VBA code for integrating negative exponent - vba

Question

I am trying to perform an integration using Excel VBA. Was wondering how I would do this using an approximation method if I need to integrate a negative exponent (since x/0 is undefined, hence such approximations could not estimate it)
My current code is:
Function Integral(sExp As String, dMin As Double, dMax As Double, lBit As Long)
Dim dU As Double
Dim lU As Long
dU = (dMax - dMin) / lBit
For lU = 1 To lBit
IntegralTemp = IntegralTemp + Evaluate(Replace(sExp, "u", dMin)) * dU + 0.5 * dU * Abs(Evaluate(Replace(sExp, "u", dMin + dU)) - Evaluate(Replace(sExp, "u", dMin)))
dMin = dMin + dU
Next lU
Integral = IntegralTemp
End Function

Answer 1

Perhaps use a Monte Carlo technique to estimate the value?
Here's a wikipedia article for this approach:
http://en.wikipedia.org/wiki/Monte_Carlo_integration

Answer 2

I think this is problem with function Evaluate.
I try solve this and that has managed to make:
Look on this test function:
Function EvaluateTest(str As String)
EvaluateTest = Evaluate(str)
EvaluateTest = CStr(EvaluateTest)
End Function
debug.Print EvaluateTest("0^(-0.05)") show Error 2007 i think this is dividing by zero -> 0^(-0.05) for VBA is 0/(0^0.05)
We can catch this error and for thats case accept results = 0
I create function using to catch this 1 error and modified a bit your function. Try this code:
Function EvaluateCheck(Exp As String)
Dim EvalCheck As Variant
EvalCheck = Evaluate(Exp)
If VarType(EvalCheck) = vbError Then 'evaluate function error
Select Case CInt(EvalCheck)
Case 2007 ' 0/x
'Debug.Print "Evaluate(" & Exp & ") error= " & CStr(EvalCheck); ""
EvaluateCheck = 0
'Case 2015 ' other problems with evaluate (power)
'Debug.Print "Evaluate(" & Exp & ") error= " & CStr(EvalCheck); ""
'try use power function not evaluate
'FindTmp = WorksheetFunction.Find("^", Exp)
'If FindTmp > 0 Then
'number_ = CDbl(Mid(Exp, 1, FindTmp - 1))
'Power_ = Mid(Exp, FindTmp + 1, Len(Exp))
'Power_ = Replace(Power_, ".", ",") '<- i have problems with CDbl function i must replace . to , maybe you dont need this line
'Power_ = Replace(Replace(Power_, "(", vbNullString), ")", vbNullString) ' we dont need parentheses
'PowerDbl = CDbl(Power_)
'EvaluateCheck = WorksheetFunction.power(number_, PowerDbl)
'Else
'Debug.Print "Evaluate(" & Exp & ") error= " & CStr(EvalCheck); " i cant handle that case"
'End If
Case Else
Debug.Print "Evaluate(" & Exp & ") error= " & CStr(EvalCheck); " i cant handle that case"
End Select
Else ' evaluate no error
EvaluateCheck = EvalCheck
End If
End Function
Your function:
Function Integral(sExp As String, dMin As Double, dMax As Double, lBit As Long)
Dim dU As Double
Dim lU As Long
Dim eval As Long
Dim EvaluateVal1 As Double 'Evaluate(Replace(sExp, "u", dMin))
Dim EvaluateVal2 As Double 'Evaluate(Replace(sExp, "u", dMin + dU))
Dim sExpTmp As String
dU = (dMax - dMin) / lBit
For lU = 1 To lBit
'check evaluate
sExpTmp = Replace(sExp, "u", dMin)
EvaluateVal1 = EvaluateCheck(sExpTmp)
sExpTmp = Replace(sExp, "u", dMin + dU)
EvaluateVal2 = EvaluateCheck(sExpTmp)
IntegralTemp = IntegralTemp + EvaluateVal1 * dU + 0.5 * dU * Abs(EvaluateVal1 - EvaluateVal2)
dMin = dMin + dU
Next lU
Integral = IntegralTemp
End Function
Some results:
debug.print Integral("u^(-0.05)", 0, 1, 500)
1,05186339856455
debug.print Integral("u^(-0.05)", 0.05, 1, 500)
0,991802803730478
debug.print Integral("u^(-0.05)", 0.05, 1.05, 500)
1,0417624389399