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:
∫abf(x)dx≜Signed area under f(x) where x∈[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:
∫abf(x)dx≈k=1∑Nf(ξk)(xk−xk−1)
where ξk are sampled in [xk−1,xk] and the increasing sequence xk is called a partition of the interval [a,b]:
a=x0≤x1≤x2≤⋯≤xN=b
To formalize Riemann integral, let's define the meshxn as the length of the maximum of interval in a partition xn:
mesh{xn}≜1≤n≤Nmax(xn−xn−1)
Then we say that some function f(x) is Riemann integrable if for all ε>0, there exists δ>0 such that when meshxn≤δ we always have
∣∣k=1∑Nf(ξk)(xk−xk−1)−∫abf(x)dx∣∣<ε
Alternatively, if some function f(x) is Riemann-integrable on [a,b], then the limit
mesh{xn}→0limk=1∑Nf(ξk)(xk−xk−1)
exists and converges to ∫abf(x)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:
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 ∫abfdg 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:
Var[a,b](g)≜supk∑∣g(xk)−g(xk−1)∣
Proof. Let's define yn=y1,y2,…,yM as another partition of [a,b] such that xn is its subsequence and ηk be the sampled abscissa in [yk,yk−1], so if we designate Pk to be the set of m such that ym's are contained in interval (xk−1,xk]:
Because f(x) is uniformly continuous within [a,b], we know that for every ε>0 there exists δ>0 such that when s,t∈[a,b] satisfy ∣s−t∣≤meshxn≤δ then ∣f(s)−f(t)∣<ε. Accordingly, if we were to take the absolute values of (1) and (2), then
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
∫abf(x)dg(x)=∫abf(x)g′(x)dx(3)
Proof. Since g(x) is differentiable, we can use mean value theorem to guarantee the existence of ξk∈[xk−1,xk] such that g(xk)−g(xk−1)=g′(ηk)(xk−xk−1), so
By the uniform continuity of g′(x), we know that for all ε>0 there exists δ>0 such that ∣g′(s)−g′(t)∣<ε whenever ∣s−t∣<δ, indicating T1→∫abfg′ and T2→0 as meshxn→0. Accordingly, we arrive at the conclusion that
∫abf(x)dg(x)=∫abf(x)g′(x)dx
thus completing the proof. □
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
∫abf(x)dg(x)=f(x)g(x)∣qb−∫abg(x)f′(x)dx
Riemann-Stieltjes Integral and Partial Summation
Let h(n) be some arithmetic function and H(x) be its summatory function
R(x)=n≤x∑r(n)(4)
Let f(t) have continuous derivative on [0,∞) and b>a>0 then we have
∫abf(x)dR(x)=k=1∑Nf(ξk)[R(xk)−R(xk−1)]
where we require that meshxn<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 kn's such that n∈(xkn−1,xkn]. Hence, this integral becomes a summation that sums over integers values in (a,b]:
It is known that ϑ(x)∼x, so plugging it into the above equation gives
π(x)∼logxx
which is the prime number theorem.
Conclusion
To sum up, in this blog, we first define and explore the Riemann-Stieltjes integral, then uses this integration techniques to solve problems via asymptotic expansion. Lastly, we provide a conditional proof for the prime number theorem.