3.2 Inversion of Dirichlet series and the integral version of the fundamental theorem.

Inversion of Dirichlet series by integrals on vertical lines

Let $a(n)$ be an arithmetic function, and let $f(s)$ be the function defined by its corresponding Dirichlet series:

$$f(s) = \sum_{n=1}^\infty \frac{a(n)}{n^s}.$$

As usual, write $A(x) = \sum_{n\leq x}a(n)$. Our aim is to deduce an estimation of $A(x)$ from information about $f(s)$. In the present section, we shall carry out this programme as far as what we shall call the ``integral version of the fundamental theorem". Once this is known, it is a fairly short step to derive the ``limit" and ``series" versions, incorporating the prime number theorem as a particular case. These versions are presented in section 3.4.

An interesting alternative route to the integral version (under slightly different conditions) is described in section 3.3. However, the method of the present section is recommended to readers intending to progress to chapter 5.

The first stage, roughly, is to ``invert" the series to express $A(x)$ in terms of $f(s)$. Actually, it is easier to find an expression for

$$\int_1^x \frac{A(y)}{y^2}\; dy$$

instead of $A(x)$ itself; this will serve just as well for our ultimate purpose. The expression will be an integral involving $f(s)$, necessarily along some path in the complex plane. Note that the corresponding step for power series is very simple: if  $g(s)= \sum_{n=0}^\infty a(n)s^n$  for $\vert s\vert < R$, then $2\pi i a(n)$ equals the integral of $g(s)/s^{n+1}$ round a circle of radius less than $R$. Given that a Dirichlet series converges on a half-plane, it is not surprising that the right kind of path for such a series is a vertical line.

This is the point at which complex analysis enters our strategy in a serious way. We will only need the really basic theorems of the subject, Cauchy's integral theorem and formula, together with the following fact: if $\ell (\Gamma )$ denotes the length of a path $\Gamma$ in the complex plane, and $\vert f(s)\vert \leq M$ for $s$ on $\Gamma$, then $\vert\int_\Gamma f(s)\; ds\vert \leq M \ell (\Gamma )$.

The notation $\int_{c-iT}^{c+iT} f(s)\; ds$ (where $c,T$ are real) means the integral of $f(s)$ along the straight line from $c-iT$ to $c+iT$, in other words,

$$\int_{-T}^T f(c+it)i\; dt.$$

The limit of this integral (if it has one) as $T \to \infty$ is normally written

$$\int_{c-i\infty }^{c+i\infty } f(s)\; ds;$$

for brevity, we shall denote it by

$$\int_{L_c} f(s) \; ds .$$

We write also

$$E(x) = \left\{ \begin{array}{ll} 1 & \mbox{if } x \geq 1 \\ 0 & \mbox{if } x < 1 \end{array} \right.$$

Our starting point is the following integral. Note that $x$ is fixed and integration is with respect to $s$.

Proposition 3.2.1. If $x > 0$ and $c> 0$, then

$$\frac{1}{2\pi i} \int_{L_c} \frac{x^s}{s^2} \; ds = E(x)\log x .$$

Proof. Let $C$ be the circle $C(0,R)$, where $R > c$. Let the line Re $s = c$ meet this circle at $c \pm it_R$. Let $L_R$ be the line segment from $c-it_R$ to $c+it_R$, and let $C_1, \; C_2$ (respectively) be the sections of $C$ to the left and right of this line.

For any path $\Gamma$ in the complex plane, write

$$I(\Gamma ) = \frac{1}{2\pi i} \int_\Gamma \frac{x^s}{s^2}\; ds.$$

Now $x^s = e^{\lambda s}$, where $\lambda = \log x$, hence

$$\frac{x^s}{s^2} = \frac{e^{\lambda s}}{s^2} = \frac{1}{s^2} (1 + \lambda s + \mbox{$\frac{1}{2}$} \lambda^2 s^2 + \cdots ),$$

and this series is uniformly convergent on any set of the form $r_1 \leq \vert s\vert \leq r_2$. Hence the series can be integrated termwise on $L_R \cup C_1$, which is a closed contour enclosing the point 0. It is elementary that the integral of $s^{-1}$ around this contour is $2\pi i$, while the integral of all other powers $s^n$ is 0. So we have

$$I(L_R \cup C_1) = \lambda .$$

Now $\vert x^s\vert = x^\sigma$, where $s = \sigma + it$. So if $x \geq 1$, then for $s$ on $C_1$, we have $\vert x^s\vert \leq x^c$, so that $\vert x^s/s^2\vert \leq x^c/R^2$ and hence

$$\vert I(C_1)\vert \leq \frac{1}{2\pi } \frac{x^c}{R^2} 2\pi R = \frac{x^c}{R},$$

which tends to 0 as $R \to \infty$. Hence

$$I(L_R) \to \lambda = \log x \quad \mbox{as } R \to \infty ,$$

which is our statement for $x \geq 1$.

If $0 < x < 1$, we consider instead $I(C_2 \cup L_R)$. The integrand $x^s/s^2$ has no poles inside this contour, so by Cauchy's theorem, the integral equals 0. The function $x^\sigma$ now decreases with $x$, so $\vert x^s\vert \leq x^c$ for $s$ on $C_2$. In the same way as before, $I(C_2) \to 0$ as $R \to \infty$, and hence $I(L_R) \to 0$.  $\Box$

Note. It follows that if $c' > 1$, then

$$\frac{1}{2\pi i} \int_{L_{c'}} \frac{x^{s-1}}{(s-1)^2}\; ds = E(x)\log x,$$

In fact, by writing $c' = c+1$ and substituting $s = c'+it$, we see that this integral is identical to the one in 3.2.1.

Proposition 3.2.2. If $x > 0$ and $c > 1$, then

$$\frac{1}{2\pi i} \int_{L_c} \frac{x^s}{s(s-1)}\; ds = (x-1)E(x).$$

Proof. Similar to 3.2.1. To show that the integrals on the circular arcs tend to 0, we use $\vert s(s-1)\vert \geq R(R-1)$ on $C$ (instead of $\vert s^2\vert = R^2$). To evaluate $I(L_R \cup C_1)$, note that

$$\frac{x^s}{s(s-1)} = \frac{x^s}{s-1} - \frac{x^s}{s}.$$

By Cauchy's integral formula, it follows that

$$I(L_R \cup C_1) = x^1 - x^0 = x-1.$$

The statements follow, as before. $\Box$

Now suppose that $f(s) = \sum_{n=1}^\infty a(n)/n^s$ for Re $s > 1$. If termwise integration can be justified, then 3.2.2 will give

$$\begin{eqnarray*} \frac{1}{2\pi i}\int_{L_c} \frac{x^s}{s(s-1)}f(s)\; ds & = & \... ...t)\\ & = & \sum_{n \leq x} a(n) \left( \frac{x}{n} -1 \right), \end{eqnarray*}$$

since $E(x/n) =0$ when $n > x$. The next theorem says that this is true (but for future use, the statement is given with both sides divided by $x$).

Theorem 3.2.3. Suppose that the Dirichlet series $\sum_{n=1}^\infty a(n)/n^s$ is absolutely convergent for Re $s > 1$, with sum $f(s)$. Let $A(x) = \sum_{n\leq x}a(n)$. Then for $c > 1$ and $x > 1$,

$$\frac{1}{2\pi i} \int_{L_c} \frac{x^{s-1}}{s(s-1)}f(s)\; ds = \sum_{n \leq x} a(n)\left(\frac{1}{n} - \frac{1}{x}\right).$$

By Abel's summation formula, this can also be written

$$\int_1^x \frac{A(y)}{y^2}\; dy.$$

Proof. Write $x^s f(s) = G(s) + H(s)$, where

$$G(s) = \sum_{n\leq x} a(n) \left( \frac{x}{n} \right)^s, \qquad H(s) = \sum_{n > x} a(n) \left( \frac{x}{n} \right) ^s .$$

For $G(s)$ (which is only a finite sum) we have as above

$$\frac{1}{2\pi i} \int_{L_c} \frac{G(s)}{s(s-1)} \; ds = \sum_{n \leq x} a(n) \left( \frac{x}{n} -1 \right).$$

Now $\sum_{n > x} \vert a(n)\vert (x/n)^c$ is convergent, say to $M$. For $n > x$ and Re $s \geq c$, we have $\vert(x/n)^s\vert \leq (x/n)^c$, hence $\vert H(s)\vert \leq M$. Now consider

$$\int_{C_2 \cup L_R} \frac{H(s)}{s(s-1)}\; ds$$

as in the proof of 3.2.1. Since $H(s)$ is differentiable for Re $s > 1$, this integral equals 0, by Cauchy's theorem, and exactly as before we see that the contribution of $C_2$ tends to 0 as $R \to \infty$. Hence

$$\frac{1}{2 \pi i} \int_{L_c} \frac{H(s)}{s(s-1)} \; ds = 0.$$

To obtain the first statement, now divide both sides by $x$. The second statement follows, by Abel's summation formula in the form 1.3.6(ii), applied to the function $1/y$. $\Box$

The Riemann-Lebesgue lemma

The following result, the ``Riemann-Lebesgue Lemma", is needed for either proof of the fundamental theorems. Let $\phi$ be a differentiable, complex-valued function on $\mathbb{R}$ such that $\int_{-\infty }^\infty \vert\phi (t)\vert\; dt$ is convergent. For such a function, we can define (for all real $\lambda$)

$$F(\lambda ) = \int_{-\infty }^\infty e^{i\lambda t} \phi (t)\; dt .$$

This integral is convergent, since $\vert e^{i\lambda t}\phi (t)\vert = \vert\phi (t)\vert$. (The function $F(-\lambda )$ is the Fourier transform of $\phi$.)

Proposition 3.2.4. Let $\phi$ be a complex-valued function on $\mathbb{R}$ with continuous derivative and such that $\int_{-\infty }^\infty \vert\phi (t)\vert\; dt$ is convergent. Let $F(\lambda )$ be as above. Then $F(\lambda ) \to 0$ as $\lambda \to \infty$.

Proof. Take $\varepsilon > 0$. There exists $T$ such that $\int_T^\infty \vert\phi (t)\vert\; dt \leq \varepsilon$ (and similarly for $(-\infty , -T]$). For any $\lambda$, we then have

$$\left\vert \int_T^\infty e^{i\lambda t} \phi (t)\; dt \right\vert \leq \varepsilon .$$

Now $\vert\phi'(t)\vert$ is bounded (since continuous), on $[-T,T]$, say by $M$. Let

$$F_T(\lambda ) = \int_{-T}^T e^{i\lambda t} \phi(t)\; dt.$$

Integrating by parts, we have

$$F_T(\lambda ) = \frac{1}{i\lambda }\left[ e^{i\lambda t} \phi... ... \frac{1}{i\lambda } \int_{-T}^T e^{i\lambda t} \phi'(t)\; dt,$$

hence

$$\vert F_T(\lambda )\vert \leq \frac{1}{\lambda } \left[ \vert... ...T)\vert + \vert\phi (-T)\vert \right] + \frac{2MT}{\lambda }.$$

So $F_T(\lambda ) \to 0$ as $\lambda \to \infty$, and for sufficiently large $\lambda$, we have $\vert F_T(\lambda )\vert \leq \varepsilon$. Adding the contributions of $[T, \infty )$ and $(-\infty , -T]$, we see that  $\vert F(\lambda )\vert \leq 3\varepsilon$  for such $\lambda$. $\Box$

Note 1. Of course, we have proved the same conclusion for $F_T(\lambda )$.

Note 2. If we know that $\int_{-\infty }^\infty \vert\phi'(t)\vert\; dt$ is also convergent (with value $I_1$, say) and $\phi (\pm t) \to 0$ as $t \to \infty$, then integration by parts as above gives

$$F(\lambda ) = -\frac {1}{i\lambda } \int_{-\infty }^\infty e^{i\lambda t} \phi '(t)\; dt ,$$

and hence $\vert F(\lambda )\vert \leq I_1/\vert\lambda \vert$. This has the advantage of giving an estimate of the rate of convergence to 0. If the integral of $\vert\phi ''\vert$ is convergent, then the estimate can be improved to $I_2/\vert\lambda \vert^2$, and so on. We return to this point later.

The integral version of the fundamental theorem

We are now ready to state this theorem. For present purposes, the word ``region" simply means a subset of the complex plane.

Theorem 3.2.5. Suppose that f is a complex function differentiable on a region including 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 absolutely 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 is a function $P(t)$ such that $\vert f(\sigma \pm it)\vert \leq P(t)$ when $\sigma \geq 1$ and $t \geq t_0$
	(for some $t_0 \geq 1$), and also  $${\int_1^\infty \frac{P(t)}{t^2} \; dt}$$ is convergent.

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

Proof. Write $\phi (s) = h(s)/s$. Note that $h(s)$ and $\phi (s)$ are differentiable at all points $s$ (including 1) with Re $s \geq 1$. Using the identity $1/(s-1) = s/(s-1) - 1$, we have

$$s(s-1)\phi (s) = (s-1)h(s) = f(s) - \frac{\alpha }{s-1} - \alpha_0 = f(s) - \alpha \frac{s}{s-1} - \alpha ',$$

where $\alpha ' = \alpha _0 - \alpha$, hence

$$\phi (s) = \frac{f(s)}{s(s-1)} - \frac{\alpha }{(s-1)^2} - \frac{\alpha'}{s(s-1)}.$$ (1)

For all $s = \sigma + it$ with $\sigma \geq 1$ and $\vert t\vert \geq t_0$, we have  $\vert s(s-1) \phi (s)\vert \leq P_1(t)$, where $P_1(t) = P(t) + \vert\alpha \vert + \vert\alpha _0\vert$. For such $s$, we have $\vert s(s-1)\vert \geq t^2$, hence

$$\vert\phi (s)\vert \leq \frac{P_1(t)}{t^2}.$$ (2)

Of course,  $\int_1^\infty [P_1(t)/t^2]\; dt$  is convergent.

For $x > 1$ and $c \geq 1$, define

$$I(x,c) = \frac{1}{2\pi i} \int_{L_c} x^{s-1} \phi (s)\; ds .$$

For any $c > 1$, we have by (1) and 3.2.1, 3.2.2, 3.2.3,

$$\begin{eqnarray*} I(x,c) & = & \frac{1}{2\pi i} \int_{L_c} \frac{x^{s-1}}{s(s-1)... ...- \alpha y}{y^2}\; dy - \alpha ' \left( 1 - \frac{1}{x} \right). \end{eqnarray*}$$

Note that this is independent of $c > 1$. We will show that $I(x,1)$ has the same value. Suppose for the moment that this is correct. Then, writing $s = 1+it$, we have

$$I(x,1) = \frac{1}{2\pi } \int_{-\infty }^\infty x^{it} \phi(1... ...2\pi } \int_{-\infty }^\infty e^{i\lambda t} \phi (1+it)\; dt,$$

where $\lambda = \log x$. By (2), $\int_{-\infty }^\infty \vert\phi (1+it)\vert\; dt$ is convergent. It follows from 3.2.4 that $I(x,1) \to 0$ as $x \to \infty$. In other words,

$$\int_1^x \frac{A(y) - \alpha y}{y^2}\; dy \to \alpha ' \quad \mbox{as } x \to \infty .$$

This is our statement.

It remains to justify the statement that $I(x,1) = I(x,c)$ for $c > 1$. Now

$$I(x,c) = \frac{1}{2\pi } \int_{-\infty}^\infty g(c,t)\; dt,$$

where

$$g(c,t) = x^{c-1+it} \phi (c+it).$$

Choose $\varepsilon > 0$. There exists $T \geq t_0$ such that

$$\int_T^\infty \frac{P_1(t)}{t^2}\; dt \leq \varepsilon .$$

If $1 \leq c \leq 2$, then  $\vert x^{c-1+it}\vert = x^{c-1} \leq x$, so by (2),

$$\int_T^\infty \vert g(c,t)\vert \; dt \leq \varepsilon x.$$

Similarly for the integral on $(-\infty , -T]$. Finally, since $g(c,t)$ is continuous on the closed rectangle $1 \leq c \leq 2, \; -T \leq t \leq T$, it is uniformly continuous there, so that if $c$ is close enough to 1, then  $\vert g(c,t) - g(1,t)\vert \leq \varepsilon /2T$  for all $t$ in $[-T,T]$. For such $c$,

$$\left\vert \int_{-T}^T g(c,t) \; dt - \int_{-T}^T g(1,t)\; dt \right\vert \leq \varepsilon .$$

Together with the contributions from outside $[-T,T]$, this gives $\vert I(x,c) - I(x,1)\vert \leq \varepsilon (1 + 4x)$. Since $I(x,c)$ is independent of $c > 1$, this shows that $I(x,1) = I(x,c)$, as required.

(An alternative method for the last part is to apply Cauchy's integral theorem on the rectangle with vertices $1\pm iT$, $c\pm iT$ and show that the contribution of the horizontal sides tends to 0 as $T \to \infty$.) $\Box$

Note. It was essential to move the line of integration to $L_1$, since otherwise the integral would contain $x^{c-1}$ as a factor, destroying its chances of tending to 0 as $x \to \infty$. If, in a particular case, it were possible to move the line to $L_c$ where $c < 1$, then of course the factor $x^{c-1}$ would accelerate the convergence to 0.

EXAMPLE. In advance of more interesting examples, let us see what the theorem says in the simplest case, $a(n) = 1$ and $f(s) = \zeta (s)$. Then $A(x) = [x]$ and $\alpha = 1, \; \alpha_0 = \gamma$, so the statement is

$$\int_1^\infty \frac{[x] - x}{x^2}\; dx = \gamma - 1,$$

which of course we know from 1.4.11.

We defer further examples of 3.2.5 until section 3.4, where they will be given alongside corresponding statements for limits and series.

Further remarks.

Two variants of 3.2.3, proved by easy modifications of the same method, are:

$$\frac{1}{2\pi i} \int_{L_c} \frac{x^s}{s^2} f(s) \; ds = \s... ...leq x} a(n) (\log x - \log n) = \int_1^x \frac{A(y)}{y}\; dy ,$$ (3)

$$\frac{1}{2\pi i} \int_{L_c} \frac{x^s}{s(s+1)} f(s)\; ds =... ...t( 1 - \frac{n}{x} \right) = \frac{1}{x} \int_1^x A(y)\; dy.$$ (4)

Many books use one or other of these. If, for example, (3) is used, then the method of 3.2.5 leads to

$$\frac{1}{x} \int_1^x \frac{A(y)}{y}\; dy \to \alpha \quad \mbox{as } x \to \infty .$$

This is sufficient to give the limit version of the fundamental theorem, but not the series version. Hence our preference for the form chosen.

There is, in fact, a more direct inversion theorem along the lines first suggested: for non-integer values of $x$ (avoiding discontinuities of $A(x)$),

$$\frac{1}{2\pi i} \int_{L_c} \frac{x^s}{s}f(s)\; ds = A(x) .$$ (5)

This is called ``Perron's inversion formula". It might be thought that it is a more obvious formula to use than any of the previous ones. However, with only $s$ in the denominator the previous proof fails, since we would be dividing by $R$ instead of $R^2$. In fact, both the proof of (5) and its application demand careful evaluations of integrals on finite intervals $[c-iT, c+iT]$ instead of the whole of $L_c$, using rectangles instead of circles. See [Est] or [Widder].

By Abel's summation formula,

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

so in all these statements, both sides can be expressed in terms of $A(x)$, and one might suspect that the statements are valid for any suitable function $A(x)$, whether or not it is the summation function of an arithmetic function $a(n)$. This is, indeed, the case, and in fact all the statements are special cases of a more general result, the Mellin inversion theorem. Given a function $B(x)$ on $[0, \infty )$, let

$$g(s) = \int_0^\infty \frac{B(x)}{x^{s+1}}\; dx$$

where this converges. Under suitable conditions, the theorem states that

$$B(x) = \frac{1}{2\pi i} \int_{L_c} x^s g(s)\; ds.$$ (6)

(In the usual terminology, the ``Mellin transform" of $B(x)$ is $g(-s)$ rather than $g(s)$.) If $s = \sigma + i\lambda$, the substitution $x = e^t$ gives

$$g(s) = \int_{-\infty }^\infty e^{-i\lambda t} C(t)\; dt ,$$

where  $C(t) = e^{-\sigma t} B(e^t)\; dt$. This is the Fourier transform of $C(t)$, and in this way the Mellin inversion theorem can be equated with the Fourier inversion theorem, which may be more familiar to some readers.

Statement (5) is clearly a case of (6), and with some work one can see that the other identities are also cases of it, with $B(x)$ taken to be the function on the right-hand side each time. This certainly places these statements in the wider mathematical landscape, but for the result we actually want, Theorem 3.2.3, it is simpler and more direct to proceed as we did.

Exercises

1. Write down the value of  $${ \frac{1}{2\pi i} \int_{L_c} \frac{x^s}{s^{k+1}}\; ds}$$ , where $c> 0$ and $k \geq 1$ is an integer.

2. Modify the proof of 3.2.3 to show that under the same conditions,

$$\frac{1}{2\pi i} \int_{L_c} \frac{x^s}{s^2}f(s)\; ds = \int_1^x \frac{A(t)}{t}\; dt.$$

(You need not bother to justify termwise integration.)

3. By considering  $f(s) - \alpha \zeta (s)$, show that theorem 3.2.5 can be deduced from the special case in which $\alpha = 0$ (thereby removing the need for 3.2.1).

4. Let $x > 0$,  $x \neq 1$ and $c> 0$. By integrating on a rectangle with $c\pm iT$ as two vertices, prove that

$$\frac{1}{2\pi i} \lim_{T \to \infty }\int_{c-iT}^{c+iT} \frac{x^s}{s}\; ds = E(x).$$

Show that formula (5) follows if termwise integration is valid. Show also by direct integration that

$$\frac{1}{2\pi i} \lim_{T \to \infty }\int_{c-iT}^{c+iT} \frac{1}{s}\; ds = \mbox{$\frac{1}{2}$}.$$