Re: Computing my own function efficiently
- To: mathgroup@smc.vnet.net
- Subject: [mg11169] Re: Computing my own function efficiently
- From: "Tom Marchioro" <tlm@u.washington.edu>
- Date: Wed, 25 Feb 1998 03:31:43 -0500
- Organization: University of Washington
- References: <6cgg8n$sg1@smc.vnet.net> <6cteh7$lgs@smc.vnet.net>
Equivalently, the answer is easily derivable if you just take Cos[b x]=(E^I b x + E^-I b x)/2 and then use Jordan's Lemma on each resulting term (after scaling x-> x/b). such integrals arise frequently in physics, usually in the context of calculating a Green's function, and numerically they can be extremely slow to converge (no doubt what prompted his original question). The possibility of "extending" these integrals into the complex plane and getting closed form solutions remains one of the wonders of mathematics. If you *insist* on doing the integral numerically, the Cos transformation is still the best way, just follow it up by using Fourier[...] on the resulting terms. Hope this proves helpful --- Tom -- Dr. Thomas L. Marchioro II Departments of Physics and Chemistry University of Washington 206-323-9599 http://borg.chem.washington.edu/~tlm mike johnson wrote in message <6cteh7$lgs@smc.vnet.net>... >Tommy Nordgren wrote: >> >> I have a function that is defined by: f[k_,b_] := Integrate[ >> Cos[b x] Exp[-x^2]/(k^2+x^2),{x,-Infinity,Infinity}] Since Mathematica >> can't solve the integral... > >Gradshteyn & Ryzhik, Table of Integrals, Series & Products, 4th ed. Sec >3.954 gives the following algebraic answer (after some change in >notation) in terms of error functions: > >Pi^(3/2) Exp[k^2]/(4k)* > (2 Cosh[b k] - Exp[-b k] Erf[k-b/2] - Exp[bk] Erf[k+b/2]) > >Sometimes the old ways are still useful. -- >Michael A. Johnson, Mail Stop L-463 >Lawrence Livermore Nat'l Lab >7000 East Ave., P. O. Box 808 >Livermore, CA 94551 >Fax: (510) 422-6007 >mike-johnson@llnl.gov >