Consider $$f(z)=\int_0^{\infty} \frac{e^{-t}}{t+z}dt.$$

I'm trying to determine where this function has branch points, define suitable branch cuts, and determine the discontinuity across the cut. First of all, I believe it has an essential singularity along the entire negative real axis, correct?

I'm not sure how to handle the branch points. Intuitively it seems that 0 is a branch point, but I'm not sure what the cut or discontinuity would be--or how to justify it.

  • $\begingroup$ This function has a logarithmic branch point. $\endgroup$ Dec 7, 2012 at 5:06

1 Answer 1


You can make a change of variables to rewrite the integral as $$\int_z^{\infty} \dfrac{e^{-t + z}}{t}dt = e^z \int_z^{\infty} \dfrac{e^{-t}}{t} dt,$$ where the integral is taken along a contour which is the translation of the positive real axis by $z$.

Now it's easy to see how the function behaves as $z$ moves around $0$: the contour over which we're integrating winds around the origin, and so by the residue theorem we add the value $2\pi i e^z$. So the singularity at the origin looks like $e^z \log z$, and there is a logarithmic branch point at the origin.

  • $\begingroup$ Thanks! That's very helpful. So how would I defined a branch cut and find the discontinuity across it? $\endgroup$
    – Alex
    Dec 7, 2012 at 4:25
  • $\begingroup$ @Alex: Dear Alex, It will be just like for $\log z$. You can take any ray from the origin as a branch cut, and when you cross it at the point $z$, the value of the function will jump by $2\pi i e^z$ (as the residue computation that I sketched shows). $\endgroup$
    – Matt E
    Dec 7, 2012 at 4:56

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