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obscure bug using FindRoot[ Integrate[...

Mike Luntz
Posted 11 years ago
POSTED BY: Mike Luntz
4 Replies
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Mike Luntz
Mike Luntz
Posted 11 years ago

Many thanks to all who replied. There were a number of great suggestions in the replies.
POSTED BY: Mike Luntz

At one of the technologies conferences, Stephen Wolfram talked about programming style, pointing out that breaking up code into small pieces can make it much easier to work with and debug. I think this problem is an example of that. Defining the integral as a function and testing that would have made the problem more easily understandable.
POSTED BY: Frank Kampas

Hello, Adding to these comments, this also illustrates those cases where NIntegrate needs to have a function with numeric argument:

expr[z_?NumericQ] := Integrate[f[x], {x, z, \[Infinity]}]
FindRoot[expr[z] == 0.5, {z, 1}]

However, if this was the problem that the original poster had in mind, then why not?

g = Integrate[DiracDelta[x] + Exp[-x], {x, z, \[Infinity]}, Assumptions -> Element[z, Reals]]
FindRoot[g == 0.5, {z, 1}

This avoids repeated calls to Integrate.
POSTED BY: W. Craig Carter

Not really a bug, notice Integrate returns a ConditionalExpression, and it is documented that ConditionalExpression propagates, i.e. ConditionalExpression[g[z], ...] == 0.5 becomes ConditionalExpression[g[z] == 0.5, ...] which returns False when z is 1 and of course False is not a number as the message says.

There are many ways around this, one could use Normal to remove the ConditionalExpression, or use Assumptions:

FindRoot[Integrate[f[x], {x, z, Infinity}, Assumptions -> Element[z, Reals]] == 0.5, {z, 1}]
POSTED BY: Ilian Gachevski
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