I've been asked to find the value of the following integrals using Cauchy Formula: I'll ask for help on the first one just to make sure i know what I'm doing.

Let $f(z) = \frac{e^{z}}{(z+1)(z+3)}$ then calculate $$\int_{S_{4}^{+}(i)} f(z)~ dz$$

Cauchy's formula is $$f(z) = \frac{1}{2 \pi i} \int_{\gamma}\frac{f(w)}{w-z} dw$$

so in the above i want to choose $f(w) = $ either $\frac{e^{w}}{w+1}$ or $\frac{e^{w}}{w+3}$

Now the thing which i'm confused about is the procedure in which to solve this problem. I understand that if the points of discontinuity (in this case, z=-1 and z=-3) were outside the contour of integration, AND f(z) (ie the integrand $\frac{e^{z}}{(z+1)(z+3)}$) is differentiable on on D itself then then overall integral for a closed contour will equal zero via the cauchy integral theorem since the contour itself is contractible to a point.

if; on the other hand, one of the points of discontinuity lies with in the contour of integration then we choose the other one to be f(w) and then compute the integral at that point.

eg for $$f(z)=\frac{1}{z^2+1}$$ along the contour $S_{1}^{+}(i)$ which is the positively orientated circle of radius 1 around the centre point of i. we know that $$f(z)=\frac{1}{(z-i)(z+i)}$$ and so the contour clearly has i in the contour, so we choose $$f(w) = \frac{1}{w+i}$$ (since that's w=-i) and use the cauchy integral theorem. giving $$\int_{S_{+}^{1}(i)} \frac{dw}{w^2+1}=\int_{S_{+}^{1}(i)}\frac{f(w)}{w-i}dw = 2 \pi i f(i)=\pi$$ we're integrating around the point i (Since that's the point of discontinuity lying with in the contour) and essentially collapsing the contour so that's the only point that we care about.

now, with my original problem, i have two points of discontinuity, -1 and -3, both of which fall with in our contour. so I'm confused as to which values to pick. I'm hoping to get a thorough enough understanding that i can apply it to the other questions asked.

and when dealing with these questions what's the procedure, i assume it's going to be along the lines of

  1. locate points of discontinuity.
  2. make sure the function is holomorphic/analytic
  3. check which points lie with in the contour

etc, etc.

I'm pretty sure had the contour been $S_{1}^{+}(i)$ then the integral would be zero. further i understand that the original function is the composition of two analytic functions and so analytic on the region $\mathbb{C}\setminus \{-3,-1\}$

Thanks for taking the time to read this. appreciate it very much.


1 Answer 1



Often, the splitting of the integrand into several parts with different singularities is helpful. For example:

$$\frac{e^z}{(z+1)(z+3)} = \frac{1}{2}\left(\frac{e^z}{(z+1)} - \frac{e^z}{(z+3)} \right)$$

Now, you can apply Cauchy-integral formula to the two parts independently.

  • $\begingroup$ This is how I would normally do it in all honesty but I’ve been led to believe that the Cauchy integral formula can be used so I don’t need to consider partial fractions. Is this true?, further in the above two integrals then I’d use $f(w)=e^w$ At z = -1 and -3 right? $\endgroup$
    – Vaas
    Commented Oct 31, 2019 at 12:29
  • 1
    $\begingroup$ Your thoughts are actually heading to the residue theorem: en.wikipedia.org/wiki/Residue_theorem And referring to your question: Yes. After decomposing the function you can just plug in the corresponding values $w$. $\endgroup$ Commented Oct 31, 2019 at 12:41
  • $\begingroup$ Ah that makes sense because I imagine that would be the next couple of chapters. Thank you for your help, appreciate it. $\endgroup$
    – Vaas
    Commented Oct 31, 2019 at 12:59

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