# Imaginary components of real integrals

by cesiumfrog
Tags: components, imaginary, integrals, real
 P: 2,050 Why does the incomplete gamma function have an imaginary component, when the exponential integral does not? $$\Gamma(0,z,\infty)\equiv\int^\infty_z \frac{e^{-t}}t dt$$ $$Ei(z)\equiv-\int^\infty_{-z} \frac{e^{-t}}t dt$$ Looking at how these integrals are usually defined I would have expected them to be nearly identical, $\Gamma(0,z,\infty)=-Ei(-z)$, for all z. Instead, while that is satisified for positive real z, the general relation is more complicated and particularly for negative real values $\Gamma(0,x,\infty)=-Ei(-x)-i\pi$ (e.g., according to MathWorld). Why is this? I understand that the integrands have a singularity at the origin of the complex plane. I think I can see how integrating a contour around this will contribute ±2πi, or half that if only skipping from the positive to the negative side of the real axis. But I can't see why this would only apply when we label it the gamma function and not the exponential integral? Plotting with Alpha/Mathematica (for real arguments) seems to confirm that the real components of $\Gamma(0,x,\infty),\Gamma(0,x),E_1(x)$ and Ei(x) all behave symmetrically, but that the imaginary component of Ei lacks the step-function which the others all exhibit. Curiously, I was able to obtain the step-function in the imaginary component of Ei by subtracting a tiny imaginary component from its argument. Should I conclude that, by some quirk, Ei alone is just being defined in a completely separate manner along the real axis (perhaps, defaulting to Cauchy principal values rather than complex analysis as the method to bypass the singularity)? More generally, for an arbitrary integrand with (multiple) singularities, is there a convention which determines which is the principal branch (e.g., why does $\Gamma(0,-2)$ involve negative rather than positive iπ)?