The Integrate function in several HP calculators calculates the integral between two values, say A and B. Are there any tricks to use the same method when integrating between A and infinity, minus infinity and A, and minus infinity and plus infinity?

I know there are special types of Gaussian quadrature methods that can handle the above cases. My question is directed at working with the algorithms for finite integrals.

Namir

Quote:

The Integrate function in several HP calculators calculates the integral between two values, say A and B. Are there any tricks to use the same method when integrating between A and infinity, minus infinity and A, and minus infinity and plus infinity?
I know there are special types of Gaussian quadrature methods that can handle the above cases. My question is directed at working with the algorithms for finite integrals.

Namir

The usual trick is to first perform a simple change of variable which will reduce any interval, including infinities at one or both extremes, to any finite interval you care for such as [0,1] or [-1,1].

Best regards from V.

Many RF engineers would like the conversion below,

s = (z - 1) / (z + 1)

that is used to plot various impedance in a Smith chart.

Lyuka

looks like:

s = (x^2 - 1) / (x^2 + 1)

is a better transformation, since it is valid for all real values of x.

Namir

Hallo namir,

when you need only numerical results, you can also use instead of infinities great/small numbers like +/- 10^6.

That's sometimes tricky, shown in the advanced handbook for the 15c.

sincerely

peacecalc

*Edited: 10 Nov 2011, 11:59 p.m. *

That's the kind of tricks I was looking for.

Putting it in pseudo-code form in the case of integrating from A to infinity:

Given f(x), A, SmallValue, RelTolerance, and DiffTolerance

InfVal=10^6

IntegralVal1 = Integral of f(x) from A to InfVal

Do

InfVal = 10 * InfVal

IntegralVal2 = Integral of f(x) from A to InfVal

if |IntegralVa2| < SmallValue then

RelativeErr = 0

DiffErr = IntegralVa2 - IntegralVal1

else

RelativeErr = (IntegralVal2 - IntegralVal1) / IntegralVal1

DiffErr = 0

end

IntegralVal1 = IntegralVal2

Until |RelativeErr| < RelTolerance OR |DiffErr| < DiffTolerance

Integral = IntegralVal1

Here is a perhaps more efficient version that uses integration by parts:

Given f(x), A, SmallValue, RelTolerance, and DiffTolerance

B = 10^6 // or any other different high value (10^3, 10^4, 10^5, 10^6, 10^7, etc

IntegralVal0 = Integral of f(x) from A to B // should calculate most of the final answer

InfVal= 10 * B

IntegralVal1 = Integral of f(x) from B to InfVal

Do

InfVal = 10 * InfVal

IntegralVal2 = Integral of f(x) from A to InfVal

if |IntegralVa2| < SmallValue then

RelativeErr = 0

DiffErr = IntegralVa2 - IntegralVal1

else

RelativeErr = (IntegralVal2 - IntegralVal1) / IntegralVal1

DiffErr = 0

end

IntegralVal1 = IntegralVal2

Until |RelativeErr| < RelTolerance OR |DiffErr| < DiffTolerance

Integral = IntegralVal0 + IntegralVal1

*Edited: 11 Nov 2011, 5:47 a.m. *

IIRC, there is a discussion worth to read in the PPC-ROM manual.

Hope this hepls

Ciao.....Mike

Tanh-Sinh transformation or "Double exponential method" may also be efficient, especially when the function is oscillating.

There is a fast implementation in C (hpgcc2) for HP50G in my package INT1D at www.hpcalc.org.

As already pointed out, there are basically two approaches for handling improper integrals:

- Use a large (finite) value where the integrand becomes negligible compared to the final result.
Example: for an 8-digit result of the left tail Normal CDF a lower limit of -6 usually is sufficient as the PDF here is mererly 6E-9 and the whole integral from -infinity to -6 is less than 1E-9.

- Apply a variable transformation to make infinity "less infinite". ;-) This is the preferred method suggested in most of the other posts in this thread.
Example: Take a look at the 15C Advanced Functions Handbook, p. 54 ff. There, this topic is discussed with several examples on how to use the 15C's Integrate function for improper integrals. On p. 60-64 you will find a very nice and illustrative application of these ideas, evaluating the Normal CDF as well as the error function, with both plus or minus infinity as the limits, giving exact results even far out in the tails. This is accomplished by using the transformation u = exp(t²) as soon as the integration limit exceeds a certain value (here: 1,6).

Dieter