Integrate wrong for absolute value of trig function

Question

I was trying to get $\int_0^1 \lvert \cos(2 \pi k x) \rvert \,\mathrm{d}x$ for $k \in \mathbb{Z}$, and was surprised by the result (using Mathematica 10.0.1.0):

Integrate[Abs[Cos[2 π k x]], {x, 0, 1}, Assumptions -> Element[k, Integers]]

(Abs[Cos[2 k π]] Tan[2 k π])/(2 k π)

Simplify[%, Assumptions->Element[k, Integers]]

 0

Table[Integrate[Abs[Cos[2 π k x]], {x, 0, 1}], {k, 0, 10}]

 {1, 2/π, 2/π, 2/π, 2/π, 2/π, 2/π, 2/π, 2/π, 2/π, 2/π}

Table[NIntegrate[Abs[Cos[2 π k x]], {x, 0, 1}], {k, 0, 10}]

{1., 0.63662, 0.63662, 0.63662, 0.63662, 0.63662, 
  0.63662, 0.63662, 0.63662, 0.63662, 0.63662}

The answer of 0 is obviously ridiculous, since the integrand is nonnegative and not identically zero. Integrate with specific integers seems to get it right; the relevant identity at the bottom of this section of Wikipedia's list of integrals agrees that the value should be $\frac{2}{\pi}$ for $k \ne 0$, and of course it should be 1 for $k = 0$.

A similar issue happens with $\int_0^1 \lvert \sin(2 \pi k x) \rvert \,\mathrm{d}x$: Integrate gives $$\frac{\text{sgn}(k) (k-\cot (2 \pi k) \left| k \sin (2 k \pi )\right| )}{2 \pi k^2}$$ which is indeterminant for integral $k$ but has limit 0 at each integer, but whose value should also be $\frac{2}{\pi}$ for $k \ne 0$ and 0 for $k = 0$. (Again, Integrate with any particular value of $k$ gets it right.)

If I use the variable k without any Assumptions, Integrate eventually gives up.

Am I doing something wrong here in specifying my assumptions or whatever, or is this a bug for a surprisingly simple integral?

Answer 1

The problem arises because of the branch-cuts of the Tan. Since Mathematica 8.0 doesn't integrate the indefinite Integral with Abs[...], do it with Sqrt[Sqr[Cos[...]]]

In[173]:= int1[x_, k_] = Integrate[Sqrt[Cos[2 \[Pi] k x]^2], x]

Out[173]= (Sqrt[Cos[2 k \[Pi] x]^2] Tan[2 k \[Pi] x])/(2 k \[Pi])

Plotting this, shows it has 2 k Branch-Cuts where the Tan is +-Infinity.

Plot[(Sqrt[Cos[2 k \[Pi] x]^2] Tan[2 k \[Pi] x])/(2 k \[Pi]) /. 
k -> 3, {x, 0, 1}]

At the first branch-cut at x= 1/(4 k) the functions upper limit is

In[129]:= Limit[(Sqrt[Cos[2 k \[Pi] x]^2] Tan[2 k \[Pi] x])/(
2 k \[Pi]), x -> 1/4/k, Direction -> -1]

Out[129]= -(Sqrt[k^2]/(2 k^2 \[Pi]))

The lower limit there is (Sqrt[k^2]/(2 k^2 [Pi])) The total jump at x=1/(4 k) is therefore 1/(k Pi). This is the same for the other branch-cuts. Multiplied with the 2 k branch-cuts you get the right result: 2/Pi

Mathematica didn't take into accout this in the definite integral.