I have a not-so-complicated piecewise cubic function, shown below as the yellow curve on the right. It's derivative is on the left; the blue lines are references. Please see the code below where I call it myF (and its derivative myf).

To my surprise, the integration with a parameter $u$ and sine in the argument gives an ugly result involving the complex $i$.

Is the result of $\int$myF$(u\sin x)$d$x$ (as shown in the screenshot below) correct? Or is it some kind of failure due to my erroneous way of using Integrate coupled withPiecewise?

As just the area under the curve modulated by sine, I for now don't see how this can be mathematically true to have involved $\sqrt{-1}$.

enter image description here

Screen shot below:

enter image description here

The code to define the curves, generate the plots, and do the integration is below. When the parameter $u < \frac12$ of course there's no problem since the integration doesn't go over the `splitting point' (where is differentiable).

I tried various ways to code the integration with assumptions, but the outcomes are basically the same. Some hint or confirmation would be greatly appreciated. Thank you.

ClearAll[myF, myf, u, v, ImgSz];
myF[v_] := Piecewise[{
    {0, v < 0},
    {4/3 v^3, 0 <= v < 1/2},
    {1/3 (1 - 6 v + 12 v^2 - 4 v^3), 1/2 <= v < 1},
    {1, v >= 1}
myf = D[myF[u], u] /. {u -> v};
ImgSz = 250;
Row[{  Plot[{ 1,  (*  symmetry reference  *)
    }, {v, -.3, 1.3}, PlotRange -> {-0.1, 2}, ImageSize -> ImgSz],
  Plot[{ v, (*  linear reference  *)
    }, {v, -.3, 1.3}, PlotRange -> {-0.1, 1}, ImageSize -> ImgSz]
Assuming[  1/2 < u < 1 ,
 Integrate[ myF[ u Sin[x]], {x, 0, Pi}] ]

Summarizing Edit

Okay, so I couldn't recognize the expression as real-disguised-in-complex-conjugate, neither did I know reliable ways to test that. I'm glad to have learned from both answer posts.

In my case, defining the function with UnitStep doesn't allow Mathematica to evaluate differently like in another post (which I cannot find right now). Nor does PrincipalValue -> True or change of variables apply here as it does sometimes. I have accepted that in my case I need to post-process the expression the way I see fit. Nonetheless, for the record, there are some known bugs, old and new, solved and unsolved.


2 Answers 2


You can convert the answer to a real-looking expression as follows:

 ComplexExpand[answer, TargetFunctions -> {Re, Im}],
 1/2 < u < 1]
  1/9 (-11 Sqrt[-1 + 4 u^2] + 16 u^2 (u - Sqrt[-1 + 4 u^2]) + 6 (1 + 6 u^2) ArcSec[2 u])

Just because an expression contains the imaginary unit $i$ doesn't mean that its imaginary part is non-zero. For example, you could naïvely think that $$ \frac{1}{x-i}+\frac{1}{x+i} $$ is complex, but actually it equals $\frac{2 x}{x^2+1}$, which is explicitly real (for real $x$).

In your particular case, you found an expression that looks complex, but it is actually real. For example, if you set u=3/4 and evaluate the integral numerically, you get 0.718597+i5.5511e-17, which has an spurious imaginary part due to numerical error.

If you evaluate

Plot[ReIm@NIntegrate[myF[u Sin[x]], {x, 0, Pi}], {u, 1/2, 1}]

you get

enter image description here

which confirms the fact that the integral is indeed real - its imaginary part vanishes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.