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    n1 Rd  0  c  i0 r   n2 c (1 n)

(2) 1 dzF(,) z  With the constants a ()  , here  means that we must integrate n  2i   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   xm1dx  (m)   2r  xm2r dx (3) 0 2 0 r1 (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 k1 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 n0  k value in the spirit of Zeta function regularization [ 4] so  n   R (k) , this is why n0 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

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)  Dn 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 aderivative of any order, there are mainly [6] 3 definitions

1 x 1  q  Dn f() x  dt( x t)n1 f() t dt Dq f( x ) lim ( 1)m f( x  (q m)) h (7)  h0 q    ()n 0 h m0 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  q1 2i  (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 b1/ 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  n1 (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 b1 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 n0 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.  h 

5  dx From Fourier analysis one can interpretate , the integral  as the following 0 x a  H() x  1  H( x a) convolution H* x1    dx  dx   dx with H(x) the ‘Heaviside  x a a x  x step function’ , using the property of the Fourier transform and the convolution theorem   1 H() x 1  i  iua H* x    dx    d u   ()u e  I() a (13)  x a 2i   u 

Here |x| is the absolute value function that takes the value x or –x depending on if ‘x’ is either positive or negative, solving Fourier transform (13) we can solve the logarithmic UV divergence. If we can solve (13) for a fixed a then for another value ‘b’ so b is  dx b   dx different from 0 or inifinite we have  log     I() a 0 x a a  0 x b

 1 Another simpler method is that if we have I() a   dx divergent integral , we 0 x a  1 1 differentiate with respect to ‘a’ I'( a )   dx   so integrating again with  2 0 (x a) a respect to ‘a’ I() a  log(a )  c() a  c1 , with c1 an universal divergent

 ( 1) 1 D I() a  dx ( 1) (  1) ( 1)   0 (14) a   1  0 (x a) a

    So I() a (  1) ( 1)Da a  (,)a  , with   0 and Da (,)a   0 , one of the problems is the apparent absurdity since due to the term 1 there is a complex contribution to an integral with real-valued integrand. An alternative formulation based on Hurwitz Zeta function is the following

   1  (0,a )  dxlog(x a)   dx and H  log (a ) log 2 a  0 (15) a 0 0 x a a

 For the sum lo g( n a)  s H (0,a ) , this is the Zeta-regularized definition for the n0 Determinant of an operator , a combination of the 2 expressions inside (15) gives the  1  ' regularized value for the Harmonic sum  dx   ()a  R (in case a=1 we get 0 x a  the Euler-Mascheroni constant) , this is the analogue result to simply using ‘Ramanujan  resummation’ for the series  ns s >0 , and s=-1 , ‘R’ stands for the Remainder term n0 1  B 2r1  1  inside Euler-Maclaurin summation formula and is R 2r    2r1   2a r1 (2r )! x  x a x0

6 Appendix: Convolution theorem

In this paper we have introduced and used the ‘Convolution theorem’ , if we have the convolution of 2 functions or distributions f(x) and g(x) defined as  h() z   dxf()( x g x z) then H() u  F()( u G u) with F, G and H the fourier  transform respectively of h(z) f(x) and g(x).

    Proof:= if we define  dzeiuz h() z   dzeiuz  dxf()( x g x z)  dxeiux f() x  F() u      and  dxeiux g() x  G() u , we make the change of variable y x  z so dx = -dz into  the first integral so we have the following identities

       dzeiuz  dxf()( x g x z)    dyeiu( x y)  dxf()() x g y    dyeiux  dxf()() x g y eiuy       (16)

The first integral on the left is just H(u) and using Fubini’s theorem to interchange the order of integration we have been able to proof that H() u  F()( u G u) .

  If we name g() x  H() x and f() x  xk , then  dxxk   H( x a) xk dx , hence a  applying convolution theorem and the Fourier transform

  k k  i  k iua  dxx  i  du  ()u   D  ()u e k > 0 (17) a   u  1 Unfortunately there are some oddities with defining product of distributions Dk  , x  Dk  ,   within distribution theory, however if the integral  f() x dx makes a sense as a Riemann integral , if we define F(u) as the Fourier transform of f(x) then the F(0) a Convolution theorem gives the regularized result    dtf() t  f() c for some 2 c 1 d| x | Real ‘c’ , for the case of f() x  with Fourier transform the integral is x dx 1 d| x | divergent in the Riemann sense but can be attached a finite value   log(a ) 2 dx x0 , since |x| is even its derivative will be odd so the mean value of the derivative near x=0

7  will be 0 and we have  x1 dx  log(a ) . The last method would be to use the Euler- a Maclaurin summation formula to get

 1  1 1  B d 2r1  1  dx    2r (18)    2r1   x0 0 x a n0 n a 2a r1 (2r )! dx  x a 

 1 The problem is that    H (1,a ) is still divergent , although using Ramanujan- n0 n a  1  summation we can attach this series the finite value    H (1,a )   ()a and n0 n a  plug this result into (17) , In both cases the approximation of the integral by a sum  1   dx   ()a and the result  log(a )  c are equivalent for a   (big   a n0 n a  0 x a a) since using the Stirling’s approximation for log (x ) and taking the derivative we get  '(a ) / (a ) the asymptotic result lim 1 a log(a )

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