Infinite region
The method can be combined with the Laurent transform Laurent.htm with the variable changes. for\BLI2.FOR
Eqn 13
Immediately this raises the question of what about t = 0.
FUNCTION FTEST(T)
IMPLICIT REAL*8
(A-H,O-Z)
T=MAX(1D-2,T)
IF(T.EQ.1)THEN
FTEST=0
RETURN
ENDIF
X=T/(1-T)
FTEST=1/X**12-1/X**6
RETURN
END
Figure 1 The Lennard Jones potential with a Laurent Transform.
Note
that while the above gives a visual view of the function,
that the integral in x from 0 to infinity is
Eqn 14
Since the regions are not uniform, this is
The changes are all in the function
FUNCTION FTEST(T)
IMPLICIT REAL*8
(A-H,O-Z)
T=MAX(1D-2,T)
IF(T.EQ.1)THEN
FTEST=0
RETURN
ENDIF
X=T/(1-T)
FTEST=1/X**12-1/X**6
FTEST=FTEST*(1+X)**2
RETURN
END
Figure 2 Integrand in integral of LJ potential. Note that integral is approximately infinite.
A more reasonable integrand is
IMPLICIT REAL*8
(A-H,O-Z)
DIMENSION XI(2001),FI(2001),ERR(2001)
EXTERNAL FTEST
DATA
AM/.7D0/,A/3.7D0/
B=0
E=1
NP=128
CALL BLI(XI,FI,ERR,B,E,NP,FTEST)
OPEN(1,FILE='BLI.OUT')
WRITE(1,'(2G12.4)')(XI(I),FI(I),I=1,NP)
ANUM=0
DO I=1,NP-1
ANUM=ANUM+(FI(I)+FI(I+1))*(XI(I+1)-XI(I))
ENDDO
ANUM=.5D0*ANUM
PRINT*,' ANUM ',ANUM
ANAL=ATAN(AM/A)
PRINT*,' ANAL ',ANAL
STOP
END
FUNCTION FTEST(T)
IMPLICIT REAL*8
(A-H,O-Z)
DATA
AM/.7D0/,A/3.7D0/
T=MAX(1D-12,T)
IF(T.EQ.1)THEN
FTEST=0
RETURN
ENDIF
X=T/(1-T)
FTEST=SIN(AM*X)*EXP(-A*X)/X
FTEST=FTEST*(1+X)**2
RETURN
END
RUN
ANUM 0.187003938004170
ANAL 0.186979269918911
The
function BLI is a linear version of findfun described in the Class2000 notes.
Assignment
- Over the range 0 to infinity
for a > 0