Selection (relational algebra): Difference between revisions

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In the selection clause the not equal sign is also allowed, a reference is in " Database Sysem Concepts" byAbraham Silberschatz Yale University Henry F. Korth Lehigh University S. Sudarshan Indian Institute of Technology, Bombay in the sixth edition P.218
 
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'''Feynman parametrization''' is a technique for evaluating [[loop integral]]s which arise from [[Feynman diagram]]s with one or more loops. However, it is sometimes useful in integration in areas of [[pure mathematics]] as well.
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[[Richard Feynman]] observed that:
 
:<math>\frac{1}{AB}=\int^1_0 \frac{du}{\left[uA +(1-u)B\right]^2}</math>
 
which simplifies evaluating integrals like:
 
:<math>\int \frac{dp}{A(p)B(p)}=\int dp \int^1_0 \frac{du}{\left[uA(p)+(1-u)B(p)\right]^2}=\int^1_0 du \int \frac{dp}{\left[uA(p)+(1-u)B(p)\right]^2}.</math>
 
More generally, using the [[Dirac delta function]]:
 
:<math>\frac{1}{A_1\cdots A_n}=(n-1)!\int^1_0 du_1 \cdots \int^1_0 du_n \frac{\delta(u_1+\dots+u_n-1)}{\left[u_1 A_1+\dots +u_n A_n\right]^n}.</math>
 
Even more generally, provided that <math> \text{Re} ( \alpha_{j} ) > 0  </math> for all <math> 1 \leq j \leq n </math>:
 
:<math>\frac{1}{A_{1}^{\alpha_{1}}\cdots A_{n}^{\alpha_{n}}}=\frac{\Gamma(\alpha_{1}+\dots+\alpha_{n})}{\Gamma(\alpha_{1})\cdots\Gamma(\alpha_{n})}\int_{0}^{1}du_{1}\cdots\int_{0}^{1}du_{n}\frac{\delta(\sum_{k=1}^{n}u_{k}-1)u_{1}^{\alpha_{1}-1}\cdots u_{n}^{\alpha_{n}-1}}{\left[u_{1}A_{1}+\cdots+u_{n}A_{n}\right]^{\sum_{k=1}^{n}\alpha_{k}}}
.</math> <ref>
{{cite web
| author= Kristjan Kannike
| title= Notes on Feynman Parametrization and the Dirac Delta Function
| url= http://www.physic.ut.ee/~kkannike/english/science/physics/notes/feynman_param.pdf
| archiveurl= http://web.archive.org/web/20070729015208/http://www.physic.ut.ee/~kkannike/english/science/physics/notes/feynman_param.pdf
|archivedate=2007-07-29
| work=
| publisher=
| date=
| accessdate=2011-07-24
}} </ref>
 
See also [[Schwinger parametrization]].
 
==Derivation==
:<math>\frac{1}{AB} = \frac{1}{A-B}\left(\frac{1}{B}-\frac{1}{A}\right)=\frac{1}{A-B}\int_B^A \frac{dz}{z^2}.</math>
Now just linearly transform the integral using the substitution,
:<math>u=(z-B)/(A-B)</math> which leads to <math>du = dz/(A-B)</math> so <math>z = uA + (1-u)B</math>
and we get the desired result:
:<math>\frac{1}{AB} = \int_0^1 \frac{du}{\left[uA + (1-u)B\right]^2}.</math>
 
==References==
{{reflist}}
 
[[Category:Quantum field theory]]
 
 
{{applied-math-stub}}
{{quantum-stub}}

Latest revision as of 15:55, 6 August 2014

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