Is Swapping Limits and Integrals Justified in Complex Analysis?

[chisigma]

Recently, for the solution of an integral I assumed, with a bit of optimism in truth, the validity of the following formula ...

$\displaystyle \lim_{r \rightarrow r_{0}} \int_{a}^{b} f(r, \theta)\ d \theta = \int_{a}^{b} \lim_{r \rightarrow r_{0}} f(r, \theta)\ d \theta\ (1)$

The question is: this is a correct procedure? ... or ... what property must have f (*, *) so that it is a proper procedure? ...

Kind regards

$\chi$ $\sigma$

 

[Ackbach]

I believe if $f$ is uniformly $r$-continuous about some finite neighborhood of $r_0$, that will be a sufficient (not sure about necessary) condition to ensure that you can do this. Some relevant theorems would be the Lebesgue Dominated Convergence Theorem, the Monotone Convergence Theorem, and Fatou's Lemma. For a more Riemannian approach, you could check out Theorem 7.16 in Baby Rudin. You haven't given much information on the $r$-dependence of $f$, so these theorems may or may not apply.

 

[chisigma]

I believe if $f$ is uniformly $r$-continuous about some finite neighborhood of $r_0$, that will be sufficient (not sure about necessary) conditions to ensure that you can do this. Some relevant theorems would be the Lebesgue Dominated Convergence Theorem, the Monotone Convergence Theorem, and Fatou's Lemma. For a more Riemannian approach, you could check out Theorem 7.16 in Baby Rudin. You haven't given much information on the $r$-dependence of $f$, so these theorems may or may not apply.

The case was...

$\displaystyle \lim_{r \rightarrow 0} i\ \int_{0}^{\pi} \frac{e^{i\ r\ e^{i\ \theta}} - 1}{r}\ e^{- i\ \theta}\ d \theta = i\ \int_{0}^{\pi} \lim_{r \rightarrow 0} \frac{e^{i\ r\ e^{i\ \theta}}- 1}{r}\ e^{- i\ \theta}\ d \theta = - \int_{0}^{\pi} \lim_{r \rightarrow 0} e^{i\ r\ e^{i\ \theta}}\ d \theta = - \int_{0}^{\pi} d \theta = - \pi\ (1) $

... where I applied l'Hopital rule...

Kind regards

$\chi$ $\sigma$

 

[Euge]

The case was...

$\displaystyle \lim_{r \rightarrow 0} i\ \int_{0}^{\pi} \frac{e^{i\ r\ e^{i\ \theta}} - 1}{r}\ e^{- i\ \theta}\ d \theta = i\ \int_{0}^{\pi} \lim_{r \rightarrow 0} \frac{e^{i\ r\ e^{i\ \theta}}- 1}{r}\ e^{- i\ \theta}\ d \theta = - \int_{0}^{\pi} \lim_{r \rightarrow 0} e^{i\ r\ e^{i\ \theta}}\ d \theta = - \int_{0}^{\pi} d \theta = - \pi\ (1) $

... where I applied l'Hopital rule...

Kind regards

$\chi$ $\sigma$

If the integrals are treated as Lebesgue integrals, then your limit procedure is valid by the dominated convergence theorem. First put the $i$ inside the integral. Let $f_r(\theta)$ be the resulting integrand. A version of the mean value theorem shows that $\sup_{r > 0}|f_r| \le 1$. Since $\int_0^\pi 1 \, d\theta$ is finite, we deduce from the dominated convergence theorem (for real variable limits) that

\(\displaystyle \lim_{r\to 0} \int_0^{\pi} f_r(\theta)\, d\theta = \int_0^{\pi} \lim_{r\to 0} f_r(\theta)\, d\theta.\)

Now

\(\displaystyle \lim_{r\to 0} f_r(\theta) = ie^{-i\theta}\frac{d}{dr}|_{r = 0}( e^{ire^{i\theta}}) = ie^{-i\theta} \cdot ie^{i\theta} = -1.\)

Hence

\(\displaystyle \int_0^{\pi} \lim_{r\to 0} f_r(\theta)\, d\theta = \int_0^{\pi} -1\, d\theta = -\pi.\)

If you try to use the monotone convergence theorem or Fatou's lemma for this problem, it'll have to be applied to the real and imaginary parts of $f_r$, separately.

 

[Opalg]

The case was...

$\displaystyle \lim_{r \rightarrow 0} i\ \int_{0}^{\pi} \frac{e^{i\ r\ e^{i\ \theta}} - 1}{r}\ e^{- i\ \theta}\ d \theta = i\ \int_{0}^{\pi} \lim_{r \rightarrow 0} \frac{e^{i\ r\ e^{i\ \theta}}- 1}{r}\ e^{- i\ \theta}\ d \theta = - \int_{0}^{\pi} \lim_{r \rightarrow 0} e^{i\ r\ e^{i\ \theta}}\ d \theta = - \int_{0}^{\pi} d \theta = - \pi\ (1) $

In general, you should not smuggle a limit past an integral without some justification such as the theorems suggested by Ackbach. But you can justify the above result (1) by a direct argument.

If $z = re^{i\theta}$ then \(\displaystyle e^{i r e^{i\ \theta}} - 1 = e^{iz}-1 = \cos z - 1 + i\sin z.\) As $z\to0,$ $\cos z = 1 + O(|z|^2)$ and $\sin z = z + O(|z|^3)$. Thus $$\frac{\cos z - 1 + i\sin z}{|z|} = \frac{iz}{|z|} + O(|z|) = ie^{i\theta} + O(|z|),$$ so that $$i\int_0^\pi \frac{e^{i r e^{i\ \theta}} - 1}r\,e^{-i\theta}\,d\theta = i\int_0^\pi \bigl(ie^{i\theta} + O(|z|) \bigr) e^{-i\theta}\,d\theta = -\pi + O(|z|).$$ As $r = |z| \to0$, this converges to $-\pi$, as required.

I think that the above argument amounts to the assertion that the limit \(\displaystyle \lim_{r\to0} \frac{e^{i r e^{i\ \theta}} - 1}r = ie^{i\theta}\) is uniform in $\theta$. That will allow one of Ackbach's theorems to apply.