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Let $f:\mathbb{C}\to\mathbb{C}$ given for $f(z)=\int_0^z \frac{1-e^t}{t} dt-\ln z$ and put $g(x,y)=\text{Re}(f(z))$. While using the computer, how to determine the curve $g(x,y)=0$?

Thanks for the help.

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    This is just too vague to answer... What do you know about $f$? Why don't you just solve $g(x,y) =0$ for $y$ and plot? What software do you use?2011-11-21
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    @Dirk: Ok, i will accept your suggestion. I tried to use maple, but had no success. Thanks2011-11-21

1 Answers 1

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Using Mathematica:

ContourPlot[With[{z = x + I y},
                  Re[EulerGamma - ExpIntegralEi[z]]] == 0,
             {x, -20, 20}, {y, -20, 20}]

exponential integral contour

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    You said you have Maple. I assume it would have the exponential integral available. Search the docs on how that function is used, as well as how to do contour plots.2011-11-21
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    Thank you.. one question: How do I install the Mathematica Program? He is free?2011-11-21
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    It's not free. There is a [trial version](http://www.wolfram.com/mathematica/trial/) however.2011-11-21
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    Another question: and analytically?2011-11-21
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    [It's sorta kinda complicated...](http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/19/01/)2011-11-21
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    I agree with you. This is complicated.2011-11-22
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    Well, [not exactly](http://functions.wolfram.com/GammaBetaErf/ExpIntegralEi/26/03/0001/).2011-11-23
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    One more question: $g(x,y)=\text{Re}[\gamma-\Gamma(0,-z)]=\text{Re}[\gamma+Ei(z)]$? I think that the contours change..2011-11-23
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    Why not try it out? I've given you the relationship between incomplete gamma and the exponential integral.2011-11-23