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Statement of the theorem

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{{Short description|Method which uses known Integrals to integrate derived functions}}
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In [[calculus]], '''integration by parametric derivatives''', also called '''parametric integration''',<ref name="Zatja1989">{{cite journal |last1=Zatja |first1=Aurel J. |title=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}}</ref> is a method which uses known [[Integrals]] to [[integral|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 that
 <math>\frac{d}{dt}\left(\int_a^b f(x,t)dx\right)=\int_a^b \frac{\partial}{\partial t} f(x,t)dx</math> 
It is also true for non-finite bounds.

==Examples==
===Example One: Exponential Integral===
For example, suppose we want to find the integral

: <math>\int_0^\infty x^2 e^{-3x} \, dx.</math>

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:

: <math>
\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}
</math>

This converges only for ''t'' > 0, which is true of the desired integral. Now that we know

: <math>\int_0^\infty e^{-tx} \, dx = \frac{1}{t},</math>

we can differentiate both sides twice with respect to ''t'' (not ''x'') in order to add the factor of ''x''2 in the original integral.

: <math>
\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}
</math>

This is the same form as the desired integral, where ''t'' = 3. Substituting that into the above equation gives the value:

: <math>\int_0^\infty x^2 e^{-3x} \, dx = \frac{2}{3^3} = \frac{2}{27}.</math>

===Example Two: Gaussian Integral===
Starting with the integral <math>\int^\infty_{-\infty} e^{-x^2t}dx=\frac{\sqrt\pi}{\sqrt t}</math>,
taking the derivative with respect to ''t'' on both sides yields
 <math>\begin{align}
&\frac{d}{dt}\int^\infty_{-\infty} e^{-x^2t}dx=\frac{d}{dt}\frac{\sqrt\pi}{\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}</math>. 
In general, taking the ''n''-th derivative with respect to ''t'' gives us
 <math>\int^\infty_{-\infty} x^{2n}e^{-x^2t}= \frac{(2n-1)!!\sqrt \pi}{2^n}t^{-\frac{2n+1}{2}}</math>.

===Example Three: A Polynomial===
Using the classical <math>\int x^t dx=\frac{x^{t+1}}{t+1}</math> and taking the derivative with respect to ''t'' we get
 <math>\int \ln(x)x^t= \frac{\ln(x)x^{t+1}}{t+1} - \frac{x^{t+1}}{(t+1)^2}</math>.

===Example Four: Sums===
The method can also be applied to sums, as exemplified below.
 Use the [[Weierstrass factorization theorem|Weierstrass factorization]] of the [[Hyperbolic sine|sinh]] function:
 <math>\frac{\sinh (z)}{z}=\prod_{n=1}^\infty \left(\frac{\pi^2 n^2 + z^2}{\pi^2 n^2}\right)</math>.
 Take the logarithm:
 <math>\ln(\sinh (z)) - \ln(z)=\sum_{n=1}^\infty \ln\left(\frac{\pi^2 n^2 + z^2}{\pi^2 n^2}\right)</math>.
 Derive with respect to ''z'':
 <math>\coth(z) - \frac{1}{z}= \sum^\infty_{n=1}\frac{2z}{z^2+\pi^2n^2}</math>.
 Let <math>w=\frac{z}{\pi}</math>:
 <math>\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}</math>.

==References==
{{reflist}}

==External links==
[https://en.wikibooks.org/wiki/Calculus/Parametric_Integration WikiBooks: Parametric_Integration]

[[Category:Integral calculus]]

{{mathanalysis-stub}}

Integration using parametric derivatives - Revision history

imported>Hpecora1: /* Statement of the theorem */