How can we prove the uniform convergence of an integrand, without using the dominated convergence theorem?

Question

$$\int_{-\pi}^\pi\frac{(\delta e^{i\theta})^{s-1}\cdot\exp(-a\delta\cdot e^{i\theta})}{1-\exp(-\delta\cdot e^{i\theta})}\cdot i\delta e^{i\theta}\,\mathrm{d}\theta$$

Edit: Sorry, forgot to add: a $\geq$ 1 is a real number and s is complex, with Re(s) > 1.

I need to prove that an "ugly" integral of $f_\delta(z)$ converges to $0$ as $\delta$ goes to $0$. $f$ is a product/composition of a bunch of exponential functions, so I feel like it should already be uniformly convergent to $f(z)$. If I could just pull the limit into the integral, I would be done because $f_\delta(z)$ clearly converges to $0$ regardless of $z$ (is that uniform convergence already or just pointwise convergence?).

And please don't suggest the dominated or bounded convergence theorems. Bounding the function is the entire issue here, and I can not find any bound for the life of me because bounding the absolute value of a weird complex function is extremely awkward.

Edit: The function is a complex one, but its a "normal" integral and not a contour integral. Or rather, its part of a contour integral that goes along the real axis.

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$If we pass back to $z=\delta e^{i\theta}$, which I suspect is the source of the integral, then we want to analyse: $$I_\delta=\oint_\gamma\frac{z^{s-1}\cdot e^{-az}}{1-e^{-z}}\d z$$Where $\gamma$ is the anticlockwise circle of radius $\delta>0$ about the origin. We can note that: $$1-e^{-z}=z-\frac{1}{2}z^2+o(z^2)$$Dividing yields: $$\frac{1}{1-e^{-z}}=\frac{1}{z}+O(1)$$And: $$I_\delta=\oint_\gamma z^{s-2}e^{-az}\d z+\oint_\gamma z^{s-1}e^{-az}\cdot O(1)\d z$$

Let's see. There's not much context here, but I'll let $a\in\Bbb C$ be arbitrary and I'll let $s\in\Bbb C$ be arbitrary too. The first integrand can be bounded from above by: $$|z^{s-2}e^{-az}|\le\delta^{\Re s-2}\exp(|\Re a|\cdot\delta)\\|z^{s-1}e^{-az}\cdot O(1)|\le\delta^{\Re s-1}\exp(|\Re a|\cdot\delta)\cdot\sup_{z\in\gamma}|O(1)|$$By the basic estimation lemma and the triangle inequality we get: $$|I_\delta|\le2\pi\delta\cdot\exp(|\Re a|\cdot\delta)\cdot[\delta^{\Re s-2}+\delta^{\Re s-1}\cdot\sup_{z\in\gamma}|O(1)|]$$Distributing the $\delta$, we see that the 'dangerous' term is $\delta^{\Re s-1}$. As long as $\Re s>1$, all the terms in $\delta$ vanish as $\delta\to0^+$. The exponential term tends to $1$ - irrelevant - and the sup-$|O(1)|$ term tends to a constant. In fact, it tends to $1/2$. The $2\pi$ constant is also irrelevant: in conclusion, the integral vanishes as $\delta\to0^+$, provided $\Re s>1$ and with no conditions on $a$. It may be that the integral vanishes for a wider domain - I don't know - but that'd require a harder analysis.

It was rigorous to use the $O$s and $o$s in this way since I always treated them as genuine functions, don't worry!

You could have conducted the very same proof in the Riemann form. However you can't see the wood for the trees with all those $e^{i\theta}$s floating around - at least, I can't. It is occasionally helpful to pass to the Riemann form to evaluate such limits, but certainly not always.

Answer 2

As noted before, the integral in the OP is \begin{align} I:=\int_{-\pi}^{\pi}\frac{(\delta e^{i\theta})^{s-1}e^{-a\delta e^{i\theta}}}{1-e^{-\delta e^{i\theta}}}\,i\delta e^{i\theta}\,d\theta &= \int_{-\pi}^{\pi}\frac{(\delta e^{i\theta})^{s-1}e^{-a\delta e^{i\theta}}}{\frac{1-e^{-\delta e^{i\theta}}}{\delta e^{-i\theta}}}\,i\,d\theta \end{align}

Since $\lim_{z\rightarrow0}\frac{1-e^{-z}}{z}=1$, there is $\delta_0>0$ such that $|z|<\delta_0$ implies that $\Big|\frac{1-e^{-z}}{z}\Big|>\frac12$.

Let $s=\sigma+i\tau$. Assuming that $\sigma>1$ and $a>0$, for $0<\delta<\delta_0$ $$|I|\leq 2\delta^{\sigma-1}\int_{-\pi}^{\pi} e^{-\theta\tau}e^{-a\delta\cos\theta}\,d\theta<4\pi \delta^{s-1}e^{\pi\tau}e^{a\delta}\xrightarrow{\delta\rightarrow0}0$$

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Comment: Here I am interpreting $(e^{i\theta})^{s-1}$ as $e^{i(s-1)\theta}$. The integral may also be interpreted as a counter integral by

1. taking the principal brach of logarithm ($\mathbb{C}\setminus\{re^{\pi i}: r>0\}$),
2. and integrating the function $$ f(z)=\frac{z^{s-1}e^{-az}}{1-e^{-z}}$$ along the open path $\gamma(\theta)=\delta e^{i\theta}$ , $|\theta|<\pi$.