Thanks for all the help everyone!

Here's another one over the reals. The idea is very similar to S M Blinder's approach - we have to deal with an integral that Mathematica does not appear to like.

Plot[Sign[Sin[Pi 2^x]], {x, 0, 10}, PlotPoints -> 5000]

This function oscillates faster and faster, and jumps from -1 to +1.

So it obviously does not converge. It turns out that Mathematica cannot directly integrate this to infinity. We can only get the first couple of steps:

Table[Integrate[Sign[Sin[Pi 2^x]], {x, 0, k}], {k, 1, 4}]

or

N[%]
{-1., -0.830075, -0.741434, -0.696551}

But it can determine the roots of the sin function.

Reduce[Sign[Sin[Pi 2^x]] == 0 && x \[Element] Reals, x]

Between the zeros the function is constant -1 or 1. Note that the zeros are between consecutive $\log(k)/\log(2)-\log(k+1)/\log(2)$ so we can calculate the Integral as the limit of the following sum:

Limit[Sum[(-1)^(k + 1) (Log[ k]/Log[2] - Log[1 + k]/Log[2]), {k, 1, j}], j -> Infinity]

which evaluates to

or -0.651496.

So, the integral is finite and the function does not converge. The example given by S M Blinder is easier, because Mathematica evaluates the integral, but I found this example also instructive, because we solve it without using direct integration. (At least if I haven't made a mistake somewhere.)

My feeling is that this is the right result. There is, however, this calculation:

Integrate[Sign[Sin[Pi 2^x]], {x, 0, k}, Assumptions -> {x \[Element] Reals && k \[Element] Reals}]

which evaluates to

and hence does not converge, i.e.

Limit[k/Sign[Sin[2^k \[Pi]]], k -> Infinity]

evaluates to

Interval[{-\[Infinity], \[Infinity]}]

So, I might have made a mistake after all. It is, however, weird that the Integral evaluates to something that is basically proportional to $k$. That does not appear to make sense. I must admit that the following baffles me a bit:

Evaluate[Integrate[Sign[Sin[Pi 2^x]], {x, 0, k}, Assumptions -> {x \[Element] Reals && k \[Element] Reals}]]
(*k/Sign[Sin[2^k \[Pi]]]*)

and

Table[Integrate[Sign[Sin[Pi 2^x]], {x, 0, k}], {k, 1, 4}]
(*{-1, -((2 (2 Log[2] - Log[3]))/Log[2]), (-11 Log[2] + 2 Log[5] + 
2 Log[7])/Log[2], -((2 (13 Log[2] - 2 Log[3] - Log[5] - Log[11] - Log[13]))/
Log[2])}*)

@Ilian Gachevski , @Vitaliy Kaurov , @Daniel Lichtblau : Am I using the assumptions wrong?

Cheers,

M

I worked with Sign[Sin[Exp[x]]] for simplicity. Integrate fails here as it does in your example, without really letting the user know.

Plot[{Evaluate[Integrate[Sign[Sin[Exp[x]]], {x, 0, k}]], Sin[Exp[k]]/Abs[Sin[Exp[k]]]}, {k, 0, 4}]

Mucking about by hand one can procure the actual integral. The partial sums in phase with Sin[Exp[x]] here take a couple of interesting forms. The one first incident at x = Log[ 2 Pi ] (after a full Sin cycle) happened to equate to OEIS A161737!

Limit[Log[2^(5 - 4 (1 + m)) m \[Pi] ((-2 + 2 (1 + m))!)^2 (m!)^-4], m -> \[Infinity]]
==
Log[2]