THE THEORY OF DISTRIBUTIONS APPLIED TO c DIVERGENT INTEGRALS OF THE FORM dx()x u a (x b) Jose Javier Garcia Moreta Graduate student of Physics at the UPV/EHU (University of Basque country) In Solid State Physics Address: Address: Practicantes Adan y Grijalba 2 5 G P.O 644 48920 Portugalete Vizcaya (Spain) Phone: (00) 34 685 77 16 53 E-mail: [email protected] MSC: 60E05, 62E99, 35Q40 ABSTRACT: In this paper we review some results on the regularization of divergent integrals of c ()x dx dx the form and in the context of distribution theory a x b a x a Keywords: Regularization, distribution theory, fractional calculus Regularization of divergent integrals: In QFT (Quantum field theroy) there are mainly two kinds of divergent integral , the UV (ultraviolet) divergence, that happens whenever an integral is divergent when x and the IR (infrared) one that occurs if the integral is divergent as x 0 ,a few examples of these integrals are dxx3 dxx2 dx dx a R or a Z (1) 2 2 3 0 x a 0 (x a) 0 x( x a) 0 x The first two integrals have an UV divergence, whereas the last ones have an IR divergence,the names IR and UV divergences come from the fact that if we use the h Wave-particle Duality in QM setting x=p we find that UV divergence is | p | 1 equivalent to have small wavelengths (ulra violet) and the IR divergence happens for big wavelengths (infrared) so the name is not casual, for more details [5] and [9] To deal with UV divergences on an integral of the form dkd F() k on Rd we simply Rd make a change of variable to d- dimensional polar coordinates , to rewrite the integral c k a () a dkd F() k d drr d 1F(,) r d dr a () ri 1 d n () i n1 Rd 0 c i0 r n2 c (1 n) (2) 1 dzF(,) z With the constants a () , here means that we must integrate n 2i z 1 over the angular variables , in case F is invariant under rotations on the d-dimensional 2 d / 2 plane the angular part of the integral can be calculated exactly as d , the (d / 2) idea of expanding our function into a Laurent series , is to isolate the UV divergencies of the form drr k so they can be ‘cured’ using the zeta regularization algorithm. 0 Choosing any c >1 , since the integral has a UV divergence , after expanding the integrand F(,) r r d 1 into a convergent Laurent series for |r| >1 there will be only a finite number of divergent integrals of the form drr k plus a logarithmic divergent 0 dr integral (this is another example of UV divergence) , now to get finite results 0 r c from the divergent integrals we will apply the recurrence deduced in our previous paper [4 ] when discussing the zeta regularization applied to integrals m B m!( m 2r 1) xm dx xm1dx (m) 2r xm2r dx (3) 0 2 0 r1 (2r )!(m 2r 1)! 0 Equation (3) is a ‘regularization’ in the sense that if we had the upper limit N-1 instead N N k1 of the expression (3) would give the well-known result dxxk k >0 or k=0 , 0 k 1 for the case dxxk , the series nk is no longer convergent and must be given a finite 0 n0 k value in the spirit of Zeta function regularization [ 4] so n R (k) , this is why n0 the term R (m) (m) apperas inside (3) , note that (3) is a recurrence formula with initial term dx 1 1 1 1 1 1 (0) 1/ 2 and allows to calculate or assign a 0 2 finite value to every divergent integral of the form drr k for ‘k’ a positive integer , (the 0 main problem for the logarithmic divergence is the pole of the Zeta funciton at s=1) the first terms are I(0, ) (0) 1/ 2 dx 0 I(0, ) I(1, ) ( 1) xdx 2 0 (4) 1 B I(2, ) I(0, ) ( 1) 2 a I(0, ) x2 dx 21 2 2 0 3 1 B I(3, ) I(0, ) ( 1) 2 a I(0, ) ( 3) B a I(0, ) x3 dx 21 2 31 2 2 2 0 k Notation I(,) m stands for lim dxx , being ‘Lambda’ a cut-off with certain 0 physical meaning. This example is valid for UV divergences but what would happen to the case of c dx dr ‘singular integrals’ a <b<c or to the logarithmic divergence ? a x b 0 r c o Regularization of divergent integrals using Distribution theory: As a first example, let us suppose we want to calculate the integral dxf() x Dn (x a) , for n=0 we know that in spite of the ‘pole’ of the delta function at x=a we have that dxf()( x x a) f() a for any test function f(x) , then we can use two method We perform integration by parts avoiding the singular point at x=a so dx( D)n f()( x x a) dxf() x Dn (x a)( D)n f() a (5) We integrate n-times (no matter if n is non-integer) with respect to ‘a’ I() a dxf() x Dn (x a) Dn I() a dxf( x )( 1)n (x a) ( 1)n f() a (6) a 3 n n Both methods are equivalent since if we recall the identity Da Da I() a I() a we yield to the same result, no matter if we integrate/differentiate with respect to the parameter ‘a’ or if we integrate by parts. c dx dx A similar idea can be applied to integrals of the form and first we define a x b a x a (x b)s for a<x<c the distribution T(,) b x , for a regular enough function ()x we 0 otherwise can consider (in the sense of distribution) our divergent integral to be the linear c dx operatorT[] ()x for any real ‘s’ , the idea is to perform a differ-integration s a (x b) -times so s 1/ 2 . First we must introduce the concept of differ-integral or fractional derivative of any order, there are mainly [6] 3 definitions 1 x 1 q Dn f() x dt( x t)n1 f() t dt Dq f( x ) lim ( 1)m f( x (q m)) h (7) h0 q ()n 0 h m0 m The first definition inside (7) is the expression for fractional integral (not valid for negative ‘n’ ) , the second one is the ‘Grunwald-Letnikov’ differintegral valid for positive ‘q’ , the third alternative for the derivative comes from the definition of (q 1) dz Cauchy’s integral formula Dq f() x f() z for any rectificable curve q1 2i (z x) on the complex plane that includes the point z=x c dx Example: Let be the singular integral I() b ()x , in order to give it a finite s a (x b) value first we differentiate -times with respect to ‘b’ so s 1/ 2 hence (1 s) c dx D I() b ()x , we make then the change of variable x b u 2 so b (1 s ) a x b 2 (1 s) c b our integral becomes D I() b du( b u2 ) , then we define the b (1 s ) a b dF function F so (b u2 ) this implies du 2 (1 s) D I() b F( c b) F( a b) finally taking the inverse operator we b (1 s ) 2 (1 s) can set I() b DF ( c b) DF ( a b) (,)b (8) (1 s ) b b d f Here ‘mu’ is a real number and D f exists in the sense of a fractional b dx derivative/integral and Db (b , ) 0 . We set the condition s 1/ 2 because with a change of variable we can avoid the pole x b1/ 2 at x=b. 4 The same strategy can be applied to singular integral equation, for example if we wished to solve the following equation dt (1 n) dt f() x g() x f() t so D f() x D g() x f() t (9) n x x a t x (1 n ) a t x Where n 1/ 2 , making the change of variable t x u 2 so (9) becomes 2 (1 n) D f() x D g() x du f( x u2 ) Gamma function (10) x x (1 n ) a x Now, equation (10) has NO poles at t=x , to solve this integral equation without singularity we could use an iterative process 2 (1 n) f() x g() x D f() x D g() x du f (x u2 ) (11) 0 x n x n n1 (1 n ) a x Where we have supposed that Dx (x , ) 0 (,)x in order to simplify the calculations of the solution of the integral. Then one could ask, what happens in the limit b , in this case the quantity x b1 becomes zero for every x , so the I(b) must tend to 0 for ‘b’ big hence we should choose the function (,)b with the conditions Db (b , ) 0 and 2 (1 s) lim DFb ( c b) DFb ( a b) (,)b 0 (12) b (1 s ) o Logarithmic divergences The case of the logarithmic integral is a bit different, since formula (4) can not handle it ,due to the pole of zeta function at s=1, one of the ideas to apply [4] is just to replace the dx 1 divergent integral by a divergent series with ‘h’ being an step (we 0 x a n0 n a/ h use Rectangle method ) ,this series is still divergent but can be assigned a finite value via ‘Ramanujan resummation’ equal to the logarithmic derivative of Gamma function ' a ,however this kind of method depends on the value of the step ‘h’ given.