Riemann-Stieltjes Integration and Asymptotics

As the title suggests, today we are going to do some integral calculus. First, let's recall the definition of Riemann integral:

Riemann Integral and Its Formal Definition

Traditionally, an integral of some function $f(x)$ over some interval $[a,b]$ is defined as the signed area of $f(x)$ over the curve:

$\int_a^bf(x)\mathrm dx\triangleq\text{Signed area under $f(x)$ where $x\in[a,b]$}$

and Riemann integral is one way to define it rigorously. Particularly, we define it as the sum of the areas of tiny rectangles:

$\int_a^bf(x)\mathrm dx\approx\sum_{k=1}^Nf(\xi_k)(x_k-x_{k-1})$

where $\xi_k$ are sampled in $[x_{k-1},x_k]$ and the increasing sequence $\{x_k\}$ is called a partition of the interval $[a,b]$:

$a=x_0\le x_1\le x_2\le\cdots\le x_N=b$

To formalize Riemann integral, let's define the $\operatorname{mesh}\{x_n\}$ as the length of the maximum of interval in a partition $\{x_n\}$:

$\operatorname{mesh}\{x_n\}\triangleq\max_{1\le n\le N}(x_n-x_{n-1})$

Then we say that some function $f(x)$ is Riemann integrable if for all $\varepsilon>0$, there exists $\delta>0$ such that when $\operatorname{mesh}\{x_n\}\le\delta$ we always have

$\left|\sum_{k=1}^Nf(\xi_k)(x_k-x_{k-1})-\int_a^bf(x)\mathrm dx\right|<\varepsilon$

Alternatively, if some function $f(x)$ is Riemann-integrable on $[a,b]$, then the limit

$\lim_{\operatorname{mesh}\{x_n\}\to0}\sum_{k=1}^Nf(\xi_k)(x_k-x_{k-1})$

exists and converges to $\int_a^bf(x)\mathrm dx$.

From Riemann Integral to Riemann-Stieltjes Integral

Although Riemann integral appears to be sufficient to integrate functions, it is not friendly to integrate piecewise continuous functions. Let's first look at its definition:

$\int_a^bf(x)\mathrm dg(x) =\lim_{\operatorname{mesh}\{x_n\}\to0}I(x_n,\xi_n)\triangleq\lim_{\operatorname{mesh}\{x_n\}\to0}\sum_{k=1}^Nf(\xi_k)[g(x_k)-g(x_{k-1})]$

In order for this it to exist, we need to set up constraints on $f(x)$ and $g(x)$:

Theorem 1: The Riemann-Stieltjes integral $\int_a^bf\mathrm dg$ exists if $f$ is continuous on $[a,b]$ and $g$ is of bounded variation on $[a,b]$

When we say $g$ is of bounded variation, we mean that the following quantity exists:

$\displaystyle{\operatorname{Var}_{[a,b]}}(g)\triangleq\sup\sum_k|g(x_k)-g(x_{k-1})|$

Proof. Let's define $\{y_n\}=\{y_1,y_2,\dots,y_M\}$ as another partition of $[a,b]$ such that $\{x_n\}$ is its subsequence and $\eta_k$ be the sampled abscissa in $[y_k,y_{k-1}]$, so if we designate $P_k$ to be the set of $m$ such that $y_m$'s are contained in interval $(x_{k-1},x_k]$:

$P_k=\{m|x_{k-1}<y_m\le x_k\}$

then we have

$\sum_{m\in P_k}[g(y_m)-g(y_{m-1})]=g(x_k)-g(x_{k-1})$

which implies

$I(x_n,\xi_n)=\sum_{k=1}^N\sum_{m\in P_k}f(\xi_k)[g(y_m)-g(y_{m-1})]\tag1$ $I(y_n,\eta_n)=\sum_{k=1}^N\sum_{m\in P_k}f(\eta_m)[g(y_m)-g(y_{m-1})]\tag2$

Because $f(x)$ is uniformly continuous within $[a,b]$, we know that for every $\varepsilon>0$ there exists $\delta>0$ such that when $s,t\in[a,b]$ satisfy $|s-t|\le\operatorname{mesh}\{x_n\}\le\delta$ then $|f(s)-f(t)|<\varepsilon$. Accordingly, if we were to take the absolute values of (1) and (2), then

$\begin{aligned} |I(x_n,\xi_n)-I(y_n,\eta_ n)| &=\sum_{n=1}^N\sum_{m\in P_k}|f(\eta_m)-f(\xi_k)||g(y_m)-g(y_{m-1})| \\ &\le\varepsilon\sum_{m=1}^M|g(y_m)-g(y_{m-1})| \le\varepsilon\operatorname{Var}_{[a,b]}(g) \end{aligned}$

Now, let $\{z_n\}$ be another partition of $[a,b]$, $\zeta_n$ be its correponding samples abscissa and $\{y_n\}$ be the union of both partitions, then

$\begin{aligned} |I(x_n,\xi_n)-I(z_n,\zeta_n)| &=|I(x_n,\xi_n)-I(y_n,\eta_n)-[I(z_n,\zeta_n)-I(y_n,\eta_n)]| \\ &\le|I(x_n,\xi_n)-I(y_n,\eta_n)|+|I(z_n,\zeta_n)-I(y_n,\eta_n)| \\ &\le2\varepsilon\operatorname{Var}_{[a,b]}(g) \end{aligned}$

which indicates the validness of this theorem. $\square$

Properties of Riemann-Stieltjes Integral

In addition to the constraint for the Riemann-Stieltjes integral to exist, we can also transform it into a Riemann integral at specific occasions:

Theorem 2: If $g'(x)$ exists and is continuous on $[a,b]$ then

$\int_a^bf(x)\mathrm dg(x)=\int_a^bf(x)g'(x)\mathrm dx\tag3$

Proof. Since $g(x)$ is differentiable, we can use mean value theorem to guarantee the existence of $\xi_k\in[x_{k-1},x_k]$ such that $g(x_k)-g(x_{k-1})=g'(\eta_k)(x_k-x_{k-1})$, so

$\begin{aligned} \sum_{k=1}^N|g(x_k)-g(x_{k-1})| &=\sum_{k=1}^N|g'(\eta_k)|(x_k-x_{k-1}) \\ &\to\int_a^b|g'(x)|\mathrm dx\quad(\operatorname{mesh}\{x_n\}\to0) \end{aligned}$

As a result, $g(x)$ is of bounded variation, implying the existence of the left hand side of (3).

$\begin{aligned} S(x_n,\xi_n) &=\sum_{k=1}^Nf(\xi_k)g'(\eta_k)(x_k-x_{k-1}) \\ &=\underbrace{\sum_{k=1}^Nf(\xi_k)g'(\xi_k)(x_k-x_{k-1})}_{T_1} +\underbrace{\sum_{k=1}^Nf(\xi_k)[g'(\eta_k)-g'(\xi_k)](x_k-x_{k-1})}_{T_2} \\ \end{aligned}$

By the uniform continuity of $g'(x)$, we know that for all $\varepsilon>0$ there exists $\delta>0$ such that $|g'(s)-g'(t)|<\varepsilon$ whenever $|s-t|<\delta$, indicating $T_1\to\int_a^bfg'$ and $T_2\to0$ as $\operatorname{mesh}\{x_n\}\to0$. Accordingly, we arrive at the conclusion that

$\int_a^bf(x)\mathrm dg(x)=\int_a^bf(x)g'(x)\mathrm dx$

thus completing the proof. $\square$

In addition, we can also apply integration by parts on Riemann-Stieltjes integrals. Particularly, if we assume $f$ has a continuous derivative and $g$ is of bounded variation on $[a,b]$, then

$\int_a^bf(x)\mathrm dg(x)=f(x)g(x)|_q^b-\int_a^bg(x)f'(x)\mathrm dx$

Riemann-Stieltjes Integral and Partial Summation

Let $h(n)$ be some arithmetic function and $H(x)$ be its summatory function

$R(x)=\sum_{n\le x}r(n)\tag4$

Let $f(t)$ have continuous derivative on $[0,\infty)$ and $b>a>0$ then we have

$\int_a^bf(x)\mathrm dR(x)=\sum_{k=1}^Nf(\xi_k)[R(x_k)-R(x_{k-1})]$

where we require that $\mathrm{mesh}\{x_n\}<1$. Recall (4), we observe that $R(x)$ is a step function that only jumps at integer values, so we only need to sum over $k_n$'s such that $n\in(x_{k_n-1},x_{k_n}]$. Hence, this integral becomes a summation that sums over integers values in $(a,b]$:

$\begin{aligned} \int_a^bf(x)\mathrm dR(x) &=\sum_{a<n\le b}f(\xi_{k_n})r(n) \\ &=\sum_{a<n\le b}f(n)r(n)+\sum_{a<n\le b}[f(\xi_{k_n})-f(n)]r(n) \\ \end{aligned}$

Since $f$ is uniformly continuous on $[a,b]$, we have that $|f(x)-f(y)|<\delta$ when ever $|x-y|<\varepsilon$, thus the second summation is of $\mathcal O(\delta)$, leaving us

$\int_a^bf(x)\mathrm dR(x)=\sum_{a<n\le b}f(n)r(n)\tag5$

Employing (5) in different situations can give us plentiful brilliant results. Let's have a look at some of them:

Asymptotic Expansions

It was well-known that the harmonic series $1+\frac12+\frac13+\cdots$ diverges, and we can provide a formal proof using Riemann-Stieltjes integral:

$\begin{aligned} \sum_{n\le N}\frac1n &=\int_{1-\varepsilon}^N{\mathrm d\lfloor x\rfloor\over x} \\ &=\left.\lfloor x\rfloor\over x\right|_{1-\varepsilon}^N+\int_{1-\varepsilon}^N{\lfloor x\rfloor\over x^2}\mathrm dx \\ &=\int_{1-\varepsilon}^N{\mathrm dx\over x}+1-\int_{1-\varepsilon}^N{\{x\}\over x^2}\mathrm dx \\ &=\int_1^N{\mathrm dx\over x}+1-\int_1^N{\{x\}\over x^2}\mathrm dx \\ &=\log N+1-\int_1^\infty{\{x\}\over x^2}\mathrm dx+\int_N^\infty{\{x\}\over x^2}\mathrm dx \\ &=\log N+\gamma+\mathcal O\left(\frac1N\right) \end{aligned}$

Since $\log N\to\infty$ as $N\to\infty$, we conclude that the harmonic series diverges.

Evaluation of an Interesting Series

$\sum_{n=1}^\infty{(-1)^n\log n\over n}$

Now, let's first consider the finite case:

$\begin{aligned} \sum_{n=1}^N{(-1)^n\log n\over n} &=2\sum_{n\le N/2}{\log(2n)\over2n}-\sum_{n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\ &=\sum_{n\le N/2}{\log2+\log n\over n}-\sum_{n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\ &=\log2\sum_{n\le N/2}\frac1n-\sum_{N/2<n\le N}{\log n\over n}+\mathcal O\left(\log N\over N\right) \\ \end{aligned}$

In fact, using Riemann-Stieltjes integral, we can show

$\begin{aligned} \sum_{N/2<n\le N}{\log n\over n} &=\int_{N/2}^N{\log x\over x}\mathrm d\lfloor x\rfloor \\ &={N\log(N)-N\log(N/2)\over N}-\int_{N/2}^N[x-\{x\}]\mathrm d\left(\log x\over x\right)+\mathcal O\left(\frac1n\right) \\ &=\log2-\int_{N/2}^N\left({1-\log x\over x}\right)\mathrm dx+\mathcal O\left(\log N\over N\right) \\ &=\int_{N/2}^N{\log x\over x}\mathrm dx+\mathcal O\left(\log N\over N\right) \\ &=\frac12[\log^2N-\log^2(N/2)]+\mathcal O\left(\log N\over N\right) \\ &=\frac12[\log N+\log(N/2)][\log N-\log(N/2)]+\mathcal O\left(\log N\over N\right) \\ &=\frac12\log2[2\log N-\log2]+\mathcal O\left(\log N\over N\right) \\ &=\log2\log N-\frac12\log^22+\mathcal O\left(\log N\over N\right) \end{aligned}$

Now, employing this obtained identity and the asymptotic formula for harmonic series yields:

$\begin{aligned} \sum_{n\le N}{(-1)^n\log n\over n} &=\log2(\log N+\gamma)-\log2\log N+\frac12\log^22+\mathcal O\left(\log N\over N\right) \\ &=\gamma\log2+\frac12\log^22+\mathcal O\left(\log N\over N\right) \end{aligned}$

Now, take the limit $n\to\infty$ on both side gives

$\sum_{n\ge1}{(-1)^n\log n\over n}=\gamma\log2+\frac12\log^22$

The Prime Number Theorem

If we were to define the prime indicator function

$\mathbf1_p(n)= \begin{cases} 1 & \text{$n$ is a prime} \\ 0 & \text{otherwise} \end{cases}$

Then the prime counting function can be defined as

$\pi(x)=\sum_{p\le x}1=\sum_{n\le x}\mathbf1_p(n)$

Now, let's also define Chebyshev's function $\vartheta(x)$:

$\vartheta(x)=\sum_{p\le x}\log p=\sum_{n\le x}\mathbf1_p(n)\log n$

Hence, we have

$\begin{aligned} \pi(x) &=\sum_{n\le x}{1\over\log n}\cdot\mathbf1_p(n)\log n\\ &=\int_{2-\varepsilon}^x{\mathrm d\vartheta(t)\over\log t} \\ &={\vartheta(x)\over\log x}+\int_2^x{\vartheta(t)\over t\log^2t} \end{aligned}$

It is known that $\vartheta(x)\sim x$, so plugging it into the above equation gives

$\pi(x)\sim{x\over\log x}$

which is the prime number theorem.

Conclusion

To sum up, in this blog, we first define and explore the Riemann-Stieltjes integral, then use this integration technique to solve problems via asymptotic expansion. Lastly, we provide a proof for the prime number theorem.