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I came across the question of what percent of numbers from 0 to 1 have a logerithm whose floor is even. What I did was I made a piecewise function: Piecewise[{{1, EvenQ[Floor[Log[x]]] == True}}, 0]. If I plot the function I get: Piecwise function plotted
so it clearly recognizes this function. However, when I try to integrate the function from 0 to 1 with Integrate[Piecewise[{{1, EvenQ[Floor[Log[x]]] == True}}, 0], {x, 0, 1}], the result that it gives me is 0. This is clearly false, as all outputs are either 0 or 1. I tried integrating from .1 to 1 and I still get 0. I even tried integrating from 0.2 to 0.3 where the only outputs are 1, and therefore the integral should be 0.1, but I get an output of 0. I tried doing it with conditionals, but I still get the output as 0. Is there something I am doing wrong or is Mathematica just failing to integrate it?

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  • $\begingroup$ EvenQ[a] evaluates to False immidiately for symbolic input, thats why you get 0, to avoid it use something without the Q, e.g. a/2 ∈ Integers $\endgroup$
    – Coolwater
    Commented Jun 16, 2022 at 17:33
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    $\begingroup$ Mod[...<Integer-valued expression>..., 2] == 0 is a standard way to "algebraically" represent the condition for being even. So Mod[Floor[Log[x]], 2] == 0. But one problem, I think, is that there are infinitely many singular points on the interval of integration. I don't know if/when Mma is/will be programmed to handle that in Integrate. $\endgroup$
    – Michael E2
    Commented Jun 16, 2022 at 18:05

2 Answers 2

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I think this gives the answer, if I didn't get mixed up about even/odd.

Sum[Integrate[Boole[Mod[n, 2] == 0], {x, Exp[-n], Exp[-n + 1]}],
 {n, Infinity}]
(*  1/(1 + E)  *)

I couldn't figure out how to get Integrate (or DSolve) to handle the condition on Log[x]. But here's a way to approximate it with NDSolve:

Block[{x0 = 1/10^6}, 
 NDSolveValue[{y'[x] == sgn[x], y[1] == 0, 
   WhenEvent[Mod[Log[x], 1] == 0, sgn[x] -> 1 - sgn[x]], 
   sgn[1] == 0}, -y[x0], {x, x0, 1}, DiscreteVariables -> sgn]
 ]
(*  0.268941  *)

1/(1 + E) // N
(*  0.268941  *)

Addendum

To evaluate to arbitrary precision:

Block[{prec = 100},
  NIntegrate[
   Piecewise[{{1, 0 <= Mod[u, 2] < 1}}, 0] Exp[u],
   {u, Log[10^-prec], Sequence @@ Range[Ceiling@Log[10^-prec], 0]},
   WorkingPrecision -> prec]
  ] - 1/(1 + E)
(*  -0.*10^-101  *)
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    $\begingroup$ A simpler version is Block[{x0 = 10^-6}, NDSolveValue[{int'[x] == Mod[Ceiling[Log[x]], 2], int[x0] == 0}, int[1], {x, x0, 1}]] $\endgroup$
    – Carl Woll
    Commented Jun 16, 2022 at 18:47
  • $\begingroup$ @CarlWoll Thanks. At some point I gave up on things like Mod@Floor@Log@x, before I got to NDSolve. $\endgroup$
    – Michael E2
    Commented Jun 16, 2022 at 19:00
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Ok I fixed it. Thank you to Coolwater for the suggestion. I did

Integrate[
 Piecewise[{{1, Floor[Floor[Log[x]]/2] == Floor[Log[x]]/2}}, 0], {x, 0, 1}]

which got rid of the EvenQ but still checked that it was even. I then used NIntegrate to get an approximate answer and it works, giving me the expected ~0.27. I am still unsure of why it could plot the function but not integrate it, but I got it working.

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  • $\begingroup$ What result do you get by executing the integral? It does not evaluate, when I run it. $\endgroup$
    – bbgodfrey
    Commented Jun 17, 2022 at 1:44

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