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{{Short description|Method which uses known Integrals to integrate derived functions}}{{multipleissues|{{refimprove|date=July 2019}}{{context|date=June 2019}}}}In calculus, integration by parametric derivatives, also called parametric integration,JOURNAL, Zatja, Aurel J., Parametric Integration Techniques {{!, Mathematical Association of America |website=www.maa.org |date=December 1989 |volume=Mathematics Magazine |url=https://www.maa.org/sites/default/files/268948443847.pdf |access-date=23 July 2019}} 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

By using The Leibniz integral rule with the upper and lower bounds fixed we get thatfrac{d}{dt}left(int_a^b f(x,t)dxright)=int_a^b frac{partial}{partial t} f(x,t)dxIt is also true for non-finite bounds.

Examples

Example One: Exponential Integral

For example, suppose we want to find the integral 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: begin{align}& 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{align}This converges only for t > 0, which is true of the desired integral. Now that we know we can differentiate both sides twice with respect to t (not x) in order to add the factor of x2 in the original integral. begin{align}& 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 (-x e^{-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{align}This is the same form as the desired integral, where t = 3. Substituting that into the above equation gives the value:

Example Two: Gaussian Integral

Starting with the integral int^infty_{-infty} e^{-x^2t}dx=frac{sqrtpi}{sqrt t},taking the derivative with respect to t on both sides yieldsbegin{align}&frac{d}{dt}int^infty_{-infty} e^{-x^2t}dx=frac{d}{dt}frac{sqrtpi}{sqrt t}&-int^infty_{-infty} x^2 e^{-x^2t} = -frac{sqrt pi}{2}t^{-frac{3}{2}}&int^infty_{-infty} x^2e^{-x^2t}= frac{sqrt{pi}}{2}t^{-frac{3}{2}}end{align}.In general, taking the n-th derivative with respect to t gives usint^infty_{-infty} x^{2n}e^{-x^2t}= frac{(2n-1)!!sqrt pi}{2^n}t^{-frac{2n+1}{2}}.

Example Three: A Polynomial

Using the classical int x^t dx=frac{x^{t+1}}{t+1} and taking the derivative with respect to t we getint ln(x)x^t= frac{ln(x)x^{t+1}}{t+1} - frac{x^{t+1}}{(t+1)^2}.

Example Four: Sums

The method can also be applied to sums, as exemplified below. Use the Weierstrass factorization of the sinh function:frac{sinh (z)}{z}=prod_{n=1}^infty left(frac{pi^2 n^2 + z^2}{pi^2 n^2}right).Take the logarithm:ln(sinh (z)) - ln(z)=sum_{n=1}^infty lnleft(frac{pi^2 n^2 + z^2}{pi^2 n^2}right).Derive with respect to z:coth(z) - frac{1}{z}= sum^infty_{n=1}frac{2z}{z^2+pi^2n^2}.Let w=frac{z}{pi}:frac{1}{2}frac{coth(pi w)}{pi w} - frac{1}{2}frac{1}{z^2}=sum^infty_{n=1}frac{1}{n^2+w^2}.

References

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External links

WikiBooks: Parametric_Integration{{mathanalysis-stub}}