Integration of e^x^2 pdf Continue Integral gaussian function equal to sqrt (π) This integral element of statistics and physics should not be confused with the Gaussian square, the method of numerical integration. Graph f (x) e x x 2 displaystyle f(x)e'-x-{2} and the area between the function and x displaystyle x -axis, which is equal to π sndisplay. Gossian Integral, also known as the Euler-Poisson integral, is an integral part of the Gaussian function f (x) , e x x 2 displaystyle f(x)-x-x-x-{2} across the real line. Named after the German mathematician Carl Friedrich Gauss, integral is ∫ - ∞ ∞ e x 2 d x π . Display-style int ---neft ifti (e'-x-{2}), dx'sqrt (pi) Abraham de Moyvre originally discovered this type of integral in 1733, while Gauss published the exact integral in 1809. Integral has a wide range of applications. For example, if variables change slightly, it is used to calculate the normalized constant of normal distribution. The same integral with the end limitations is closely related to both the error function and the cumulative distribution function of normal distribution. In physics, this type of integral often appears, for example, in quantum mechanics to find the density of the probability of the earth's state of the harmonic oscillator. This integral is also used in the way of integral formulation to find the propactor of the harmonic oscillator, and in statistical mechanics to find its section function. Although no elementary function exists for the error function, as can be proven by the Risch algorithm, the Gaussian integral can be solved analytically through multivariable calculus methods. That is, there is no elementary perpetual integral for ∫ e - x 2 d x, ∞ ∞ display style int e-x'{2},dx, but a certain integral ∫ {2} a certain integral element of the arbitrary Gaussian function is the ∫ th ∞ ∞ e a (x) 2 d x π a. «Дисплей стиль »int » - »infty» (e'-a(x'b) {2},dx'sqrt (фрак )pi s'a. Вычисление полярными координатами Стандартный способ вычисления госсианского интеграла, идея которого восходит к Пуассону, заключается в том, чтобы использовать свойство, которое: ( ∫ - ∞ ∞ е х 2 х х х ) 2 ∫ - ∞ ∞ ∞ ∞ ∫ ∞ ∞ е х 2 х х ∫ - ∞ ∞ й й 2 д й й й й й й ∫ «Дисплей стиль (слева)»-инфти »инфти»-х-х-{2} ,dx'right){2}{2}-int Infty e-yo {2}, di-int (x'{2}'y'{2}), dx, dy. Consider the function e q (x 2 y 2) - e - r 2 {2} {2} (display e'-(x'{2}'y'{2}) In two ways: on the one hand, by double integration in the Cartesian coordinate system, its integral part is the square: ( ∫ x 2 d x) 2 ; Display style (left) (int e-x'{2}, dxright) {2}; on the other hand, by integrating the shell (a case of double integration in polar coordinates), its integral is calculated to be π displaystyle pi Comparison of these two calculations gives integral, although you need to take care of the wrong integration of the integrate participation. ∫∫ R 2 e (x 2 q y 2) d x y ∫ 0 2 π ∫ 0 ∞ e - r 2 r d d θ 2 π ∫ 0 ∞ r e R 2 d r 2 π ∫ ∞ 0 1 2 e s ∞ π ∫ d's π ∞ π s {2}e- (x'{2}'y'y'{2} {2} {0} {0}) {0} 6pt-r'{2}2'2'pi (int)-intrafti ({0} tfrac {1}{2}'e's) , ds'r'{2}'6pt'{0} pi (e-e-e's) {0}-e'-infty) 6pt'pi , 'end'aligned' where the r factor is the Jacobian determinant, which appears due to the conversion to polar coordinates (r dr dθ is a standard measure on a plane, Expressed in the polar coordinates of Wikibooks:Calculus/Polar IntegrationGeneralizations), and replacement includes the adoption of s s , r2, so ds No 2r Dr. The combination of these crops (∫ - ∞ ∞ e - x 2 d x ) 2 x π, display style (left) - infty yenfti e-x-{2}, dx'right) - {2} Pi, so ∫ - ∞ ∞ e - x 2 x x π display style (int) - infty e-x-{2}, dx'sqrt. Full proof to justify incorrect double integrals and equating the two expressions, we start with the near-uximing function: I (a) - ∫ a e and x 2 d x . Displaystyle I(a)int (a-a-a-h-{2}'dx.) If a one-by-∫ - ∞ ∞ e - x 2 d x displaystyle (int)- infty,e'-x'{2},dx, were absolutely converged, we would say that its main value is Cauchy, that is, the limit of lim a → ∞ I (a) (a) display style lim to infty I (a) would coincide with ∫ and ∞ ∞ e x 2 d x. «Дисплейстайл »int»---инфти »нефти» (e'-x'{2},dx.) Чтобы увидеть, что это так, учти, что ∫ - ∞ ∞ e х 2 d x < ∫ - ∞ 1 х х х х 2 х х х х ∫ 1 е х х 2 х х ∫ 1 ∞ х х х 2 х < ∞. «Дисплейстайл »int»---инфти »ифти»-х-х{2},dx<-in {2}t х-инт -1{1}e'-x-{2}, dx'int ({1}'infty 'xe'-x'{2}, dx<'infty. так что мы можем вычислить ∫ - ∞ ∞ е х х х х дисплей стиль int-инфти infty-x'{2},dx, просто взяв предел lim a → ∞ I (a) Взяв квадрат I (a) (a) ,displaystyle I(a) дает I 2 (a ) ( ∫ - e q x 2 d x) ( ∫ - e q y y 2 d y) ∫ a a ( ∫ a e q y 2 d y ) e q x 2 d x ∫ - a ∫ a e q ( x 2 q y 2 ) d y d x . displaystylestart aligned I'{2} {2} (a) int q-a'e'y'y'y {2},dy'right) a'ae'y'y{2},dy'right),e-x'{2},dx's (x{2}'y'y {2}) , the above double integral can be seen as an area of ∫∫ - a, a ×, a, e q (x 2 q y 2) d (x, y) , iint-a,a'times, a'e(x'{2}'y'y'{2}),'d(x,y), took over the square with vertices (a, a) (a, a) (a, a) (a, a) (for, a) on a Xi plane. Since the exponential function is larger than 0 for all real numbers, it follows that the integral taken over the square's round should be smaller than I (a) 2 displaystyle I(a) {2}, and similarly {2} the integral taken over the square should be larger than i (a) 2 display I(a) display I(a) The integrals on the two discs can be easily calculated, Switching from The Cartesian coordinates to polar coordinates: x x x y-kos ⁡ θ th sin ⁡ θ display style beginning J (r, θ) - ∂ x ∂ g ∂ x ∂ θ ∂g ∂ mr ∂g ∂ θ ⁡ θ ⁡ θ ⁡ θ ⁡ θ (∂ θ ⁡ θ ⁡ θ ⁡ θ ⁡ θ (∂ θ ⁡ θ ⁡ θ ⁡ θ ⁡ θ r,'theta) beginning 'bmatrixdfrac (partial x'partial r'dfrac partial xpartial partial partial theta 1em dfrac (partial y' partial r'dfrac (partial y'partial theta endbmatrix'start'bmatrix'cos (h'bmatrix'cos Y J (g, θ) g (g, θ) y d (g, θ). Display style d(x,y) J (r,'theta) d(r,'theta) d'r, d'r, 'theta'. ∫ 0 2 π ∫ 0 e r 2 d d θ qlt; I 2 (a ) qlt; ∫ 0 2 π ∫ 0 2 r e and r 2 d d d θ . Display style int {0}2'pi int {0}a're'-r'{2}, dr', d'theta qlt;I'{2} (a)) qlt;-ent ({0})2'pi ({0})a-sqt {2}'re'r'{2}, dr.d'{2}. (See the polar coordinates from the Cartesian coordinates to help in the polar transformation.) Integration, π (1 - e - 2 ) i 2 (a ) zlt; π (1 - e q 2 a 2 ). Display style (1-e-a-{2}); I {2} (1-e-2a-{2}). The theorem of compression, it gives the Gaussian integral ∫ - ∞ ∞ e x 2 d x π. Display-style int ---neftifti (e'x-{2}), dx'sqrt (p) Pocart coordinates Another technique that goes back to Laplace (1812), is the next. Let the th x x s y y and x d s. Displaystyle beginning alignedy'xs'dy'x, ds. Since the limits on s as y → ±∞ depend on the X sign, it simplifies the calculation to use the fact that e'x2 is a pretty function, and therefore an integral part over all real numbers is only twice as inalienable from zero to infinity. A what ∫ - ∞ ∞ е х 2 х х 2 ∫ 0 ∞ е х 2 д х х . «Дисплей стиль »int» и «инфти» (e'-x-x-{2}),dx-2'int ({0} {2}) Таким образом, в диапазоне интеграции x ≥ 0, а переменные y и s имеют одинаковые пределы. Это дает: I 2 и 4 ∫ 0 ∞ ∫ 0 ∞ e q (x 2 и y 2 ) d y d x 4 ∫ 0 ∞ ( ∫ 0 ∞ e ) х 2 й й 2 ) г г г х х 4 ∫ 0 ∞ (∫ 0 ∞ е х 2 ( 1 х 2 ) х д с ) х х 4 ∫ 0 ∫ ∞ (∫ 0 ∞ е х х 2 ( 1 х 2) х х х х х х 4 ∫ 0 ∞ х 1 х 2 ( 1 х 2 ) х 2 ( 2 ) 1 с 2 ) х 0 х 0 х х й ∞ й й 4 ( 1 2 ∫ 0 ∞ d s 1 й 2 π ∞ ⁡ ) {\displaystyle {\begin{aligned}I^{2}&=4\int _{0}^{\infty }\int _{0}^{\infty }e^{-(x^{2}+y^{2})}dy\ ,dx\\[6pt]&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-(x^{2}+y^{2})}\,dy\right)\,dx\\ [6pt]&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-x^{2}(1+s^{2})}x\,ds\right)\,dx\\[6pt]&=4\int _{0}^{\infty }\left(\int _{0}^{\infty }e^{-x^{2}(1+s^{2})}x\,dx\right)\,ds\\[6pt]&=4\int _{0}^{\infty }\left[{\frac {1}{-2(1+s^{2})}}e^{-x^{2}(1+s^{2})}\right]_{x=0}^{x=\infty }\,ds\\[6pt]&=4\left({\frac {1}{2}}\int _{0}^{\infty }{\frac {ds} {1+s^{2}}}\right)\\[6pt]&=2{\Big [}\arctan s{\Big ]}_{0}^{\infty }\\[6pt]&=\pi .\end{aligned}}} Таким образом, π, как и ожидалось, «displaystyle I»sqrt «pi».