4
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(Edited to emphasize the point about assumptions).

Consider a simple integral:

Integrate[Exp[I x/2], {x, 0, 2 Pi}]

Mathematica produces the following answer: 4 I.

Consider now a different form of this integral:

Integrate[Sqrt[Exp[I x]], {x, 0, 2 Pi}]

This changes the answer to 4. This is presumably consistent with Mathematica's definition of the square root.

However, providing an assumption,

Integrate[Exp[I x/2], {x, 0, 2 Pi}, Assumptions -> x > 0]
Integrate[Sqrt[Exp[I x]], {x, 0, 2 Pi}, Assumptions -> x > 0]

makes both integrals produce the same answer, 4 I.

How come?

This was tested in Mathematica 13, Mathematica 14.2, and Mathematica 14.3.

This behavior was reported to Wolfram Support and confirmed to be a bug by them.

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6
  • $\begingroup$ Try NIntegrate[Sqrt[Exp[I x]], {x, 0, 2 Pi}] - it confirms it is 4. But we have branch cut at Pi so if you treat this branch cut differently than default behavior of Mathematica then you may come to a different result. $\endgroup$ Commented Aug 21 at 15:48
  • $\begingroup$ So I think it is Integrate[Sqrt[Exp[I x]], {x, 0, 2 Pi}, Assumptions -> x > 0] that gives wrong result. $\endgroup$ Commented Aug 21 at 15:49
  • $\begingroup$ Indeed, in the problem where I stumbled upon this behavior I had a differently chosen branch cut for the square root. But the part about the assumption still remains a mystery. I updated my post to emphasize this point. $\endgroup$ Commented Aug 21 at 16:21
  • $\begingroup$ It is not a mystery it is a bug. But not in Integrate[Sqrt[Exp[I x]], {x, 0, 2 Pi}] as you thought but in Integrate[Sqrt[Exp[I x]], {x, 0, 2 Pi}, Assumptions -> x > 0]. $\endgroup$ Commented Aug 21 at 16:36
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    $\begingroup$ Yes, the Sqrt+Assumptions case is befouled. $\endgroup$ Commented Aug 21 at 17:50

2 Answers 2

Reset to default
3
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expr1 = Exp[I x/2];
expr2 = Sqrt[Exp[I x]];

ReImPlot[expr1, {x, 0, 2 Pi}]
ReImPlot[expr2, {x, 0, 2 Pi}]

NIntegrate[expr1, {x, 0, 2 Pi}]
NIntegrate[expr2, {x, 0, 2 Pi}]

enter image description here

So my conclusion is that this is correct:

Integrate[expr2, {x, 0, 2 Pi}]

4

and this is wrong (a bug):

Integrate[expr2, {x, 0, 2 Pi}, Assumptions -> x > 0]

4 I

And the reason of the bug is probably this another bug:

FunctionDiscontinuities[Sqrt[Exp[I x]], x, Reals]

True

But there is discontinuity at x==Pi.

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2
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This is a problem of multivalued functions. Square root is 2 valued. Sqrt uses the so called principal value defined as having -Pi/2 < arg <= Pi/2, where arg=argument(z)= angle between real axis and line origin to z. Whereas Exp[I x/2] x in (0,2Pi) has an argument of: 0< arg <= Pi.You can see this by plotting the arguments of the integrands:

Plot[Arg[Exp[I x/2]], {x, 0, 2 Pi}]

enter image description here

Plot[Arg@Sqrt[Exp[I x]], {x, 0, 2 Pi}]

enter image description here

As the absolute value of both integrands are the same you are actually integrating in the first case:

ParametricPlot[ReIm[Exp[I x/2]], {x, 0, 2 Pi}]

enter image description here

and in the second case:

ParametricPlot[ReIm[Sqrt[Exp[I x]]], {x, 0, 2 Pi}]

enter image description here

Therefore, in the first case, the real parts cancel, whereas in the second case the imaginary parts cancel.

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  • $\begingroup$ Very confusing answer. $\endgroup$ Commented Aug 21 at 19:13

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