# Show that $\frac{1}{2 \pi i} \oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\cdot \frac{\mathrm{d} \xi}{\xi^{n+1}} \to 0$ as $n \to \infty$

Let $$f = (f_0,f_1,\ldots,f_n,\ldots) \in \mathcal{P}(\mathbb N)$$ be a probability distribution on $$\mathbb N$$ and denote by $$\hat{f}(z) = \sum_{n\geq 0} z^n f_n$$ for its probability generating function. It has been shown in this post that we have the following Bromwich's inversion formula (using complex integration over the positively oriented unit circle $$\mathbb{S}^1$$): $$\sum\limits_{i > n} f_i = \frac{1}{2 \pi i} \oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\cdot \frac{\mathrm{d} \xi}{\xi^{n+1}} \label{1}\tag{1}$$ for each $$n\geq 0$$. Note that the left hand side of equation \eqref{1} represents the probability $$\mathbb{P}(X > n)$$ for a $$f$$-distributed random variable $$X$$. Now my question is, since we have the obvious asymptotic limit $$\lim\limits_{n \to \infty} \mathbb{P}(X > n) = 0.$$ How can we show that the right hand side of equation \eqref{1} also tends to zero as $$n \to \infty$$ (without using the probabilistic interpretation of the contour integration)?

• I am failing to understand the content of this question, if $A_n = B_n$ and $A_n\to 0$, then $B_n \to 0$ more or less trivially? Commented Aug 13, 2023 at 16:00
• Yes, but the goal is to use complex integration to show that the integral converges to zero, WITHOUT using the probabilistic interpretation. I am sorry if this causes you some confusion. Commented Aug 13, 2023 at 16:42

$$\newcommand\R{\mathbb R}$$Assume that the integral $$I_n:=\oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\,\frac{d\xi}{\xi^{n+1}} =\int_0^{2\pi} \frac{1-\hat{f}(e^{it})}{1-e^{it}}\,\frac{e^{it}\,i\,dt}{e^{i(n+1))t}}$$ exists in the Lebesgue sense -- that is, $$\int_0^{2\pi} \Big|\frac{1-\hat{f}(e^{it})}{1-e^{it}}\Big|\,dt<\infty.$$ Then $$I_n=\int_0^{2\pi} \frac{1-\hat{f}(e^{it})}{1-e^{it}}\,\frac{\,i\,dt}{e^{int}}\to0$$ (as $$n\to\infty$$) by the Riemann–Lebesgue lemma.
Detail: The Riemann–Lebesgue lemma states that $$\int_\R g(t)e^{-itx}\,dt\to0$$ as $$|x|\to\infty$$ if $$\int_\R |g(t)|\,dt<\infty$$. Here the Riemann–Lebesgue lemma is used with $$g(t):=\frac{1-\hat{f}(e^{it})}{1-e^{it}}\,i\,1(0.
• Hi Professor, I think your claimed identity $\oint_{\mathbb{S}^1} \left|\frac{1-\hat{f}(\xi)}{1-\xi}\,\frac{d\xi}{\xi^{n+1}}\right| =\int_0^{2\pi} \left|\frac{1-\hat{f}(e^{it})}{1-e^{it}}\right|\,dt<\infty$ is wrong as the notation for $|d\xi|$ is problematic when $\xi \in \mathbb{C}$ (or at least you should define what $d\xi$ means when $\xi \in \mathbb{S}^1 \subset \mathbb{C}$). Also, you invoked the RL lemma but the statement in the wiki is not the same one as in the situation here (the range of integration is not infinity and the integrand is not of the form in the wiki). Commented Aug 13, 2023 at 18:08
• @FeiCao : I see nothing wrong in what was written here. Yet, I have rewritten the answer in a way that will hopefully be clearer. As for the Riemann–Lebesgue lemma, here it is used with $n=1$ and $\frac{1-\hat{f}(e^{it})}{1-e^{it}}\,i\,1(0<t<2\pi)$ in place of $f(t)$ for a general integrable function $f$. Let me know if anything still seems unclear to you. Commented Aug 13, 2023 at 18:51
• The thing is that you have two different $I_n$ which confuses me a lot. What bothers me the most is the application of Riemann-Lebesgue lemma. The interval of integration is $[0,2\pi]$ instead of $\mathbb R$ (or $\mathbb{R}_+$) so I do not see that the claim is a direct consequence of the lemma. Commented Aug 13, 2023 at 18:57
• There was a typo there, which is now fixed. The second expression for $I_n$ just spells out the definition of the complex integral. That the integration is over $(0,2\pi)$ rather than $\mathbb R$ is taken care of by the factor $1(0<t<2\pi)$, as explained in the added Detail. Let me know if anything still seems unclear to you. Commented Aug 13, 2023 at 19:09