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double precision function exprjh(x)
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C$Id: integ.F,v 1.1 2005-03-08 23:58:03 d3g293 Exp $
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c dumb solution to underflow problems on sun
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implicit double precision (a-h,o-z)
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common/values/fm(2001,5),rdelta,delta,delo2
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dimension t(2001),et(2001)
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c initalize common block for computation of f0 by recursion down
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fm(ii,maxm+1)=(et(ii)+t(ii)*fm(ii,maxm+1))*rr
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fm(ii,i)=(et(ii)+t(ii)*fm(ii,i+1))*rr
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subroutine f0(value, t)
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implicit real*8 (a-h,o-z)
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common/values/fm(2001,5),rdelta,delta,delo2
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parameter(fac0=0.88622692545276d0,
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$ rhalf=0.5d0,rthird=0.3333333333333333d0,rquart=0.25d0)
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c computes f0 to a relative accuracy of better than 4.e-13 for all t.
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c uses 4th order taylor expansion on grid out to t=28.0
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c asymptotic expansion accurate for t greater than 28
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value = fac0 / sqrt(t)
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n = idint((t+delo2)*rdelta)
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value = fm(n,1)+x*(fm(n,2)+rhalf*x*(fm(n,3)+
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$ rthird*x*(fm(n,4)+rquart*x*fm(n,5))))
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subroutine addin(g, i, j, k, l, fock, dens, iky)
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implicit double precision (a-h, o-z)
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dimension fock(*), dens(*), iky(*)
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c add (ij|kl) into the fock matrix
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jk = iky(max(j,k)) + min(j,k)
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jl = iky(max(j,l)) + min(j,l)
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aij = g4*dens(kl)+fock(ij)
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fock(kl) = g4*dens(ij)+fock(kl)
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if(i.eq.k.or.j.eq.l) gg = g2
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ajk = fock(jk) - gil*dens(il)
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ail = fock(il) - gil*dens(jk)
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aik = fock(ik) - gg*dens(jl)
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fock(jl) = fock(jl) - gg*dens(ik)
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subroutine dfill(n,val,a,ia)
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implicit real*8 (a-h,o-z)
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c initialise double precision array to scalar value
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do 20 i = 1,(n-1)*ia+1,ia