I want to find a lower bound for $\left|\int_\gamma f(z)dz\right|$. I know of the estimation lemma and Jordan's lemma for an upper bound, but I don't know of any for a lower bound.

The motivation is that I want to prove that a certain integral diverges on a given contour, and I'm looking for ways to do that. I think that for a given smooth contour $\gamma_R$ that depends on a parameter $R$, and a given function $f(z)$ such that $|f(z)|\rightarrow\infty$ as $R\rightarrow\infty$, one can conclude that $\left|\int_{\gamma_R}f(z)dz\right|\rightarrow\infty$ as $R\rightarrow\infty$. In an attempt to prove that, I use a naive approximation for $\left|\int_{\gamma_R}f\right|$: $$\left|\int_{\gamma_R}f\right|=\left|\int_If(\gamma_R(t))\dot{\gamma_R}(t)\right|\geq \left|\sum_{j=1}^nf\left(\gamma_R\left(\frac{j}{n}\right)\right)\dot\gamma_R\left(\frac{j}{n}\right)\right|$$ here I use a lower bound of the Riemann integral after the choice of parameterization for $\gamma_R$, without loss of generality and for ease of writing I assumed that the interval of the parameterization is $[0,1]$ and I used the partition of $I$ to $n$ subintervals of length $\frac{1}{n}$. And then using the triangle inequality and the assumption we have that $\left|\int_{\gamma_R}f(z)dz\right|\rightarrow\infty$ (On second thought, we might also need that the derivative of $\gamma_R$ behaves nice enough).

So, to be very specific, the 3 questions I have here: a) is there an "estimation lemma" for a lower bound? b) is my reasoning correct in the above proof? c) are there criteria for contour integral divergence or similar tests?


  • $\begingroup$ Do you have an example of $f$ in mind? If $f$ does not oscillate too much it might be manageable. $\endgroup$ – timur Aug 4 '12 at 2:09
  • $\begingroup$ For some $0<x<1$, let $f(z)=\frac{x^{iz+\frac{1}{2}}}{iz+\frac{1}{2}}$. Then, unless my calculations are incorrect we have $|f(z)|\rightarrow\infty$ for $z$ on a half circle of radius $R$ and $R$ tends to infinity. $\endgroup$ – Donjim Aug 14 '12 at 17:26

I don't see how this is possible. Just take $f(z) = z$ and $\gamma(t) = Re^{2\pi it}$. Then $\left|f(z)\right| = R$, but $\int_{\gamma_R} f(z)dz = 0$ for every $R$. Note that it is not essential that the curve is closed: it will work just as well with half circles.

That means for question (a): no lower bound in terms of $\left|f(z)\right|$ exists. The closest thing I can think of is $$ \left|\int_\gamma f(z)dz\right| = \left|\int_\gamma wf(z)dz\right| \ge \Re \int_\gamma wf(z)dz = \int_0^1 \Re (wf(\gamma(t))\gamma'(t))dt \ge \min \Re((wf\circ\gamma)\gamma') $$ for any constant $|w| = 1$.

For question (b): something must be wrong; I don't really see how you can apply the triangle inequality, so maybe that is the problem.

To question (c) I can only say that I don't know any.

  • $\begingroup$ Thanks for your answer, and I apologize for taking so long to comment. Of course, your example is excellent. I totally missed that trivial example. About (b), I guess your are right. I should put some more thought into it. $\endgroup$ – Donjim Aug 14 '12 at 17:28

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