Estimation lemma

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In mathematics, the estimation lemma gives an upper bound for a contour integral. If f is a continuous function on the contour Γ and if f(z) ≤ M for all z on Γ, then

\left|\int_\Gamma f(z) \, dz\right| \leq Ml(\Gamma),

where l(Γ) is the length of Γ. In particular, it is the case that

\left|\int_\Gamma f(z) \, dz\right| \leq l(\Gamma) \max_{z\in\Gamma}|f(z)|.

Intuitively, the lemma is very simple to understand. If a contour is thought of as many smaller contour segments connected together, then there will be a maximum |f(z)| for each segment. Out of all the maximum |f(z)|'s for the segments, there will be an overall largest one. Hence, if the overall largest |f(z)| is summed over the entire path then the integral of f(z) over the path must be less than or equal to it.

The estimation lemma is most commonly used as part of the methods of contour integration with the intent to show that the integral over part of a contour goes to zero as | z | goes to infinity. An example of such a case is shown in the example below.

[edit] Example

the contour
Enlarge
the contour

Problem. Find an upper bound for

\left|\int_\Gamma \frac{1}{(z^2+1)^2} \, dz\right|,

where Γ is the upper half-circle | z | = a traversed once in the counterclockwise direction.


Solution. First observe that the path of integration has length

l(\Gamma)=\frac{1}{2}(2\pi a)=\pi a.

Next we seek an upper bound M for the integrand when | z | = a. By the triangle inequality we see that

|z^2+1|\geq |z^2| - 1 = a^2 - 1

since | z | = a on Γ. Hence

\left|\frac{1}{(z^2+1)^2}\right|\leq \frac{1}{(a^2-1)^2}.

Therefore we apply the estimation lemma with M = 1 / (a2 − 1)2. The resulting bound is

\left|\int_\Gamma \frac{1}{(z^2+1)^2}\,dz\right| \leq \frac{\pi a}{(a^2-1)^2}.

[edit] References

  • Saff, E.B, Snider, A.D., Fundamentals of Complex Analysis for Mathematics, Science, and Engineering (Prentice Hall, 1993).
  • Howie J.M., Complex Analysis (Springer, 2003).
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