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A: Contour plot after numerical integration involving branch cut

azerbajdzanAs far as I know Mathematica does not have a function for identification of branch cuts. What can be done is to use RegionFunction to remove unwanted regions but it assumes you know positions of branch cuts. For your example it can be done the following way using threshold 0.01. If you provide a ...

user64494
user64494
This is a fake, not a solution How about Num[x_?NumericQ] := NIntegrate[Sqrt[y + (y)^(1/2)], {y, 0, x}]; ComplexContourPlot[Im[sNum[x]] == 0, {x, -2 - I, 2 + I}] and Num[x_?NumericQ] := NIntegrate[Sqrt[y + (y + I)^(1/2)], {y, 0, x}]; ComplexContourPlot[Im[sNum[x]] == 0, {x, -2 - I, 2 + I}]?
azerbajdzan
azerbajdzan
@user64494 Exactly the same RegionFunction works also for your code. Refrain from commenting my posts in the future. Your comments have no sense whatsoever.
user64494
user64494
Can you kidly support your claim " Exactly the same RegionFunction works also for your code" by a code and a plot? And the same with Num[x_?NumericQ] := NIntegrate[Sqrt[y + (y + I)^(1/5)], {y, 0, x}]; ComplexContourPlot[Im[sNum[x]] == 0, {x, -2 - I, 2 + I}].
azerbajdzan
azerbajdzan
@user64494 If you do not know how to work with RegionFunction study the documentation or ask your own question.
user64494
user64494
Sorry, you wrote "If you provide a more complicated example I may try to do the same with it".
xzczd
xzczd
00:39
@user64494 Directly add azerbajdzan's RegionFunction to your first sample gives a reasonable result: 
sNum[x_?NumericQ] := NIntegrate[Sqrt[y + (y)^(1/2)], {y, 0, x}]; ComplexContourPlot[Im[sNum[x]] == 0, {x, -2 - I, 2 + I},
 RegionFunction -> (Abs@Im[#1] > 0.01 || Re[#1] > 0 &)]
user image
This output is the same as that given by the corresponding Integrate[...] output.
Which part do you find this output incorrect? If you think it's incorrect, what's the correct one in your mind?
@user64494 As to sNum[x_?NumericQ] := NIntegrate[Sqrt[y + (y + I)^(1/5)], {y, 0, x}];, I think it doesn't involve branch cut issue? (Corresponding Integrate[...] involves ConditionalExpression[...], which seems to be a bit troublesome, but it doesn't seem to be related to this post. ) If you think it involves branch cut(s), then again, can you show us the desired output in your mind?
 
15 hours later…
user64494
user64494
16:08
@xczd: In 14.1 `Integrate[Sqrt[y + (y)^(1/2)], {y, 0, x}]` performs `1/12 (Sqrt[Sqrt[x] + x] (-3 + 2 Sqrt[x] + 8 x) +
3 ArcCoth[Sqrt[Sqrt[x] + x]/Sqrt[x]])` without any condition. The command of Maple 2024 `FunctionAdvisor(branch_cuts, sqrt(sqrt(x) + x)*(-3 + 2*sqrt(x) + 8*x)/12 + arccoth(sqrt(sqrt(x) + x)/sqrt(x))/4, plot = 2.)` produces the following branch cuts `[sqrt(sqrt(x) + x)*(-3 + 2*sqrt(x) + 8*x)/12 + arccoth(sqrt(sqrt(x) + x)/sqrt(x))/4, x < 0, And(x = alpha + 1/2 - sqrt(4*alpha + 1)/2, alpha in RealRange(-infinity, Open(0))), And(x = alpha + 1/2 + sqrt(4*alpha + 1)/2, alpha in …
@xzczd: In 14.1 `Integrate[Sqrt[y + (y)^(1/2)], {y, 0, x}]` performs `1/12 (Sqrt[Sqrt[x] + x] (-3 + 2 Sqrt[x] + 8 x) +
3 ArcCoth[Sqrt[Sqrt[x] + x]/Sqrt[x]])` without any condition. The command of Maple 2024 `FunctionAdvisor(branch_cuts, sqrt(sqrt(x) + x)*(-3 + 2*sqrt(x) + 8*x)/12 + arccoth(sqrt(sqrt(x) + x)/sqrt(x))/4, plot = 2.)` produces the following branch cuts `[sqrt(sqrt(x) + x)*(-3 + 2*sqrt(x) + 8*x)/12 + arccoth(sqrt(sqrt(x) + x)/sqrt(x))/4, x < 0, And(x = alpha + 1/2 - sqrt(4*alpha + 1)/2, alpha in RealRange(-infinity, Open(0))), And(x = alpha + 1/2 + sqrt(4*alpha + 1)/2, alpha in …

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