# Line integral with Wolfram|Alpha?

Posted 4 months ago
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 Hey there! I would like to calculate a Line integral such as e^zcos(piz)/(z^2+z) through to |z|=1How can I do that with Wolfram? Answer
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Posted 4 months ago
 Please use spaces and parentheses to clarify the complex function. For example, does piz mean Pi z? And does the formula involve Exp[z] or Exp[z Cos[Pi z]]?Also, do you mean to integrate around the circle Abs[z]==1?Did you try to parametrize the circle and reduce the line integral to an ordinary integral (of a complex-valued function of a real variable)? Answer
Posted 4 months ago
 Hey! thanks for trying to help me. I will clarify the function: exp(z)*cos(pi * z)/(z^2+2) around abs(z), How would you insert that in the Wolfram!Alpha? Answer
Posted 4 months ago
 Is the denominator z^2 + 2 or z^2 + z? Answer
Posted 4 months ago
 I suspect that Wolfram|Alpha cannot directly evaluate a contour integral, but that you will have to do the first step yourself, namely, parameterizing the contour (in the standard way?), etc. Answer
Posted 4 months ago
 I did do the usual parameterization to evaluate the integral, but using Mathematica and found it was divergent. Answer
Posted 4 months ago
 Integrate works with complex numbers. For this region you can simply let $z=e^{it}$ for $t\in[0,2\pi)$. Does this work? Integrate[(E^z Cos[\[Pi] z])/(z^2 + z) /. z -> Exp[I t],{t,0,2\[Pi]}] Your integrand is divergent at $z=-1$, and Mathematica correctly finds that it is divergent. If you're looking for the residue, try Integrate[(E^z Cos[\[Pi] z])/(z^2 + z) /. z -> .5 Exp[I t],{t,0,2\[Pi]}] where I have replaced $z$ with $.5e^{it}$ to avoid any other poles. Answer
Posted 4 months ago
 Am I missing something -- didn't you neglect to insert the dz multiplier after the parameterization? Answer
Posted 4 months ago
 Effectively I'm letting $z(t)=.5e^{it}$, and computing $\int_0^{2\pi}\frac{\exp(z(t))\cos(\pi z(t))}{z(t)^2+z(t)}dt$. The parameter of integration and its bounds are specified by {t,0,2\[Pi]}, just notation. Mathematica can typeset it nicely too, though with With[{z=Exp[I t]}, Inactive@Integrate[(E^z Cos[\[Pi] z])/(z^2 + z),{t,0,2\[Pi]}]] On the subject of contour integrals, I suppose Mathematica can't handle complex regions--most of their Region related functions are explicitly $\mathbb R^n$. That being said, it handles arbitrary real regions as integration bounds no problem, and even some basic complex bounds (ha) like piecewise linear or circular, consider (from the docs) Integrate[1/(z + 1/2), {z, 1, E^((I \[Pi])/3), E^((2 I \[Pi])/3), -1, E^(-((2 I \[Pi])/3)), E^(-((I \[Pi])/3)), 1}] Answer
Posted 4 months ago
 Definition (or formula from a theorem) of contour integral of complex function $f(z)$ over a contour $C$: let $\gamma: [a, b] \rightarrow \mathbb{C}$ be a parameterization of $C$. Then $\int_C f(z) dz = \int _a^b f(\gamma(t)) \gamma'(t) dt$.Your expression for the integral seems to be missing the factor $\gamma'(t)$, which for $\gamma(t) = 0.5 e^{i t}$ should be $0.5 i e^{i t}$. Answer
Posted 4 months ago
 Yes, I forgot this! My bad. Proper code is Integrate[Unevaluated@D[z,t](E^z Cos[\[Pi] z])/(z^2+z) /. z -> .5 Exp[I t],{t,0,2\[Pi]}] though it appears to not be able to find the closed form and just approximates $2\pi i$. And the previous code yielding 0 is a dead giveaway. Answer
Posted 4 months ago
 I guess this is a standard example for the use of residues! The function f[z_] := Exp[z]*Cos[Pi*z]/(z^2 + z) has two poles: poles = Solve[z^2 + z == 0, z] (* Out: {{z\[Rule]-1},{z\[Rule]0}} *) The one at z=-1 is on the contour: ComplexPlot3D[f[z], {z, -1.1 - 1 I, 1 + 1 I}, RegionFunction -> (Abs[#] <= 1 &), BoxRatios -> {Automatic, Automatic, 1}] The respective residues are: Residue[f[z], {z, #}] & /@ {0, -1} (* Out: {1,1/\[ExponentialE]} *) So - according to residue theorem - we find for the line integral along Abs[z]==1 (the residue on the pole counts half!): 2 Pi I (1 + 1/(2 E)) But Mathematica can actually handle those integrals directly; because of the pole on the contour the option PrincipalValue -> True is needed: Integrate[f[Exp[I \[Phi]]] I Exp[I \[Phi]], {\[Phi], 0, 2 Pi}, PrincipalValue -> True] (* Out: \[ImaginaryI] (2+1/\[ExponentialE]) \[Pi] *) Answer
Posted 4 months ago
 Counting the residue at the pole on the contour itself seems to require a bit of justification, as in answers to: https://math.stackexchange.com/questions/1075962/pole-on-the-contour-using-the-residu-theorem-what-is-this-formula-of-plemelj Answer
Posted 4 months ago
 Counting the residue at the pole on the contour itself seems to require a bit of justification, ... Well, yes! But the pole in question is of first order, the path of integration is sufficiently well behaving, and in those cases the calculation (I think / I learned at school) is that simple. Or do you disagree with the result? Answer
Posted 4 months ago
 I do not disagree with the result, but rather meant to suggest that some kind of rigorous justification is really needed for such things, as in the math.stackexchange reference I cited. Answer
Posted 4 months ago
 It should be emphasized that this path integral actually is divergent, and only exists in the sense of a principal value integration. Answer
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