I encountered some days before an integral representation for a modified Bessel function and should differentiate it. But in this representation :

$$K(\omega,a)=\int_0^{\infty} \frac{ds}{s} e^{-i\omega \ln{s} + ia(s-\frac{1}{s})},$$

it doesn't seem possible to define a proper differentiation with respect to $a$ without distribution theory. However in this representation:

$$K(\alpha,x) = \int_0^{\infty} \cosh(\alpha t)e^{-x\cosh{t}}\,dt $$

we can differentiate as much as we want with respect to $x$, and we can relate the two representations with the following steps :

  • $\alpha \rightarrow i\alpha $
  • expand the $\cosh$, and make the integral on $t$ goes to all reals
  • define $y=e^t$
  • rotate the "contour" : $y\rightarrow iy$

and except a multiplicative factor (depending on $\alpha$) we're done.

But I was wondering where the problems are gone ? How by changing our variables, etc. (not ill-defined operations in this case, at least it doesn't seem to me where they could be so) we could in some sense "regularize" our function ?

I think this is related to the others scheme of regularization, renormalization we encounter in quantum field theory, where by redefinition, differentiation and integration, analytic continuation, etc., we could better define some functions (usually absorbing infinities).

Nevertheless, where are the problems hiding (in my particular case or in the others regularization schemes)? Why does this work? Does that mean that the problems were not really there, or that we just put them in some parenthesis and concentrate on the non-divergent part (I vote for the last option in the case of renormalization with the "effective" variables)? If it is the last case for the modified Bessel function, where are these parenthesis ?

I would appreciate if someone could help me see better this problem.

  • $\begingroup$ should it be $a\to i \alpha$? $\endgroup$ – innisfree Jun 3 '15 at 14:15
  • $\begingroup$ Actually, $\alpha$ will be $\omega$ and $x$ will be $a$, sorry for the bad chosen notation... $\endgroup$ – faero Jun 3 '15 at 15:40
  • $\begingroup$ I dont think you can differentiate the second expression wrt $x$ under the integral. I dont see any uniform integrable bound in $x$, and I dont even think there is one. Are you sure you can just differentiate that guy? $\endgroup$ – Daniel Jun 9 '15 at 12:19

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