# The whole of analytic number theory rests on one marvellous formula due to Leonhard Euler ( ): n s = primes p. 1 p

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1 Chapter Euler s Product Formula. The Product Formula The whole of analytic number theory rests on one marvellous formula due to Leonhard Euler ( ): n N, n>0 n s = primes p ( p s ). Informally, we can understand the formula as follows. By the Fundamental Theorem of Arithmetic, each n is uniquely expressible in the form n = 2 e 2 3 e 3 5 e5, where e 2, e 3, e 5,... are non-negative integers (all but a finite number being 0). Raising this to the power s, Adding for n =, 2, 3,..., n s = 2 e 2s 3 e 3s 5 e 5s. n s = ( + 2 s + 2 2s + ) ( + 3 s + 3 2s + ) ( + 5 s + 5 2s + ), each term on the left arising from just one product on the right. But for each prime p, + p s + p 2s + = ( p s), and the result follows. Euler s Product Formula equates the Dirichlet series n s on the left with the infinite product on the right.

2 .2. INFINITE PRODUCTS 2 To make the formula precise, we must develop the theory of infinite products, which we do in the next Section. To understand the implications of the formula, we must develop the theory of Dirichlet series, which we do in the next Chapter..2 Infinite products.2. Definition Infinite products are less familiar than infinite series, but are no more complicated. Both are examples of limits of sequences. Definition.. The infinite product c n n N is said to converge to l 0 if the partial products P n = c m l as n. 0 m n We say that the infinite product diverges if either the partial products do not converge, or else they converge to 0 (as would be the case for example if any factor were 0). Proposition.. If c n is convergent then c n. Proof We have c n = P n. P n Since P n l and P n l, it follows that c n l l =. It is usually more convenient to write the factors in the form c n = + a n. In these terms, the last Proposition states that ( + an ) convergent = a n 0.

3 .2. INFINITE PRODUCTS The complex logarithm The theory of infinite products requires some knowledge of the complex logarithmic function. Suppose z 0. Let z = re iθ, where r > 0. We are interested in solutions w C of If w = x + iy then ie e w = z. e x = r, e iy = e iθ, x = log r, y = θ + 2nπ for some n Z. Just one of these solutions to e w = z satisfies π < y = I(w) π. We call this value of w the principal logarithm of z, and denote it by Log z. Thus e Log z = z, π < I(z) π. Then The general solution of e w = z is Now suppose It follows that w = Log z + 2nπi (n Z). w = Log z, w 2 = Log z 2. e w +w 2 = z z 2 = e Log(z z 2 ) Log(z z 2 ) = Log z + Log z 2 + 2nπi, where it is easy to see that n = 0, or. If R(z) > 0 then z = re iθ with π/2 < θ < π/2. It follows that and so Thus R(z ), R(z 2 ) > 0 = π/2 < I(Log z ), I(Log z 2 ) < π/2; π < I(Log z + Log z 2 ) < π. R(z ), R(z 2 ) > 0 = Log(z z 2 ) = Log z + Log z 2. In particular, this holds if z, z 2 < (Fig.).

4 .2. INFINITE PRODUCTS 4 r z θ Figure.: z <, Log z = log r + iθ.2.3 Convergence Proposition.2. Suppose a n for n N. Then ( + an ) converges Log( + a n ) converges. Proof Suppose Log( + a n ) converges to S. Let S n = m n Log( + a m ). Then But e Sn = m n( + a m ). S n S = e Sn e S. Thus ( + a n ) converges to e s. Conversely, suppose ( + a n ) converges. Let P n = ( + a n ). m N Given ɛ > 0 there exists N such that if m, n N. P n P m < ɛ

5 .2. INFINITE PRODUCTS 5 It follows that if m, n N then Log(P n /P N ) = Log(P m /P N ) + Log(P n /P m ). In particular (taking m = n ), Log(P n /P N ) = Log(P n /P N ) + Log( + a n ). Hence Log(P n /P N ) = Log( + a m ). N<m m Since P n l = Log(P n /L N ) Log(l/P N ), we conclude that n>n( + a n ) converges to Log(l/P N ); and in particular, n 0 Log( + a n ) is convergent. Proposition.3. Suppose a n for n N. Then an convergent = ( + a n ) convergent. Proof The function Log( + z) is holomorphic in z <, with Taylor expansion Log( + z) = z z 2 /2 + z 3 /3. Thus if z < /2 then Log( + z) z + z 2 + z 3 + = z z 2 z. Now suppose a n converges. Then a n 0; and so for n N. It follows that a n /2 Log( + a n ) 2 a n for n N. Hence Log( + an ) converges.

6 .3. PROOF OF THE PRODUCT FORMULA 6.3 Proof of the product formula Proposition.4. For R(s) >, n s = n N, n>0 primes p ( p s ), in the sense that each side converges to the same value. Proof Let σ = R(s). Then Thus Now and so N M+ N M+ n s = n σ. n σ n σ n s n N n M N M+ n σ. x σ dx; x σ dx = ( M σ N σ) σ 0 as M, N. Hence n s is convergent, by Cauchy s criterion. On the other hand, ( p s ) is absolutely convergent, since p s = p σ n σ, which we just saw is convergent. Hence ( p s ) is convergent, by Proposition.3; and so therefore is ( p s ). To see that the two sides are equal note that p N ( ) χ(p)p s = χ(n)n s + n N χ(n)n s, where the second sum on the right extends over those n > N all of whose prime factors are N. As N, the right-hand side n s, since this sum is absolutely convergent; while by definition, the left-hand side ( p s ). We conclude that the two sides converge to the same value.

7 .4. EULER S THEOREM 7.4 Euler s Theorem Proposition.5. (Euler s Theorem) primes p p =. Proof Suppose /p is convergent. Then ( ) p is absolutely convergent, and so converges to l say, by Proposition?? It follows that ( ) l. p But p N N n p N ( p since each n on the left is expressible in the form with p,..., p r N. Hence /n is convergent. But n = p e p er r ), Thus N N+ n n+ n > dx n x. dx x = log(n + ). Since log N as N it follows that /n is divergent. Our hypothesis is therefore untenable, and p diverges. This is a remarkably accurate result; only just diverges. For it follows p from the Prime Number Theorem, π(x) x log x,

8 .4. EULER S THEOREM 8 that if p n denotes the n th prime (so that p 2 = 3, p 5 =, etc) then p n n log n. To see that, note that π(p n ) = n (ie the number of primes p n is n). Thus setting x = p n in the Prime Number Theorem, n p n, log p n ie Taking logarithms, hence ie p n n log p n. log p n log n log log p n 0; log n log p n, We conclude that log p n log n. p n n log p n n log n. Returning to Euler s Theorem, we see that /p behaves like /n log n. The latter diverges, but only just, as we see by comparison with dx = log log x. x log x On the other hand, converges for any ɛ > 0, since p n p log ɛ p n log +ɛ n converges by comparison with dx x log +ɛ x = ɛ log ɛ x. What is perhaps surprising is that it is so difficult to pass from Euler s Theorem to the Prime Number Theorem.

9 Chapter 2 Dirichlet series 2. Definition Definition 2.. A Dirichlet series is a series of the form a s + a 2 2 s + a 3 3 s +, where a i C. Remarks.. For n N we set n s = e s log n, taking the usual real-valued logarithm. Thus n s is uniquely defined for all s C. Moreover, m s n s = (mn) s, n s n s = n (s+s ) ; while for all s. s = 2. The use of s rather than s is simply a matter of tradition. The series may of course equally well be written a + a 2 2 s + a 3 3 s The term Dirichlet series is often applied to the more general series a 0 λ s 0 + a λ s + a 2 λ s 2 +, 2

10 2.2. CONVERGENCE 2 2 where and 0 < λ 0 < λ < λ 2 <, λ s s log λ = e Such series often appear in mathematical physics, where the λ i might be, for example, the eigenvalues of an elliptic operator. However, we shall only make use of Dirichlet series of the more restricted type described in the definition above; and we shall always use the term in that sense, referring to the more general series (if at all) as generalised Dirichlet series. 4. It is perhaps worth noting that generalised Dirichlet series include power series f(x) = c n x n, in the sense that if we make the substitution x = e s then f(e s ) = c n e ns = c n (e n ) s. 2.2 Convergence Proposition 2.. Suppose f(s) = a s + a 2 2 s + converges for s = s 0. Then it converges for all s with R(s) > R(s 0 ). Proof We use a technique that might be called summation by parts, by analogy with integration by parts. Lemma. Suppose a n, b n (n N) are two sequences. Let A n = a m, B n = b m. m n m n Then N a n B n = A N B N+ A M B M M N A n b n+. M

11 2.2. CONVERGENCE 2 3 Proof Substituting a n = A n A n, N N a n B n = (A n A n )B n M M N = A n (B n B n+ ) + A N B N+ A M B M M N = A n b n+ + A N B N+ A M B M. M Lemma 2. Suppose a n converges and b n converges absolutely. Then an B n converges. Proof By the previous Lemma, N N a n B n = A N B N+ A M B M A n b n+ M M = A N (B N+ B M ) + (A N A M )B M N A n b n+. M Let a n = A, bounded; say Then bn = B. The partial sums of both series must be A n C, B n D. N a n B n C B N+ B M + D A N A M + C M N b n+. M As M, N, B N+ B M 0, A N A M 0, N b n+ 0. M Hence N a n B n 0 M as M, N ; and therefore a n B n converges, by Cauchy s criterion.

12 2.2. CONVERGENCE 2 4 Let s = s s 0. Then R(s ) > 0. We apply the last Lemma with a n n s 0 for a n, and n s for B n. Thus Hence b n = B n B n = n s (n ) s = s n n x s dx x. n b n s n x s dx x = s n n x σ dx x, where σ = R(s ). Summing, It follows that N N b n s x dx σ M M x = s ( (M ) σ N σ ). σ N b n 0 as M, N. M Thus b n is convergent, and so the conditions of the last Lemma are fulfilled. We conclude that an n s 0 n s = a n n (s 0+s ) = a n n s is convergent. Corollary 2.. A Dirichlet series either. converges for all s, 2. diverges for all s, or 3. converges for all s to the right of a line R(s) = σ 0, and diverges for all s to the left of this line.

13 2.3. ABSOLUTE CONVERGENCE 2 5 σ + it X + it σ 0 σ X Figure 2.: Uniform convergence Definition 2.2. We call σ 0 the abscissa of convergence, setting σ 0 = if the series always converges, and σ 0 = if the series never converges. Proposition 2.2. The function f(s) = a s + a 2 2 s + is holomorphic in the half-plane R(s) > σ 0. Proof Suppose σ > σ 0. The argument in the proof of the last Proposition actually shows that a n n s converges uniformly in any rectangle {S = x + it : σ x X; T t T } strictly to the right of R(s) = σ 0 (Fig 2.), since N M b n s σ M σ s s 0 σ σ 0 M (σ σ 0) in this region. Thus f(s) is holomorphic in this rectangle. holomorphic in the half-plane R(s) > σ 0. We conclude that f(s) is 2.3 Absolute convergence Absolute convergence is simpler than convergence, since an n s = a n n σ, where σ = R(s). Thus a Dirichlet series converges absolutely at all, or none, of the points on the line R(s) = σ.

14 2.3. ABSOLUTE CONVERGENCE 2 6 Proposition 2.3. If f(s) = a s + a 2 2 s + converges absolutely for s = s 0 then it converges absolutely for all s with R(s) R(s 0 ). Proof This follows at once from the fact that each term a n n s = a n n σ is a decreasing function of σ. Corollary 2.2. A Dirichlet series either. converges absolutely for all s, 2. does not converge absolutely for any s, or 3. converges absolutely for all s to the right of a line R(s) = σ, and does not converge absolutely for any s to the left of this line. Definition 2.3. We call σ the abscissa of absolute convergence, setting σ = if the series always converges absolutely, and σ = if the series never converges absolutely. Proposition 2.4. We have Proof Suppose Then is convergent. Hence In particular, a n n s is bounded, say σ 0 σ σ 0 +. R(s) > σ 0. f(s) = a s + a 2 2 s + a n n s 0 as n. a n n s C.

15 2.4. THE RIEMANN ZETA FUNCTION 2 7 But then a n n (s++ɛ) Cn (+ɛ) for any ɛ < 0. Since n (+ɛ) converges, it follows that f(s + + ɛ) converges absolutely. We have shown therefore that σ > σ 0 = σ + + ɛ σ for any ɛ > 0, from which it follows that σ 0 + σ. Proposition 2.5. If a n 0 then σ = σ 0. Proof This is immediate, since in this case an n σ = a n n σ. 2.4 The Riemann zeta function Although we have already met the function ζ(s), it may be best to give a formal definition. Definition 2.4. The Riemann zeta function ζ(s) is defined by the Dirichlet series ζ(s) = s + 2 s +. Remarks.. We shall often refer to the Riemann zeta function ζ(s) simply as the zeta function. This is slightly inaccurate, since the term zeta function is applied to a wide range of related functions. However, the Riemann zeta function is the only such function we shall use; so it will cause no confusion if we use the unadorned term zeta function to describe it. 2. For example, there is a zeta function ζ k (s) corresponding to each number field k, defined by ζ k (s) = a N(a) s,

16 2.4. THE RIEMANN ZETA FUNCTION 2 8 where a runs over the ideals in k (or rather, in the ring of integers I(k) = k Z), and N(a) is the number of residue classes moda. Since the unique factorisation theorem holds for ideals, the analogue of Euler s product formula holds: ζ k (s) = p ( N(p) s ), where the product runs over all prime ideals in I(k). This allows the Prime Number Theorem to be extended to give an approximate formula for the number of prime ideals p in the number field k with N(p) n. 3. In another direction, the zeta function ζ E (s) of an elliptic differential (or pseudo-differential) operator E is defined by ζ E (s) = λ s n, where λ n (n = 0,, 2,... ) are the eigenvalues of E (necessarily positive, if E is elliptic). The Riemann zeta function ζ(s) can be interpreted in this sense as the zeta function of the Laplacian operator on the circle S. Proposition 2.6. The abscissa of convergence of the Riemann zeta function is σ 0 =. Proof This follows at once from the fact that n σ < σ >. Let us recall how this is established, by comparing the sum with the integral x σ dx. If n x n, Integrating, n σ < n σ x σ (n ) σ. n n Summing from n = M + to N, N M+ n σ < x σ dx < (n ) σ, N M x σ dx < N+ M n σ,

17 2.4. THE RIEMANN ZETA FUNCTION 2 9 It follows that n σ and x σ dx converge or diverge together. But we can compute the integral directly: if n = then Y x dx = log X log Y, X and so the integral diverges; while if n then Y x σ dx = X σ (M σ N σ ), and so the integral converges if σ > and diverges if σ <. Corollary 2.3. The zeta function ζ(s) is holomorphic in the half-plane R(s) >. We can continue ζ(s) analytically to the half-plane R(s) > 0 in the following way. Proposition 2.7. The Dirichlet series f(s) = s 2 s + 3 s has abscissa of convergence σ 0 = 0, and so defines a holomorphic function in the half-plane R(s) > 0. Proof Suppose σ > 0. Then f(σ) = σ 2 σ + 3 σ converges, since the terms alternate in sign and decrease to 0 in absolute value. It follows, by Proposition 2., that f(s) converges for R(s) > 0. The series certainly does not converge for R(s) < 0, since the terms do not even 0. Thus σ 0 = 0. The abscissa of absolute convergence σ of f(s) is since the terms have the same absolute value as those of ζ(s). Proposition 2.8. If R(s) > then f(s) = ( 2 s )ζ(s). Proof If R(s) > then the Dirichlet series for f(s) converves absolutely, so we may re-arrange its terms: f(s) = s 2 s + 3 s = ( s + 2 s + 3 s + ) 2(2 s + 4 s + ) = ζ(s) 2 2 s ( s + 2 s + ) = ζ(s) 2 2 s ζ(s) = ( 2 s )ζ(s).

18 2.4. THE RIEMANN ZETA FUNCTION 2 0 Proposition 2.9. The zeta function ζ(s) extends to a meromorphic function in R(s) > 0, with a single simple pole at s = with residue. Proof We have ζ(s) = f(s) 2 s for R(s) >. But the right-hand side is meromorphic in R(s) > 0, and so defines an analytic continuation of ζ(s) (necessarily unique, by the theory of analytic continuation) to this half-plane. Since f(s) is holomorphic in this region, any pole of ζ(s) must be a pole of /( 2 s ), ie a zero of 2 s. But 2 s = e ( s) log 2. Hence 2 s = ( s) log 2 = 2nπi for some n Z. Thus /( 2 s ) has poles at s = + 2nπ log 2 i (n Z). At first sight this seems to give an infinity of poles of ζ(s) on the line R(s) =. However, the following argument shows that f(s) must vanish at all these points except s =, thus cancelling out all the poles of /( 2 s ) except that at s =. Consider g(s) = s + 2 s 2 3 s + 4 s + 5 s 2 6 s +. Like f(s), this converges for all σ > 0. For if we group g(σ) in sets of three terms g(σ) = ( σ + 2 σ 2 3 σ ) + (4 σ + 5 σ 2 6 σ ) + we see that each set is > 0. Thus the series either converges (to a limit > 0), or else diverges to +. On the other hand, we can equally well group g(σ) as g(σ) = σ + 2 σ (2 3 σ 4 σ 5 σ ) (2 6 σ 7 σ 8 σ ) +. Now each group is < 0, if we omit the terms σ + 2 σ. Thus g(σ) either converges (to a limit < σ + 2 σ ), or else diverges to. We conclude the g(σ) converges (to a limit between 0 and σ + 2 σ ).

19 2.4. THE RIEMANN ZETA FUNCTION 2 Hence g(s) converges for R(s) > 0. But if R(s) > we can re-write g(s) as g(s) = ( s + 2 s + 3 s + ) 3(3 s + 6 s + 9 s + ) = ( 3 s )ζ(s). Thus ζ(s) = g(s). 3 s The right hand side is meromorphic in the half-plane R(s) > 0, giving a second analytic continuation of ζ(s) to this region, which by the theory of analytic contination must coincide with the first. But the poles of /( 3 s ) occur where ie ( s) log 3 = 2mπi, s = + 2πm log 3 i. Thus ζ(s) can only have a pole where s is expressible in both forms But this implies that ie s = + 2πn log 2 i = + 2πm log 3 i m log 2 = n log 3, 2 m = 3 n, (m, n Z). which is of course impossible (by the Fundamental Theorem of Arithmetic) unless m = n = 0. We have therefore eliminated all the poles except s =. At s =, f() = = log 2. 4 (This follows on letting x from below in log( + x) = x x 2 /2 +.) On the other hand, if we set s = + s then 2 s = e s log 2 = s log 2 s 2 /2! log = s log 2( s /2 log 2 +

20 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 2 Thus 2 = s 2 s = s ( + log 2 s 2 log 2 + ) = log 2 s + h(s), where h(s) is holomorphic. Hence ζ( + s ) = s + h(s)f(s). We conclude that ζ(s) has a simple pole at s = with residue. In Chapter 7 we shall see that ζ(s) can in fact be analytically continued to the whole of C. It has no further poles; its only pole is at s =. 2.5 The Riemann-Stieltjes integral It is helpful (although by no means essential) to introduce a technique which allows us to express sums as integrals, and brings summation by parts into the more familiar guise of integration by parts. Let us recall the definition of the Riemann integral b a f(x) dx of a continuous function f(x) on [a, b]. Note that f(x) is in fact uniformly continuous on [a, b], ie given ɛ > 0 there exists a δ > 0 such that x y < δ = f(x) f(y) < ɛ for x, y [a, b]. By a dissection of [a, b] we mean a sequence We set : a = x 0 < x < < x n = b. = max 0 i<n x i+ x i. The dissection is said to be a refinement of, and we write if the set of dissection-points x i of is a subset of the set of dissection-points of. Evidently =.

21 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 3 Let S(f, ) = 0 i<n f(x i )(x i+ x i ). Then S(f, ) is convergent as 0, ie given ɛ > 0 there exists δ > 0 such that, 2 < δ = S(f, ) S(f, 2 ) < ɛ. This follows from 2 lemmas (each more or less immediate).. Suppose Then x y < δ = f(x) f(y) < ɛ. < δ, 2 = S(f, ) S(f, 2 ) < (b a)ɛ. 2. Given 2 dissections, 2 of [a, b] we can always find a common refinement 3, ie 3, 3 2. These in turn imply 3., 2 < δ = S(f, ) S(f, 2 ) < 2(b a)ɛ. Thus, by Cauchy s criterion, S(f, ) converges as 0, ie there exists an I C such that ie given ɛ > 0 there exists δ > 0 such that S(f, ) I < ɛ if < δ. Even if f(x) is not continuous, we say that it is Riemann-integrable on [a, b] with b f(x) dx = I if a S(f, ) I as 0. Now suppose M(x) is an increasing (but not necessarily strictly increasing) function on [a, b], ie x y = f(x) f(y). Then we set S M (f, ) = 0 i<n f(x i )(M(x i+ ) M(x i )).

22 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 4 Proposition 2.0. If f(x) is continuous and M(x) is increasing then S M (f, ) converges as 0. Proof The result follows in exactly the same way as for the Riemann integral above, with () replaced by. Given ɛ > 0, suppose δ > 0 is such that Then if, 2 < δ, Definition 2.5. We call x y < δ = f(x) f(y) < ɛ. S(f, ) S(f, 2 ) < (M(b) M(a))ɛ. I = lim S M(f, ) 0 the Riemann-Stieltjes integral of f(x) with respect to M(x), and write b a f(x) dm = I Functions of bounded variation Definition 2.6. A (real- or complex-valued) function f(x) is said to be of bounded variation on the interval [a, b] if there exists a constant C such that for all dissections of [a, b]. A(f, ) = f(x i ) f(x i ) C Proposition 2.. Any linear combination f(x) = µ f (x) + + µ r f r (x) (µ,..., µ r C) of functions f (x),..., f r (x) of bounded variation is itself of bounded variation. Proof This follows at once from the fact that A(f, ) µ A(f, ) + + µ r A(f r, ).

23 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 5 Proposition 2.2. Any monotone increasing or decreasing function f(x) is of bounded variation. Proof If f(x) is increasing then and so f(x i ) f(x i = f(x i ) f(x i ; A(f, ) = f(b) f(a). If f(x) is decreasing then f(x) is increasing, so the result follows from the last Proposition. Proposition 2.3. A function f(x) of class C [a, b], ie with continuous derivative f (x) on [a, b], is of bounded variation. Proof Since f (x) is continuous, it is bounded: say Also, by the Mean Value Theorem, where x i < ξ < x i. Hence and so f (x) C. f(x i ) f(x i = (x i x i )f (ξ), f(x i ) f(x i C(x i x i ); A(f, ) C(b a). Proposition 2.4. A real-valued function f(x) is of bounded variation on [a, b] if and only if it can be expressed as the difference of two increasing functions: f(x) = M(x) N(x), where M(x), N(x) are monotone increasing. Proof If f(x) is expressible in this form then it is of bounded variation, by Propositions 2.2 and 2.. For the converse, let P (f, ) = (f(x i ) f(x i )), N(f, ) = i:f(x i ) f(x i ) i:f(x i )<f(x i ) (f(x i ) f(x i )).

24 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 6 for each dissection of [a, b]. Then P (f, ), N(f, ) 0; and P (f, ) N(f, ) = f(b) f(a), P (f, ) + N(f, ) = A(f, ). It follows that 0 P (f, ), N(f, ) A(f, ). Hence are defined. P (f) = sup P (f, ), N(f) = sup N(f, ) Lemma 3. We have P (f) N(f) = f(b) f(a). Proof Given ɛ > 0 we can find dissections, 2 such that P (f) P (f, ) > P (f) ɛ, N(f) N(f, 2 ) > N(f) ɛ. If now is a common refinement of, 2 then P (f) P (f, ) P (f, ) > P (f) ɛ, N(f) N(f, ) N(f, 2 ) > N(f) ɛ. But P (f, ) N(f, ) = f(b) f(a). It follows that P (f) N(f) ɛ f(b) f(a) P (f) N(f) + ɛ. Since this is true for all ɛ > 0, P (f) N(f) = f(b) f(a). Now suppose a x b. We apply the argument above to the interval [a, x]. Let p(x), n(x) be the functions P (f), N(f) for the interval [a, x]. By the last Lemma, p(x) n(x) = f(x) f(a). It is easy to see that p(x), n(x) are increasing functions of x. For suppose a x < y b.

25 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 7 To each dissection : a = x 0 < x < x n = x of [a, x] we can associate the dissection : a = x 0 < x < x n < x n+ = y of [a, y]; and then P (f, ) P (f, ), N(f, ) N(f, ). It follows that p(y) p(x), n(y) n(x), ie p(x), n(x) are monotone increasing. Since f(x) = (f(a) + p(x)) n(x), this establishes the result. Proposition 2.5. The function f(x) is of bounded variation on [a, b] if and only if it can be expressed as a linear combination of increasing functions: f(x) = µ M (x) + + µ r M r (x), where M (x),..., M r (x) are monotone increasing, and µ,..., µ r C. Proof It follows from Propositions 2. and 2.2 that a function of this form is of bounded variation. For the converse, note that if f(x) is complex-valued then it can be split into its real and imaginary parts: f(x) = f R (x) + if I (x) where f R (x), f I (x) are real-valued functions. It is easy to see that if f(x) is of bounded variation then so are f R (x) and f I (x). Hence each is expressible as a difference of increasing functions, say f R (x) = M R (x) N R (x), f I (x) = M I (x) N I (x). But then which is of the required form. f(x) = M R (x) N R (x) + im I (x) in I (x),

26 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 8 This result allows us to extend the Riemann-Stieltjes integral to functions of bounded variation. Suppose U(x) is a function of bounded variation on [a, b]. We set for any dissection of [a, b]. S U (f, ) = f(x i )(U(x i+ ) U(x i )) Proposition 2.6. If f(x) is continuous and U(x) is of bounded variation then S U (f, ) converges as 0. Proof By the last Proposition, we can express U(x) as a linear combination of increasing functions. The result then follows from Proposition 2.0. Definition 2.7. We call I = lim S U(f, ) 0 the Riemann-Stieltjes integral of f(x) with respect to U(x), and write b a f(x) du = I. We extend the Riemann-Stieltjes integral to non-continuous functions f(x) as we do the familiar Riemann integral. Thus if S U (f, ) I as 0 then we say that f(x) is Riemann-Stieltjes integrable on [a, b], with b a f(x) du = I. Similarly, we extend the Riemann-Stieltjes integral to infinite ranges in the same was as the Riemann integral. Thus we set a X f(x)du = lim f(x)du, X a if the limit exists. In one important case the Riemann-Stieltjes integral reduces to the familiar Riemann integral.

27 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 9 Proposition 2.7. Suppose U(x) is of class C [a, b], ie U(x) has continuous derivative U (x) on [a, b]; and suppose f(x)u (x) is Riemann integrable on [a, b]. Then f(x) is Riemann-Stieltjes integrable, and b a f(x)du = b a f(x)u (x)dx. Proof Suppose is a dissection of [a, b]. S(fU, ). By the Mean Value Theorem, We compare S U (f, ) with U(x i+ ) U(x i ) = U (ξ i ) where x i < ξ i < x i+. Moreover, since U (x) is continuous on [a, b], it is absolutely continuous; so given any ɛ > 0 we can find δ > 0 such that U (x i ) U (ξ i ) < ɛ if x i+ x i < δ. Hence S U (f, ) S(fU, ) max f (b a)ɛ if < δ, from which the result follows Discontinuities Proposition 2.8. If f(x) is a function of bounded variation on [a, b] then the left limit f(x 0) = lim t x 0 f(t) exists for all x [a, b); and the right limit exists for all x (a, b]. f(x + 0) = lim t x+0 f(t) Proof The result is (almost) immediate if f(x) is increasing. It follows for any function f(x) of bounded variation by Proposition 2.5, since f(x) = µ M (x)+ +µ r M r (x) for all x and similarly for the right limit. = f(x 0) = µ M (x 0)+ +µ r M r (x 0),

28 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 20 The function f(x) is continuous at x = ξ if f(ξ 0) = f(ξ) = f(ξ + 0). Otherwise f(x) has a discontinuity at ξ. Proposition 2.9. The discontinuities of a function f(x) of bounded variation are enumerable. Proof It is sufficient to prove the result for an increasing function; for if f(x) = µ M (x) + + µ r M r (x) then the discontinuities of f(x) lie in the union of the discontinuities of M (x),..., M r (x); and a finite union of enumerable sets is enumerable. Let us define the jump at a discontinuity ξ to be Note that for an increasing function j(ξ) = f(ξ + 0) f(ξ 0). f(ξ 0) f(ξ) f(ξ + 0). Thus f(x) is discontinuous at ξ if and only if j(ξ) > 0. Lemma 4. Suppose M(x) is increasing on [a, b]; and suppose a ξ < ξ 2 < < ξ n b. Then i n j(ξ i ) f(b) f(a). Proof Choose a dissection x 0, x,..., x n of [a, b] with a = x 0 ξ < x < ξ 2 < x 2 < < x n < ξ n x n = b. Then it is easy to see that f(x i ) f(x i ) j(ξ i ); and so, on addition, f(b) f(a) j(ξ i ).

29 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 2 Corollary 2.4. Suppose M(x) is increasing on [a, b]; Then the number of discontinuities with is j(x) = f(x + 0) f(x 0) 2 r 2 r (f(b) f(a)). Using the Lemma, we can enumerate the discontinuities of M(x) by first listing those with j(x), then those with > j(x) 2, then those with 2 > j(x) 2 2, and so on. In this way we enumerate all the discontinuities: ξ 0, ξ, ξ 2,.... Remarks.. Note that we are not claiming that the discontinuities can be enumerated in increasing order, ie so that ξ 0 < ξ < ξ 2 <. That is not so, in general; f(x) could, for example, have a discontinuity at every rational point. 2. The discontinuity at ξ can be divided into two parts: f(ξ) f(ξ 0) and f(ξ + 0) f(ξ). However, if f(x) is right-continuous, ie f(x + 0) = f(x) for all x [a, b), then the second contribution vanishes, and the discontinuity is completely determined by j(ξ) = f(ξ + 0) f(ξ 0) = f(ξ) f(ξ 0). To simplify the discussion, the functions we use have all been chosen to be right-continuous; for example, we set π(x) = {p : p x}, although we could equally well have taken the left-continuous function π (x) = {p : p < x}.

30 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 22 (From a theoretical point of view, it might have been preferable to have imposed the symmetric condition f(x) = (f(x + 0) + f(x 0)). 2 However, for our purposes the added complication would outweigh the theoretical advantage.) Definition 2.8. The step function H ξ (x) is defined by 0 if x < ξ, H ξ (x) = if x ξ. Proposition Suppose U(x) is a right-continuous function of bounded variation on [a, b]. Then j(ξ) is absolutely convergent. ξ Proof It is sufficient to prove the result when U(x) is increasing, by Proposition 2.5. But in that case j(ξ) > 0, and j(ξ) f(b) f(a), ξ by Lemma 4. Proposition 2.2. Suppose U(x) is a right-continuous function of bounded variation on [a, b]. Then U(x) can be split into two parts, where f(x) is continuous, and U(x) = J(x) + f(x), J(x) = j(ξ)h ξ (x), the sum extending over all discontinuities ξ of f(x) in [a, b]. Proof It is sufficient to prove the result in the case where U(x) is increasing, by Proposition 2.5. Let f(x) = U(x) J(x). We have to show that f(x) is continuous.

31 2.5. THE RIEMANN-STIELTJES INTEGRAL 2 23 The step function H ξ (x) is right-continuous. Hence J(x) is right-continuous; and since U(x) is right-continuous by hypothesis, it follows that f(x) is rightcontinuous. We have to show that f(x) is also left-continuous. Suppose x < y. Then J(y) J(x) = j(ξ) by Proposition 4. Thus ie f(x) is increasing. Moreover, Hence x<ξ y U(y) U(x), f(x) = U(x) J(x) U(y) J(y) = f(y), 0 f(y) f(x) U(y) U(x). 0 f(y) f(y 0) U(y) U(y 0). In particular, if U(x) is left-continuous at y then so is f(x). Now suppose U(x) has a discontinuity at y. If x < y then Hence ie J(y) J(x) j(y) = U(y) U(y 0). J(y) J(y 0) U(y) U(y 0), f(y 0) = U(y 0) J(y 0) f(y) = U(y) J(y). Since f(x) is increasing, it follows that ie f(x) is left-continuous at y. f(y 0) = f(y), Definition 2.9. We call f(x) the continuous part of U(x), and J(x) the purely discontinous part. Remarks.. This is our own terminology; there do not seem to be standard terms for these two parts of a function of bounded variation. That is probably because they are more generally studied through the measure or distribution du, with the step function H ξ (x) replaced by the Dirac delta function δ ξ (x) = dh ξ.

32 2.5. THE RIEMANN-STIELTJES INTEGRAL Our definition of J(x) entails that J(a) = 0. With that condition, the splitting of U(x) is unique. If we drop the condition then J(x) and f(x) are defined up to a constant. Proposition Suppose U(x) = j(ξ)h ξ (x) is a purely discontinuous (but right-continuous) function of bounded variation on [a, b]; and suppose f(x) is continuous on [a, b]. Then b a f(x)du = j(ξ)f(ξ). Proof Since j(ξ) is absolutely convergent, it is sufficient to prove the result for a single step function H ξ (x). Suppose is a dissection of [a, b]; and suppose Then x i < ξ x i+. S Hξ (f, ) = f(x i )(H ξ (x i+ ) H ξ (x i )) = f(x i ). Since the result follows. f(x i ) f(ξ) as 0, In practice we shall encounter the Riemann-Stieltjes integral b a f(x)du in just two cases: the case above, where f(x) is continuous and U(x) is purely discontinuous; and the case where U(x) C [a, b], when (as we saw in Proposition 2.7) f(x)du = f(x)u (x)dx Integration by parts Proposition Suppose U(x), V (x) are of bounded variation on [a, b]; and suppose either U(x) or V (x) is continuous. Then b a U(x) dv + b a V (x) du = U(b)V (b) U(a)V (a).

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