Top View Method which uses known Integrals to integrate derived functions
In calculus , integration by parametric derivatives , also called parametric integration , is a method which uses known Integrals to integrate derived functions. It is often used in Physics, and is similar to integration by substitution .
Statement of the theorem [ edit ] By using the Leibniz integral rule with the upper and lower bounds fixed we get that d d t ( ∫ a b f ( x , t ) d x ) = ∫ a b ∂ ∂ t f ( x , t ) d x {\displaystyle {\frac {d}{dt}}\left(\int _{a}^{b}f(x,t)dx\right)=\int _{a}^{b}{\frac {\partial }{\partial t}}f(x,t)dx}
Example One: Exponential Integral [ edit ] For example, suppose we want to find the integral
∫ 0 ∞ x 2 e − 3 x d x . {\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}\,dx.} Since this is a product of two functions that are simple to integrate separately, repeated integration by parts is certainly one way to evaluate it. However, we may also evaluate this by starting with a simpler integral and an added parameter, which in this case is t = 3:
∫ 0 ∞ e − t x d x = [ e − t x − t ] 0 ∞ = ( lim x → ∞ e − t x − t ) − ( e − t 0 − t ) = 0 − ( 1 − t ) = 1 t . {\displaystyle {\begin{aligned}&\int _{0}^{\infty }e^{-tx}\,dx=\left[{\frac {e^{-tx}}{-t}}\right]_{0}^{\infty }=\left(\lim _{x\to \infty }{\frac {e^{-tx}}{-t}}\right)-\left({\frac {e^{-t0}}{-t}}\right)\\&=0-\left({\frac {1}{-t}}\right)={\frac {1}{t}}.\end{aligned}}} This converges only for t > 0, which is true of the desired integral. Now that we know
∫ 0 ∞ e − t x d x = 1 t , {\displaystyle \int _{0}^{\infty }e^{-tx}\,dx={\frac {1}{t}},} we can differentiate both sides twice with respect to t (not x ) in order to add the factor of x in the original integral.
d 2 d t 2 ∫ 0 ∞ e − t x d x = d 2 d t 2 1 t ∫ 0 ∞ d 2 d t 2 e − t x d x = d 2 d t 2 1 t ∫ 0 ∞ d d t ( − x e − t x ) d x = d d t ( − 1 t 2 ) ∫ 0 ∞ x 2 e − t x d x = 2 t 3 . {\displaystyle {\begin{aligned}&{\frac {d^{2}}{dt^{2}}}\int _{0}^{\infty }e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d^{2}}{dt^{2}}}e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d}{dt}}\left(-xe^{-tx}\right)\,dx={\frac {d}{dt}}\left(-{\frac {1}{t^{2}}}\right)\\[10pt]&\int _{0}^{\infty }x^{2}e^{-tx}\,dx={\frac {2}{t^{3}}}.\end{aligned}}} This is the same form as the desired integral, where t = 3. Substituting that into the above equation gives the value:
∫ 0 ∞ x 2 e − 3 x d x = 2 3 3 = 2 27 . {\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}\,dx={\frac {2}{3^{3}}}={\frac {2}{27}}.} Example Two: Gaussian Integral [ edit ] Starting with the integral ∫ − ∞ ∞ e − x 2 t d x = π t {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {\sqrt {\pi }}{\sqrt {t}}}} t on both sides yields d d t ∫ − ∞ ∞ e − x 2 t d x = d d t π t − ∫ − ∞ ∞ x 2 e − x 2 t = − π 2 t − 3 2 ∫ − ∞ ∞ x 2 e − x 2 t = π 2 t − 3 2 {\displaystyle {\begin{aligned}&{\frac {d}{dt}}\int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {d}{dt}}{\frac {\sqrt {\pi }}{\sqrt {t}}}\\&-\int _{-\infty }^{\infty }x^{2}e^{-x^{2}t}=-{\frac {\sqrt {\pi }}{2}}t^{-{\frac {3}{2}}}\\&\int _{-\infty }^{\infty }x^{2}e^{-x^{2}t}={\frac {\sqrt {\pi }}{2}}t^{-{\frac {3}{2}}}\end{aligned}}} n -th derivative with respect to t gives us ∫ − ∞ ∞ x 2 n e − x 2 t = ( 2 n − 1 ) ! ! π 2 n t − 2 n + 1 2 {\displaystyle \int _{-\infty }^{\infty }x^{2n}e^{-x^{2}t}={\frac {(2n-1)!!{\sqrt {\pi }}}{2^{n}}}t^{-{\frac {2n+1}{2}}}}
Example Three: A Polynomial [ edit ] Using the classical ∫ x t d x = x t + 1 t + 1 {\displaystyle \int x^{t}dx={\frac {x^{t+1}}{t+1}}} t we get ∫ ln ( x ) x t = ln ( x ) x t + 1 t + 1 − x t + 1 ( t + 1 ) 2 {\displaystyle \int \ln(x)x^{t}={\frac {\ln(x)x^{t+1}}{t+1}}-{\frac {x^{t+1}}{(t+1)^{2}}}}
The method can also be applied to sums, as exemplified below. Weierstrass factorization of the sinh function: sinh ( z ) z = ∏ n = 1 ∞ ( π 2 n 2 + z 2 π 2 n 2 ) {\displaystyle {\frac {\sinh(z)}{z}}=\prod _{n=1}^{\infty }\left({\frac {\pi ^{2}n^{2}+z^{2}}{\pi ^{2}n^{2}}}\right)} ln ( sinh ( z ) ) − ln ( z ) = ∑ n = 1 ∞ ln ( π 2 n 2 + z 2 π 2 n 2 ) {\displaystyle \ln(\sinh(z))-\ln(z)=\sum _{n=1}^{\infty }\ln \left({\frac {\pi ^{2}n^{2}+z^{2}}{\pi ^{2}n^{2}}}\right)} z : coth ( z ) − 1 z = ∑ n = 1 ∞ 2 z z 2 + π 2 n 2 {\displaystyle \coth(z)-{\frac {1}{z}}=\sum _{n=1}^{\infty }{\frac {2z}{z^{2}+\pi ^{2}n^{2}}}} w = z π {\displaystyle w={\frac {z}{\pi }}} 1 2 coth ( π w ) π w − 1 2 1 z 2 = ∑ n = 1 ∞ 1 n 2 + w 2 {\displaystyle {\frac {1}{2}}{\frac {\coth(\pi w)}{\pi w}}-{\frac {1}{2}}{\frac {1}{z^{2}}}=\sum _{n=1}^{\infty }{\frac {1}{n^{2}+w^{2}}}}
WikiBooks: Parametric_Integration
From Wikipedia the free encyclopedia
![{\displaystyle {\begin{aligned}&\int _{0}^{\infty }e^{-tx}\,dx=\left[{\frac {e^{-tx}}{-t}}\right]_{0}^{\infty }=\left(\lim _{x\to \infty }{\frac {e^{-tx}}{-t}}\right)-\left({\frac {e^{-t0}}{-t}}\right)\\&=0-\left({\frac {1}{-t}}\right)={\frac {1}{t}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60fed983dfedc8f42fc574bb09526dd1946d287e)
![{\displaystyle {\begin{aligned}&{\frac {d^{2}}{dt^{2}}}\int _{0}^{\infty }e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d^{2}}{dt^{2}}}e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d}{dt}}\left(-xe^{-tx}\right)\,dx={\frac {d}{dt}}\left(-{\frac {1}{t^{2}}}\right)\\[10pt]&\int _{0}^{\infty }x^{2}e^{-tx}\,dx={\frac {2}{t^{3}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/893ef2820b42fbc3d670042c03842a8445e3814d)
