# Numerically integrate complex valued function of a real variable and a complex variable

I need to numerically integrate a complex valued function of a real variable and a complex variable by the real variable. My integral:

$$\int^1_{-1}\frac{e^{ix}}{x-z}dx ~~~\text{where} ~~x \in \mathbb{R}, z \in \mathbb{C}$$

I'm trying to do this:

z := x + I y; NIntegrate[Exp[I x]/(x - z), {x, -1, 1}]


But I get the message:

NIntegrate::inumr: The integrand (I E^(I x))/y has evaluated to non-numerical values for all sampling points in the region with boundaries {{0,1}}.

How can this integration be done numerically in Mathematica?

• Welcome to Mathematica.SE DJNZ! You either need to give z a numerical value or use Integrate instead of NIntegrate. – Thies Heidecke Mar 22 at 3:36
• Very thanks! I solve a more general problem: I have to integrate a large complex function over several real variables (this is such a parametrization), and I try to understand how to do it in general, and for this I try to make a prototype on a Mathematica. My target - make a calculation without decomposition into imaginary and real parts. – DJNZ Mar 22 at 3:49
• If your general problem involves Cauchy principal values, you might want to arrange things so that "normal" integrals and principal value integrals are done separately. – J. M. will be back soon Mar 22 at 12:13

## 2 Answers

z := x + I*y;

Integrate[Exp[I*x]/(x - z), {x, -1, 1}]


(2 I Sin[1])/y

If you insist on using NIntegrate then you should provide numerical value to y?

• Thanks for answer, but I wrote "I need to integrate numerically complex... ". Yes, but how to provide numerical value for y? How to set complex variable in general? – DJNZ Mar 22 at 3:42

I will assume that $$z = a+ib$$, since otherwise you are assuming that $$x-z=iy$$ which is always imaginary in your question. However, you do not even need to actually specify $$z$$ in that form. You just need to do this:

F[z_?NumberQ] := Integrate[Exp[I*x]/(x - z), {x, -1, 1}, PrincipalValue -> True]
F[1 + 2 I]
F[-I/2]
F[I/3]
F[0]


This gives

\begin{align} &e^{-2+i} (\text{Ei}(-2)-\text{Ei}(2-2 i))\\ &\sqrt{e} \left(-\text{Ei}\left(-\frac{1}{2}-i\right)+\text{Ei}\left(-\frac{1}{2}+i\right)-2 i \pi \right)\\ &\frac{\text{Ei}\left(\frac{1}{3}+i\right)-\text{Ei}\left(\frac{1}{3}-i\right)}{\sqrt[3]{e}}\\ &2 i \text{Si}(1) \end{align}

The last example is meant to show how it works when $$z$$ is real number, in which you need principal value integral. You could also change this to numerical case

G[z_?NumberQ] :=
NIntegrate[Exp[I*x]/(x - z), {x, -1, z, 1}, Method -> PrincipalValue]


where you can account for when $$z\in (-1,1)$$ and you invoke PrincipalValue as "Method" (slightly different syntax than the Integrate command).