − x 2 | d x < ∫ − ∞ − 1 − x e − x 2 d x + ∫ − 1 1 e − x 2 d x + ∫ 1 ∞ x e − x 2 d x < ∞ .
Untuk melihat itu memang terjadi, pertimbangkan bahwa ∫ − ∞ ∞ | e − x 2 | d x < ∫ − ∞ − 1 − x e − x 2 d
So we can compute ∫ − ∞ ∞ e − x 2 d x ^{\infty }e^{-x^{2}}\,dx} by just taking the limit lim a → ∞ I ( a ) .
Ini mengartikan integral ∫ − ∞ ∞ e − x 2 d x ^{\infty }e^{-x^{2}}\,dx} dapat dihitung cukup dengan menghitung limit dari lim a → ∞ I ( a ) .
I(a).} Taking the square of I ( a ) yields I ( a ) 2( ∫ − a a e − x 2 d x ) ( ∫ − a a e − y 2 d y )∫ − a a ( ∫ − a a e − y 2 d y ) e − x 2 d x∫ − a a ∫ − a a e − ( x 2 + y 2 )
I ( a ) akan menghasilkan bentuk I ( a ) 2( ∫ − a a e − x 2 d x ) ( ∫ − a a e − y 2 d y )∫ − a a ( ∫ − a a e − y 2 d y ) e − x 2 d x∫ − a a ∫ − a a e − ( x 2 + y 2 )
I(a)^{2}&=\left(\int _{-a}^{a}e^{-x^{2}}\,dx\right)\left(\int _{-a}^{a}e^{-y^{2}}\,dy\right)\\[6pt]&=\int _{-a}^{a}\left(\int _{-a}^{a}e^{-y^{2}}\,dy\right)\,e^{-x^{2}}\,dx\\[6pt]&=\int _{-a}^{a}\int _{-a}^{a}e^{-\left(x^{2}+y^{2}\right)}\,dy\,dx.\end{aligned}}} Using Fubini's theorem, the above double integral can be seen as an area integral ∬ [ − a , a ] × [ − a , a ] e − ( x 2 + y 2 )
I(a)^{2}&=\left(\int _{-a}^{a}e^{-x^{2}}\,dx\right)\left(\int _{-a}^{a}e^{-y^{2}}\,dy\right)\\[6pt]&=\int _{-a}^{a}\left(\int _{-a}^{a}e^{-y^{2}}\,dy\right)\,e^{-x^{2}}\,dx\\[6pt]&=\int _{-a}^{a}\int _{-a}^{a}e^{-\left(x^{2}+y^{2}\right)}\,dy\,dx.\end{aligned}}} Menggunakan teorema Fubini, integral lipat dua di atas dapat dianggap sebagai integral luas ∬
[ ∂ x ∂ r ∂ x ∂ θ ∂ y ∂ r ∂ y ∂ θ ][ cos ⁡ θ − r sin ⁡ θ sin ⁡ θ − r cos ⁡ θ ] (r,\theta )={\begin{bmatrix}{\dfrac {\partial x}{\partial r}}&{\dfrac {\partial x}{\partial \theta }}\\[1em]{\dfrac {\partial y}{\partial r}}&{\dfrac {\partial y}{\partial \theta }}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-r\sin \theta \\\sin \theta &{\hphantom {-}}r\cos \theta \end{bmatrix}}}
xr cos ⁡ θ yr sin ⁡ θ . x&=r\cos \theta \\y&=r\sin \theta .\end{aligned}}} Transformasi ini memiliki determinan Jacobi J ( r , θ )[ ∂ x ∂ r ∂ x ∂ θ ∂ y ∂ r ∂ y ∂ θ ][ cos ⁡ θ − r sin ⁡ θ sin ⁡ θ r cos ⁡ θ ]r (r,\theta )={\begin{bmatrix}{\dfrac {\partial x}{\partial r}}&{\dfrac {\partial x}{\partial \theta }}\\[1em]{\dfrac {\partial y}{\partial r}}&{\dfrac {\partial y}{\partial \theta }}\end{bmatrix}}={\begin{bmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{bmatrix}}=r} dan elemen luas d ( x , y )| J ( r , θ ) | d ( r , θ )r d ( r , θ ) .
} ∫ 0 2 π ∫ 0 a r e − r 2 d r d θ <
I 2 ( a ) < ∫ 0 2 π ∫ 0 a 2 r e − r 2 d r d θ .
That is, ∫ − ∞ ∞ e − x 2 d x2 ∫ 0 ∞ e − x 2 d x .
x + ∫ 1 ∞ x e − x 2 d x < ∞ .
^{\infty }e^{-x^{2}}\,dx=2\int _{0}^{\infty }e^{-x^{2}}\,dx.}
^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}.}
I 24 ∫ 0 ∞ ∫ 0 ∞ e − ( x 2 + y 2 )
x + ∫ 1 ∞ x e − x 2 d x < ∞ .
d x I^{2}&=4\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x^{2}+y^{2}\right)}dy\,dx\\[6pt]&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-\left(x^{2}+y^{2}\right)}\,dy\right)\,dx\\[6pt]&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-x^{2}\left(1+s^{2}\right)}x\,ds\right)\,dx\\[6pt]\end{aligned}}} Then, using Fubini's theorem to switch the order of integration: I 24 ∫ 0 ∞ ( ∫ 0 ∞ e − x 2 ( 1 + s 2 ) x d x )
d x I^{2}&=4\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x^{2}+y^{2}\right)}dy\,dx\\[6pt]&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-\left(x^{2}+y^{2}\right)}\,dy\right)\,dx\\[6pt]&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-x^{2}\left(1+s^{2}\right)}x\,ds\right)\,dx\\[6pt]\end{aligned}}} Menggunakan teorema Fubini untuk menukar urutan integrasi: I 24 ∫ 0 ∞ ( ∫ 0 ∞ e − x 2 ( 1 + s 2 )
I^{2}&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-x^{2}\left(1+s^{2}\right)}x\,dx\right)\,ds\\[6pt]&=4\int _{0}^{\infty }\left[{\frac {e^{-x^{2}\left(1+s^{2}\right)}}{-2\left(1+s^{2}\right)}}\right]_{x=0}^{x=\infty }\,ds\\[6pt]&=4\left({\frac {1}{2}}\int _{0}^{\infty }{\frac {ds}{1+s^{2}}}\right)\\[6pt]&=2\arctan(s){\Big |}_{0}^{\infty }\\[6pt]&=\pi .\end{aligned}}} Therefore, Iπ }} , as expected.
I^{2}&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-x^{2}\left(1+s^{2}\right)}x\,dx\right)\,ds\\[6pt]&=4\int _{0}^{\infty }\left[{\frac {e^{-x^{2}\left(1+s^{2}\right)}}{-2\left(1+s^{2}\right)}}\right]_{x=0}^{x=\infty }\,ds\\[6pt]&=4\left({\frac {1}{2}}\int _{0}^{\infty }{\frac {ds}{1+s^{2}}}\right)\\[6pt]&=2\arctan(s){\Big |}_{0}^{\infty }\\[6pt]&=\pi .\end{aligned}}} Alhasil, terbukti
[-1,1]}(1+x^{2})^{-n}dx} That is, 2 n ∫ [ 0 , 1 ] ( 1 − x 2 ) n d x ≤ ∫ [ − n , n ] e − x 2 d x ≤ 2 n ∫ [ 0 , 1 ] ( 1 + x 2 ) − n d x }\int _{[0,1]}(1-x^{2})^{n}dx\leq \int _{[-{\sqrt {n}},{\sqrt {n}}]}e^{-x^{2}}dx\leq 2{\sqrt {n}}\int _{
2 n + 1 ^{\infty }x^{2n}e^{-{\frac {x^{2}}{a^{2}}}}\,dx={\sqrt {\pi }}{\frac {a^{2n+1}(2n-1)!!}{2^{n+1}}}} ∫ 0 ∞ x 2 n + 1 e − x 2 a 2 d
x2 ∫ 0 ∞ 1 2 e − t t − 1 2 d tΓ ( 1 2 )π ^{\infty }e^{-x^{2}}dx=2\int _{0}^{\infty }{\frac {1}{2}}\ e^{-t}\ t^{-{\frac {1}{2}}}dt=\Gamma {\left({\frac {1}{2}}\right)}={\sqrt {\pi }}} where Γ ( z )∫ 0 ∞ t z − 1 e − t d t {\textstyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}dt} is the gamma function.
π ^{\infty }e^{-x^{2}}dx=2\int _{0}^{\infty }{\frac {1}{2}}\ e^{-t}\ t^{-{\frac {1}{2}}}dt=\Gamma \left({\frac {1}{2}}\right)={\sqrt {\pi }}} dengan Γ ( z )∫ 0 ∞ t z − 1 e − t d t {\textstyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}dt} adalah fungsi gamma.
More generally, ∫ 0 ∞ x n e − a x b d xΓ ( ( n + 1 ) / b ) b a ( n + 1 ) / b , ^{\infty }x^{n}e^{-ax^{b}}dx={\frac {\Gamma {\left((n+1)/b\right)}}{ba^{(n+1)/b}}},} which can be obtained by substituting ta x b } in the integrand of the gamma function to get Γ ( z )
2 a n + 1 ^{\infty }x^{2n+1}e^{-ax^{2}}\,dx={\frac {n!}{2a^{n+1}}}} ∫ 0 ∞ x n e − a x 2 d xΓ ( n + 1 2 ) 2 a n + 1 2 ^{\infty }x^{n}e^{-ax^{2}}\,dx={\frac {\Gamma ({\frac {n+1}{2}})}{2a^{\frac {n+1}{2}}}}} Sebuah cara mudah untuk menghasilkan bentuk-bentuk tersebut adalah dengan mendiferensialkannya terhadap variabel integrasi: ∫ − ∞ ∞ x 2 n e − α x 2 d x( − 1 ) n ∫ − ∞ ∞ ∂ n ∂ α n e − α x 2 d x( − 1 ) n ∂ n ∂ α n ∫ − ∞ ∞ e − α x 2 d xπ ( − 1 ) n ∂ n ∂ α n α − 1 2π α ( 2 n − 1 ) ! !
∫ − ∞ ∞ e 1 2 i t 2 d te i π / 4 2 π ^{\infty }e^{{\frac {1}{2}}it^{2}}dt=e^{i\pi /4}{\sqrt {2\pi }}} and more generally, ∫ R N e 1 2 i x T A x d xdet ( A ) − 1 2 ( e i π / 4 2 π ) N ^{N}}e^{{\frac {1}{2}}i\mathbf {x} ^{T}A\mathbf {x} }dx=\det(A)^{-{\frac {1}{2}}}{\left(e^{i\pi /4}{\sqrt {2\pi }}\right)}^{N}} for any positive-definite symmetric matrix A .
π ^{\infty }e^{-x^{2}}dx=2\int _{0}^{\infty }{\frac {1}{2}}\ e^{-t}\ t^{-{\frac {1}{2}}}dt=\Gamma \left({\frac {1}{2}}\right)={\sqrt {\pi }}} dengan Γ ( z )∫ 0 ∞ t z − 1 e − t d t {\textstyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}dt} adalah fungsi gamma.
d n xdet ( 2 π A − 1 ) exp ⁡ ( 1 2 b T A − 1 b + c ) ^{n}}\exp {\left(-{\tfrac {1}{2}}\mathbf {x} ^{\mathsf {T}}A\mathbf {x} +\mathbf {b} ^{\mathsf {T}}\mathbf {x} +c\right)}\,d^{n}\mathbf {x}{\sqrt {\det \left(2\pi A^{-1}\right)}}\exp \left({\tfrac {1}{2}}\mathbf {b} ^{\mathsf {T}}A^{-1}\mathbf {b} +c\right)} This fact is applied in the study of the multivariate normal distribution.
d n xdet ( 2 π A − 1 ) e 1 2 b T A − 1 b + c ^{n}}\exp {\left(-{\frac {1}{2}}x^{\mathsf {T}}Ax+b^{\mathsf {T}}x+c\right)}\,d^{n}x={\sqrt {\det(2\pi A^{-1})}}e^{{\frac {1}{2}}b^{\mathsf {T}}A^{-1}b+c}}
Also, ∫ x k 1 ⋯ x k 2 N exp ⁡ ( − 1 2 ∑ i , j1 n A i j x i x j )
Selain itu, ∫ x k 1 ⋯ x k 2 N exp ⁡ ( − 1 2 ∑ i , j1 n A i j x i x j )
A}}}\,{\frac {1}{2^{N}N!}}\,\sum _{\sigma \in S_{2N}}(A^{-1})_{k_{\sigma (1)}k_{\sigma (2)}}\cdots (A^{-1})_{k_{\sigma (2N-1)}k_{\sigma (2N)}}} where σ is a permutation of {1, …, 2N} and the extra factor on the right-hand side is the sum over all combinatorial pairings of {1, …, 2N} of N copies of A−1.
A}}}\,{\frac {1}{2^{N}N!}}\,\sum _{\sigma \in S_{2N}}(A^{-1})_{k_{\sigma (1)}k_{\sigma (2)}}\cdots (A^{-1})_{k_{\sigma (2N-1)}k_{\sigma (2N)}}} dengan σ menyatakan permutasi dari { 1 , ⋯ , 2 N } } dan faktor tambahan di ruas kanan adalah hasil penjumlahan atas semua pasangan kombinatorik { 1 , ⋯ , 2 N } } dari N salinan A − 1 .
Alternatively, ∫ f ( x ) exp ⁡ ( − 1 2 ∑ i , j1 n A i j x i x j )
Alternatif lain, ∫ f ( x → ) exp ⁡ ( − 1 2 ∑ i , j1 n A i j x i x j )
f ( x ) | x0 )\exp {\left(-{\frac {1}{2}}\sum _{i,j=1}^{n}A_{ij}x_{i}x_{j}\right)}d^{n}\mathbf {x}{\sqrt {\frac {{\left(2\pi \right)}^{n}}{\det A}}}\,\left.\exp \left({\frac {1}{2}}\sum _{i,j=1}^{n}\left(A^{-1}\right)_{ij}{\partial \over \partial x_{i}}{\partial \over \partial x_{j}}\right)f(\mathbf {x} )
i j ∂ ∂ x i ∂ ∂ x j ) f ( x → ) | x →0 })\exp {\left(-{\frac {1}{2}}\sum _{i,j=1}^{n}A_{ij}x_{i}x_{j}\right)}d^{n}x={\sqrt {(2\pi )^{n} \over \det A}}\,\left.\exp {\left({1 \over 2}\sum _{i,j=1}^{n}\left(A^{-1}\right)_{ij}{\partial \over \partial x_{i}}{\partial \over \partial x_{j}}\right)}f({\vec {x}})\right|_{{\vec {x}}=0}} Untuk suatu fungsi analitik f , mengasumsikan pertumbuhan fungsi tersebut memenuhi suatu batasan yang masuk akal, dan beberapa kriteria lainnya.
f ( x 2 N + 1 ) f ( x 2 N + 2 )
x + ∫ − 1 1 e − x 2 d
d d x 2 N + 1 d d x 2 N + 2 ] D f1 2 N N !
d n x( 2 π ) n det A exp ⁡ ( 1 2 ∑ i , j1 n ( A − 1 )
∑ σ ∈ S 2 N A − 1 ( x σ ( 1 ) , x σ ( 2 ) )
d n x( 2 π ) n det A 1 2 N N ! ∑ σ ∈ S 2 N ( A − 1 ) k σ ( 1 ) k σ ( 2 )
⋯ A − 1 ( x σ ( 2 N − 1 ) , x σ ( 2 N ) ) .
d n x( 2 π ) n det A 1 2 N N ! ∑ σ ∈ S 2 N ( A − 1 ) k σ ( 1 ) k σ ( 2 )
&{\frac )\cdots f(x_{2N})\exp \left[{-\iint {\frac {1}{2}}A(x_{2N+1},x_{2N+2})f(x_{2N+1})f(x_{2N+2})\,d^{d}x_{2N+1}\,d^{d}x_{2N+2}}\right]{\mathcal {D}}f}{2}}A(x_{2N+1},x_{2N+2})f(x_{2N+1})f(x_{2N+2})\,d^{d}x_{2N+1}\,d^{d}x_{2N+2}}\right]{\mathcal {D}}f}}\\[6pt]={}&{\frac {1}{2^{N}N!}}\sum _{\sigma \in S_{2N}}A^{-1}(x_{\sigma (1)},x_{\sigma (2)})\cdots A^{-1}(x_{\sigma (2N-1)},x_{\sigma (2N)}).\end{aligned}}} In the DeWitt notation, the equation looks identical to the finite-dimensional case.
I^{2}&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-x^{2}\left(1+s^{2}\right)}x\,dx\right)\,ds\\[6pt]&=4\int _{0}^{\infty }\left[{\frac {e^{-x^{2}\left(1+s^{2}\right)}}{-2\left(1+s^{2}\right)}}\right]_{x=0}^{x=\infty }\,ds\\[6pt]&=4\left({\frac {1}{2}}\int _{0}^{\infty }{\frac {ds}{1+s^{2}}}\right)\\[6pt]&=2\arctan(s){\Big |}_{0}^{\infty }\\[6pt]&=\pi .\end{aligned}}} Alhasil, terbukti
d n x∫ exp ⁡ ( − 1 2 x T A x + b T x )
d n x∫ R n exp ⁡ ( − 1 2 x T A x )
d n x( 2 π ) n det A exp ⁡ ( 1 2 b T A − 1 b ) .
d n x( 2 π ) n det A exp ⁡ ( 1 2 ∑ i , j1 n ( A − 1 )
\int \exp \left(-{\frac {1}{2}}\sum _{i,j=1}^{n}A_{ij}x_{i}x_{j}+\sum _{i=1}^{n}b_{i}x_{i}\right)d^{n}\mathbf {x} &=\int \exp \left(-{\tfrac {1}{2}}\mathbf {x} ^{\mathsf {T}}A\mathbf {x} +\mathbf {b} ^{\mathsf {T}}\mathbf {x} \right)d^{n}\mathbf {x} \\&={\sqrt {\frac {(2\pi )^{n}}{\det
I^{2}&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-x^{2}\left(1+s^{2}\right)}x\,dx\right)\,ds\\[6pt]&=4\int _{0}^{\infty }\left[{\frac {e^{-x^{2}\left(1+s^{2}\right)}}{-2\left(1+s^{2}\right)}}\right]_{x=0}^{x=\infty }\,ds\\[6pt]&=4\left({\frac {1}{2}}\int _{0}^{\infty }{\frac {ds}{1+s^{2}}}\right)\\[6pt]&=2\arctan(s){\Big |}_{0}^{\infty }\\[6pt]&=\pi .\end{aligned}}} Alhasil, terbukti
A}}}\exp \left({\tfrac {1}{2}}\mathbf {b} ^{\mathsf {T}}A^{-1}\mathbf {b} \right).\end{aligned}}} ∫ 0 ∞ x 2 n e − x 2 / a 2 d xπ a 2 n + 1 ( 2 n − 1 ) ! !
Dengan mensubtitusi ta x b } di integran, dapat dihasilkan bentuk integral yang lebih umum, ∫ 0 ∞ x n e − a x b d xΓ ( ( n + 1 ) / b ) b a ( n + 1 ) / b ^{\infty }x^{n}e^{-ax^{b}}dx={\frac {\Gamma \left((n+1)/b\right)}{ba^{(n+1)/b}}}}
2 n + 1 ^{\infty }x^{2n}e^{-{x^{2}}/{a^{2}}}\,dx={\sqrt {\pi }}{\frac {a^{2n+1}(2n-1)!!}{2^{n+1}}}} ∫ 0 ∞ x 2 n + 1 e − x 2 / a 2 d xn !
2 n + 1 ^{\infty }x^{2n}e^{-{\frac {x^{2}}{a^{2}}}}\,dx={\sqrt {\pi }}{\frac {a^{2n+1}(2n-1)!!}{2^{n+1}}}} ∫ 0 ∞ x 2 n + 1 e − x 2 a 2 d
2 a 2 n + 2 ^{\infty }x^{2n+1}e^{-{x^{2}}/{a^{2}}}\,dx={\frac {n!}{2}}a^{2n+2}} ∫ 0 ∞ x 2 n e − b x 2 d x( 2 n − 1 ) ! !
2 n + 1 ^{\infty }x^{2n}e^{-{\frac {x^{2}}{a^{2}}}}\,dx={\sqrt {\pi }}{\frac {a^{2n+1}(2n-1)!!}{2^{n+1}}}} ∫ 0 ∞ x 2 n + 1 e − x 2 a 2 d
∫ − ∞ ∞ x 2 n e − α x 2 d x( − 1 ) n ∫ − ∞ ∞ ∂ n ∂ α n e − α x 2 d x( − 1 ) n ∂ n ∂ α n ∫ − ∞ ∞ e − α x 2 d xπ ( − 1 ) n ∂ n ∂ α n α − 1 2π α ( 2 n − 1 ) ! ! ( 2 α ) n \int _{-\infty }^{\infty }x^{2n}e^{-\alpha x^{2}}\,dx&=\left(-1\right)^{n}\int _{-\infty }^{\infty }{\frac {\partial ^{n}}{\partial \alpha ^{n}}}e^{-\alpha x^{2}}\,dx\\[1ex]&=\left(-1\right)^{n}{\frac {\partial ^{n}}{\partial \alpha ^{n}}}\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\,dx\\[1ex]&={\sqrt {\pi }}\left(-1\right)^{n}{\frac {\partial ^{n}}{\partial \alpha ^{n}}}\alpha ^{-{\frac {1}{2}}}\\[1ex]&={\sqrt {\frac {\pi }{\alpha }}}{\frac {(2n-1)!!}{\left(2\alpha \right)^{n}}}\end{aligned}}} One could also integrate by parts and find a recurrence relation to solve this.
2 a n + 1 ^{\infty }x^{2n+1}e^{-ax^{2}}\,dx={\frac {n!}{2a^{n+1}}}} ∫ 0 ∞ x n e − a x 2 d xΓ ( n + 1 2 ) 2 a n + 1 2 ^{\infty }x^{n}e^{-ax^{2}}\,dx={\frac {\Gamma ({\frac {n+1}{2}})}{2a^{\frac {n+1}{2}}}}} Sebuah cara mudah untuk menghasilkan bentuk-bentuk tersebut adalah dengan mendiferensialkannya terhadap variabel integrasi: ∫ − ∞ ∞ x 2 n e − α x 2 d x( − 1 ) n ∫ − ∞ ∞ ∂ n ∂ α n e − α x 2 d x( − 1 ) n ∂ n ∂ α n ∫ − ∞ ∞ e − α x 2 d xπ ( − 1 ) n ∂ n ∂ α n α − 1 2π α ( 2 n − 1 ) ! !
Applying a linear change of basis shows that the integral of the exponential of a homogeneous polynomial in n variables may depend only on SL(n)-invariants of the polynomial.
Menerapkan perubahan basis linier memperlihatkan bahwa integral dari eksponensial dari polinomial homogen pada n variabel hanya dapat bergantung pada invarian-SL(n) dari polinomial.
List of integrals of Gaussian functions Common integrals in quantum field theory Normal distribution List of integrals of exponential functions Error function Berezin integral
Daftar integral dari fungsi Gauss Integral yang umum dalam teori medan kuantum Distribusi normal Daftar integral dari fungsi eksponensial Fungsi galat Integral Berezin (Inggris) Weisstein, Eric W. Integral.html