obscure bug using FindRoot[ Integrate[...

Ideas
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1

There is a minor bug in Mathematica 10.0.2 when using FindRoot to find a limit of integration yielding a given integral value when the integrand contains a Dirac delta function. In this case the syntax FindRoot[f[x]==a,{x,x0}] does not work but FindRoot[f[x]-a,{x,x0}] does. The code below is an illustration of the bug.

Clear["Global`*"]
f[x_] = DiracDelta[x] + Exp[-x];

t = FindRoot[Integrate[f[x], {x, z, \[Infinity]}] == .5, {z, 1}]

FindRoot::nlnum: The function value {False} is not a list of numbers with dimensions {1} at {z} = {1.}. >>

FindRoot::nlnum: The function value {False} is not a list of numbers with dimensions {1} at {z} = {1.}. >>

FindRoot[\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(z\), \(\[Infinity]\)]\(f[
     x] \[DifferentialD]x\)\) == 0.5, {z, 1}]

t = FindRoot[Integrate[f[x], {x, z, \[Infinity]}] - .5, {z, 1}]

{z -> 0.693147}

t = FindRoot[Integrate[f[x], {x, 0, z}] == 1.5, {z, 1}]

FindRoot::nlnum: The function value {False} is not a list of numbers with dimensions {1} at {z} = {1.}. >>

FindRoot::nlnum: The function value {False} is not a list of numbers with dimensions {1} at {z} = {1.}. >>

FindRoot[\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(z\)]\(f[
     x] \[DifferentialD]x\)\) == 1.5, {z, 1}]

t = FindRoot[Integrate[f[x], {x, 0, z}] - 1.5, {z, 1}]

{z -> 0.693147}
2

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}]
1 Like
3

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.

1 Like
4

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.

2 Likes
5

Many thanks to all who replied. There were a number of great suggestions in the replies.