Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
This function has a logarithmic branch point. –  Mhenni Benghorbal Dec 7 '12 at 5:06
add comment

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
Thanks! That's very helpful. So how would I defined a branch cut and find the discontinuity across it? –  Alex Dec 7 '12 at 4:25
    
@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). –  Matt E Dec 7 '12 at 4:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.