3.3. An alternative method: Newman's proof

In this section we present an interesting alternative proof of a version of theorem 3.2.5, in which (FT3) is replaced by a different condition. The method was devised by Newman in 1980, and further simplified by Korevaar in 1982. It actually delivers a prior result from which our 3.2.5 is an easy deduction. In fact, this prior result was already proved by Ingham in 1935, but by a much more complicated method. It is sometimes called the Ingham-Newman Tauberian theorem: it is ``Tauberian" in the sense that it provides information about the integral of a function $B(x)$, given information about its Mellin (or Laplace) transform.

Like our first method, the Newman-Korevaar proof is still based on complex integration and the Riemann-Lebesgue lemma (so the reader will need this result from section 3.2), but the inversion process is by-passed in favour of an ingeniously chosen integral.

As before, we shall not try to formulate the weakest possible integrability conditions. We simply assume the following condition, which is clearly satisfied by summation functions:

(int) $B(x)$ is continuous except at integers, and has left and right limits at each integer.

We now state the Ingham-Newman theorem. Given that it implies our fundamental theorems, the statement seems remarkably innocuous: a function $g(s)$ is defined by a certain formula for Re $s > 0$ and is assumed holomorphic at 0: the conclusion is simply that the formula is valid at $s = 0$. The real message is that Tauberian theorems are far from trivial! The term ``region" will now be used to mean ``open set".

Theorem 3.3.1. Suppose that $B(x)$ is a function (real or complex-valued) that satisfies condition (int), and that for some M, we have  $\vert B(x)\vert \leq M/x$ on $[1, \infty )$. Suppose that

$$g(s) = \int_1^\infty \frac{B(x)}{x^s}\; dx$$

for ${\rm Re }\;\: s > 0$, and that $g(s)$ can be extended to a function (still denoted by $g(s)$) holomorphic on a region $E$ including $\{ s : \mbox{Re } s \geq 0 \}$. Then

$$\int_1^\infty B(x)\; dx = g(0).$$

Proof. We show first that it is sufficient to prove the case when $g(0) = 0$. Suppose this done, and that $B(x)$ and $g(s)$ are given. Let  $B_0(x) = B(x) - c/x^2$,  where $c = g(0)$. Then for Re $s > 0$,

$$\int_1^\infty \frac {B_0(x)}{x^s}\; dx = g(s) - \frac{c}{s+1}.$$

Denote this by $g_0(s)$. Then  $g_0(0) = g(0) - c = 0$, so we can conclude that

$$\int_1^\infty B_0(x)\; dx = 0.$$

The required statement follows, since

$$\int_1^\infty B_0(x)\; dx = \int_1^\infty B(x)\; dx - c \int_1^\infty \frac{1}{x^2}\; dx = \int_1^\infty B(x)\; dx - c .$$

Assume, then, that $g(0) = 0$, so that  $g(s)/s$  is holomorphic at 0. Fix $X > 1$, and let

$$g_X(s) = \int_1^X x^{-s} B(x) \; dx .$$

Then $g_X$ is holomorphic on the whole complex plane (for a proof, see Appendix D) and  $g_X(0) = \int_1^X B(x)\; dx$. We have to show that  $g_X(0) \to 0$  as $X \to \infty$.

Let $\varepsilon > 0$ be given. Choose $R$ such that  $M/R \leq \varepsilon$. Let $C$ be the circle $C(0,R)$. Now comes the inventive step. Consider the function

$$J(s) = \frac{1}{s} \left( 1 + \frac{s^2}{R^2} \right) X^s.$$

Recall that $X^s = e^{as}$, where $a = \log X$. Hence $J(s)$ is of the form

$$\left( \frac{1}{s} + \frac{s}{R^2} \right) (1 + as + \mbox{$\frac{1}{2}$} a^2 s^2 + \cdots ) = \frac{1}{s} + K(s),$$

where $K(s)$ is holomorphic everywhere. (In other words, $J(s)$ has a simple pole at 0, with residue 1). Hence by Cauchy's integral theorem and formula,

$$\frac{1}{2\pi i} \int_C J(s) g_X(s) \; ds = g_X(0).$$

Now let $C^+$, $C^-$ respectively be the portions of $C$ with Re $s \geq 0$ and Re $s \leq 0$. Also, let $L$ be the line segment from $-iR$ to $iR$. Now $J(s)g(s)$ is holomorphic on $E$ (since $g(s)/s$ is holomorphic at 0). By Cauchy's thorem, it follows that

$$\int_L J(s)g(s)\; ds - \int_{C^+} J(s)g(s)\; ds = 0.$$

By adding this zero quantity to the previous expression for $g_X(0)$, we obtain

$$g_X(0) = I_1(X) + I_2(X) + I_3(X),$$

where

$$I_1(X) = \frac{1}{2\pi i} \int_{C^+} J(s)[g_X(s) - g(s)]\; ds,$$

$$I_2(X) = \frac{1}{2\pi i} \int_{C^-} J(s) g_X(s)\; ds,$$

$$I_3(X) = \frac{1}{2\pi i} \int_L J(s) g(s)\; ds.$$

We now estimate these three integrals separately. The subtlety is that the $X^\sigma$ occurring in $\vert J(s)\vert$ is cancelled by  $g_X(s)-g(s)$  on $C^+$ and by $g_X(s)$ on $C^-$. Note first that if $s = \sigma + it$ and $\vert s\vert = R$ (so that $s\overline s = R^2$), then

$$\frac {1}{s} + \frac{s}{R^2} = \frac{1}{R^2} (\overline s + s) = \frac{2 \sigma }{R^2}.$$

If $\sigma > 0$, then  $\vert x^{-s}B(x)\vert \leq M/x^{\sigma + 1}$, so

$$\vert g(s) - g_X(s)\vert \leq \int_X^\infty \frac {M}{x^{\sigma +1}}\; dx = \frac{M}{\sigma X^\sigma }.$$

So when $\vert s\vert = R$ and $\sigma > 0$, we have

$$\vert J(s)[g(s) - g_X(s)]\vert \leq \frac{2 \sigma X^\sigma }{R^2} \: \frac{M}{\sigma X^\sigma } = \frac{2M}{R^2}.$$

By continuity, this inequality also holds at the points $\pm iR$. Hence

$$\vert I_1(X)\vert \leq \frac{\pi R}{2\pi }\; \frac{2M}{R^2} = \frac{M}{R}.$$

If $\sigma < 0$, then

$$\vert g_X(s)\vert \leq M \int_1^X x^{-\sigma -1}\; dx < \frac{M X^{-\sigma }}{\vert\sigma \vert},$$

so when $\vert s\vert = R$ and $\sigma < 0$, we have

$$\vert J(s) g_X(s)\vert \leq \frac{2\vert\sigma \vert X^\sigma... ...\: \frac{M X^{-\sigma }}{\vert\sigma \vert} = \frac{2M}{R^2}.$$

Hence, as before,

$$\vert I_2(X)\vert \leq \frac{M}{R}.$$

Finally,

$$I_3(X) = \frac{1}{2\pi } \int_{-R}^R \frac{g(it)}{it} \left( 1 - \frac{t^2}{R^2} \right) X^{it}\; dt .$$

Since  $g(it)/it$  is continuous at 0, the Riemann-Lebesgue lemma 3.2.4 (for a bounded interval) shows that  $I_3(X) \to 0$  as $X \to \infty$.

So for large enough $X$, we have  $\vert g_X(0)\vert \leq 3\varepsilon$, as required. $\Box$

Corollary 3.3.2. Under the same conditions, if

$$g_1(s) = \int_1^\infty \frac{B(x)}{x^{s-1}}\; dx$$

for Re $s > 1$, and $g_1(s)$ can be extended to be holomorphic on a region including
$\{ s: {\rm Re}\;\: s \geq 1 \}$, then

$$\int_1^\infty B(x)\; dx = g_1(1).$$

Proof. This is simply 3.3.1 applied to the function $g(s) = g_1(s+1)$. $\Box$

Before giving our new variant of 3.2.5, we give the corresponding statement for Dirichlet integrals (alias Mellin transforms) rather than series, which takes a rather simpler form.

Proposition 3.3.3. Suppose that f is a complex function differentiable on a region including $\{ s : \mbox{Re }s \geq 1 \}$ except possibly at the point 1, and that:

(i)	$${f(s) = \int_1^\infty \frac{A(x)}{x^{s+1}}\; dx}$$   for Re $s > 1$;
(ii)	$${f(s) = \frac{\alpha }{s-1} + g(s)}$$ , where g is differentiable at 1;
(iii)	there exists $M$ such that  $\vert A(x)\vert \leq Mx$  for all $x \geq 1$.

Then   $${ \int_1^\infty \frac{A(x) - \alpha x}{x^2}\; dx}$$   converges to  $g(1)$.

Proof. Let

$$B(x) = \frac{A(x)}{x^2} - \frac{\alpha }{x}.$$

Then  $\vert B(x)\vert \leq (M + \vert\alpha \vert)/x$  for $x \geq 1$. For Re $s > 1$, we have

$$\begin{eqnarray*} \int_1^\infty \frac{B(x)}{x^{s-1}}\; dx & = & \int_1^\infty \... ...right)\; dx \\ & = & f(s) - \frac{\alpha }{s-1} \\ & = & g(s). \end{eqnarray*}$$

The statement follows, by 3.3.2. $\Box$

Finally, we derive our variant of 3.2.5.

Theorem 3.3.4. Suppose that f is a complex function differentiable on a region including $\{ s : \mbox{Re }s \geq 1 \}$ except possibly at the point 1, and that:

(FT1)	the series $${\sum_{n=1}^\infty \frac{a(n)}{n^s}}$$ converges to $f(s)$ when Re $s > 1$;
(FT2)	$${f(s) = \frac{\alpha }{s-1} + \alpha_0 + (s-1)h(s)}$$ , where h is differentiable at 1;
($FT3'$)	there exists $M$ such that  $\vert A(x)\vert \leq Mx$  for all $x \geq 1$.

Then   $${ \int_1^\infty \frac{A(x) - \alpha x}{x^2}\; dx}$$   converges to  $\alpha_ 0 - \alpha$.

Proof. When Re $s > 1$, we have $x^{-s}A(x) \to 0$ as $x \to \infty$, so by AS8, $f(s) = sf_1(s)$, where

$$f_1(s) = \int_1^\infty \frac{A(x)}{x^{s+1}}\; dx.$$

(Note that condition (FT$3'$) is used again here.) Then

$$\begin{eqnarray*} f_1(s) = \frac{f(s)}{s} & = & \alpha \left( \frac{1}{s-1} - \f... ...s} + (s-1) \frac{h(s)}{s} \\ & = & \frac{\alpha }{s-1} + g(s), \end{eqnarray*}$$

where $g$ is differentiable at 1 and  $g(1) = \alpha_0 - \alpha$. The statement follows, by 3.3.3. $\Box$

The difference between 3.3.4 and 3.2.5 is that (FT3), a condition on $f(s)$, has been replaced by (FT$3'$), a condition on $A(x)$. In this sense, 3.3.4 is less purely a result deducing properties of $A(x)$ from those of $f(s)$. If  $\vert a(n)\vert \leq 1$  for all $n$, then (FT$3'$) is trivial, while (FT3) may require some work (as in the case of $\mu (n)$ and $\/\zeta (s)$). Conversely, there are cases where (FT3) follows from our earlier results, while (FT$3'$) is not easy to verify, for example

$$\frac{\zeta'(s)}{\zeta (s)^2} = \sum_{n=1}^\infty \frac{\mu (n) \log n}{n^s}.$$

For the purpose of error estimates, one needs the expression found in the proof of 3.2.5 for the integral of $A(x)/x^2 - \alpha /x$ on bounded intervals. Exercise 2 shows how to recover this expression by Newman's method.

Further notes. (1) We have presented the Ingham-Newman theorem in terms of the Mellin transform $g(s) = \int_1^\infty x^{-s} B(x)\; dx$. Such theorems are often stated in terms of the Laplace transform

$$G(s) = \int_0^\infty e^{-st} C(t)\; dt,$$

(assumed valid for Re $s > 0$). The function $C(t)$ is assumed to be bounded, and the conclusion is that

$$\int_0^\infty C(t)\; dt = G(0).$$

(Condition (int) must be modified to allow the discontinuities to be at points other than integers.) Either version is readily obtained from the other by substitution (see exercise 1). Also, routine modifications to the proof of 3.3.1 will give a direct proof of the Laplace version.

(2) Historically, the approach to the prime number theorem via Tauberian theorems became established in the 1930's. The Ingham-Newman theorem is, roughly, a weak version of the more general ``Wiener-Ikehara" Tauberian theorem, known proofs of which are quite hard.

(3) In the accounts of Newman's method known to the author, the integral statement of 3.3.4 is not presented as a theorem in its own right, with the result that there is no mention of the derivation of the series version of our fundamental theorem.

Exercises

1. Derive the ``Laplace" version of the Ingham-Newman theorem from the ``Mellin" version by substituting $x = e^t$ and choosing the function $B(x)$ suitably.

2. (The expression needed for error estimates) In 3.3.1, suppose that $g(0) = 0$ and (in addition to the other conditions),  $\vert g(\pm it)\vert \leq Ct^{-r}$ for all $t \geq 1$, where $r > 0$. By letting $R$ tend to infinity with $X$ fixed, show that

$$\int_1^X B(x)\; dx = \frac {1}{2\pi } \int_{-\infty }^{\infty } \frac{X^{it}}{it} g(it)\; dt.$$

Formulate the corresponding variant of 3.3.2 (but not 3.3.3). In 3.3.4, assume the additional condition  $\vert f(1 \pm it)\vert \leq Ct^{1-r}$  for $t \geq 1$. By taking $B(x)$ to be  $[A(x) - \alpha x - (\alpha_0 - \alpha )]/x^2$, derive the identity

$$\int_1^X \frac{A(x) - \alpha x}{x^2}\; dx = (\alpha_0 - \alp... ...t) + \frac{1}{2\pi i} \int_{L_1} X^{s-1} \frac{h(s)}{s}\; ds$$

(as found, under different conditions, in the proof of 3.2.5).