I’m trying to figure out if there is a function which is defined on the interval [1,?) for which the limit of f(x) as x->? does not exist, and the definite integral from 1 to ? converges to a finite value. Does such a function exist? Thanks in advance!
Consider E^(I x^2). The limit x->Infinity doesn’t exist because of the oscillation at infinite frequency. However:
In[22]:= Integrate[E^(I x^2), {x, 1, Infinity}]
Out[22]= (1/2 + I/ 2) Sqrt[\[Pi]/2] (1 - (1 - I) (FresnelC[Sqrt[2/\[Pi]]] + I FresnelS[Sqrt[2/\[Pi]]]))
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
Marco,
Your assumption use shouldn’t do much damage, however it does appear that Integrate[] fails to utilise assumptions as well as the Simplify[] gold standard.
Pre integral assumption simplification can miss effects on the results of the subsequent integral, whilst post integral assumption simplification can incur performance cost as the core work is done in complex suitable logic, and then discarded.
(* Making a clearer test function *)
litmus = (3^k)^(1/k);
Simplify[litmus]
Simplify[litmus, Assumptions -> {k \[Element] Reals}]
(* Seeing if Integrate leverages specified assumptions *)
Integrate[litmus, {x, 0, 5}, Assumptions -> {k \[Element] Reals}]
(* Pre and post integral assumption simplifications *)
Integrate[Simplify[litmus, Assumptions -> {k \[Element] Reals}], {x, 0, 5}]
Simplify[Integrate[litmus, {x, 0, 5}], Assumptions -> {k \[Element] Reals}]
I guess that this behaviour is documented:
It is one of these branch cut problems, I guess. Mathematica defines indefinite integrals of piecewise function up to constants at the discontinuities. It would be nice if somebody new an easy way of “removing” the discontinuities in these antiderivatives to get a continuous antiderivative.
Cheers,
Marco
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]
Thanks for all the help everyone!