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Some Applications of the Riemann Zeta, and Allied Functions,

by

S. D. CHOWLA, Lahore, India.

This paper begins with a small contribution to a long list of results(1) obtained by Dirichlet, Liouville, Cantor, Kronec- ker, Cesaro and Ramanujan. Arithmetical identities and 'order' results are deduced. The conclusion contains applications of the Zeta Functions to the Theory of Infinite Series and to the Theory of Elliptic Functions.

1. The following notation is used in this paper:

where the p's are distinct primes, and the ƒ¿'sare either zero or

positive integers.

ƒÓ

(n) is the number of positive integers less than n and prime

to n.

ƒÐ a(n) is the sum of the a-th. powers of the divisors of n.

w(n) is the number of distinct prime factors ofn.

ƒÊ

(n) is +1 or -1 according as n has an oven oran odd number

of unequal prime factors; it is O if n has a repeated factor.

If n and m are relatively prime of which m is odd, then ()

m denotes the Legendre- Jacobi symbol If n and mhave a com-

mon factor it is convenient to suppose that nIf . m is even and equal to 2.h while n is odd, we have the generalized symbol

of Kronecker(2) defined by

where h is odd and the symbols on the right side are the ibegendre- Jacobi symbols.

(1) See for examples, Cantor, Math. Ann. 16 (1881), S. 583 et seq. and Ramanujan, Mess. of Maths. 45 (1916), 81-84.

(2) Sitzungsber. Akad. Wiss. Berlin. (1885) II,768-780.

(2)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION. 203

and its analytical continuation, k being any integer, positive or negative.

k (n) is equal to ƒ¿1.ƒ¿2.ƒ¿3.... ƒ¿v

ă

(n) is equal to (-1)a1+a2+...+av

k is called the 'modulo ' of Lk(s); in what follows it is con-

venient to suppose that k has no odd square factors, and is odd and

positive. If n is not an integer we may supposethat ƒÓ(n), ƒÐa(n),

w(n), ƒÊ(n), k(•n), and ƒÉ(n) become zero(1)•

ƒÓ1(n) is equal to

2. Formulae.

(2.1)

according as n, which is a positive integer, isor is not a perfect square.

(2.2)

according as n is or is not a perfect square, d(n) being the number of divisors of n, and being O if n is not a positive integer.

(2.3)

according as n is, or is not a perfect square.

These results bear a close resemblance to Ram anujan's identity (loc. cit):

where m and n are any two positive intogers(2 ).

Hereafter, let

Zk(s) denote and Įa(n) denote,

where p runs through the different prime factors of k.

(1) By convention, etc.

(2) For a proof of this and other remarkable results stated in Itamanujan's paper, see: B. M. Wi1son : Proc. Lond. Math. Soc. 21 (1923), 235-251.

(3)For the symbols and see Wilson's paper, or Landau's " Hand-

buch.... Primzahlen " V. 2.

(3)

204• S. D. CHOWLA:

Several of the following results are the generalizations of the statements made by Ramanujan (loc. cit).

(2.4)

(2.5) (2.6)

(2.7)

In these four formulae Ā(s) denote as usual the Riemann Zeta-

Function (1) .

(2.8)

(2.9)

(2.10)

(2.11)

(1) The convergence conditions (see Wilson's paper) are the same as for Ramanujan's identity

in the formulae 2.4, 2.5, 2.6 and 2.7.

(4)

205 SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION.

Let

where the p's and q's are distinct primes, and every p is •ß1(mod. 4),

and every q is •ß3 (mod. 4).

Now,

if all the ƒÀ's are even,

and r(n)=0,

if any of the ƒÀ's is odd.

If follows that if R(s)>1, then

Since

it follows that:

(2.12)

Ramanujan(1) has proved that

From the last two results it follows that:

Equating the coefficients of n-8 on both sides,we get (1) Messenger of Math. V. 54 (1916).

(5)

206 S. D. CHOWLA :

(2.13.1)

if n is not a perfect square.

Similarly we can prove that (2.13.2)

for all integral positive values of n.

Finally I will mention the result:

(2.13.3)

where u, v and w are any positive integers.

In the special case w=1, u = v, this becomes:-

(2.13.4)

(e.g. when u is 12, both sides are equal to 648).

The formula (2.12) and the formula of Ramanuj an, last men- tioned, are particular cases of the following results :

(2.14)

(2.15)

where rk(n) is defined by the equation

(2.16)

according as n is or is not a perfect square.

(2.17)

(2.18)

(6)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION . 207

(2.19)

(2.20)

(2.21)

(2.22)

where [y] denotes the integer nearest to y , and where the p's are distinct primes.

(2.23)

(2.24)

where n and p are any positive integers.

(2.25) (2.26)

(2.27)(1)

(2.28) (2.29)(2.)

2.30)

(1) The equivalence of the first expressions,was discovered independently by Lionville and Ramanujan (See Ramanujan's paperin the Mess, of Math. 1916

. referred to before).

(2) Previously stated by Cantor. in his paper,referred to In the Introduction.

(7)

208 S. D, CHOWLA:

(2.31)

(2.32)

where

(2.33)

(2.34)

according as n is or is not a perfect square.

(2.35)

according as n is or is not a perfect square.

(2.36)

where p runs through all the prunes (2, 3, 5, 7, ....) and R(s) > 2.

(2.37)

where R(s) >4.

(2.38)

where r(x) is O if x is not a positive integer.

(2.39)

(2.40)

(2.41)

where p runs through all the. primes and O is the well-known '

order ' symbol.

(8)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION. 209

(2.42)

where 1, 3, 6,10. ... are the triangular numbers, and p runs through all the odd primes.

(2.43) when n>1.

3. We will now prove, in outline, the most important of the above formulae.

It is easy to verify that(1) :

In this put c=pa, d= pb, x=p-3 and we get:

where

Thus (2.4) has been proved.

Again when R(s)> 1,

(Ramanuj an).

If we put a=b=0 in (2.31) we get when R(s)>1

(1) This proof, is essentially the same as Wil son's proof of a statement of Ramanujat. See Proc. Lond. Math. Soc. 21 (1923), p. 236.

(9)

210 S. D. CHOWLA:

Hence when R(s)>1,

Equating the coefficients of on both sides, we get

according as n is or is not.a perfect square, the sum on the left

extending over all divisors d1 of n.

Thus we have proved a part of (2.2).

From the definition of Įa(n), it is easy to prove that: Į0

(n) =+1

if n is a perfect square, and

Į0(n)=0 if n is not a perfect square.

Now put a=b= 0 in (2.6) and use this fact. We get when

R(s ) >1

Multiplying both sides by Ā(2s), we get

Thus we have established the equivalence of thesecond and third expressions in (2.27).

The following proof, of the equivalence of the first and second expressions in (2.27), seems to have escaped notice.

It is easy to prove that when R(s) > 1,

whence

Again, when „ x„ <1, we have

(10)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION. 211 Hence

Again

From the last two results, it follows that

Comparing the two infinite series for

obtained above,we arrive at the Liouville- Ram anujan identity.

The equivalence of the three expressions in (27.2) has now been completely established.

Let us put a=b=0 and k=7 in (2.17). We get, when R(s)> 1,

Remembering that

where s is an arbitrarily small positive number, we can prove by classical methods duo to Stieltjes and Landa u(1), that the left-hand side of the above result, converges when s=1. Hence by a well

known theorem in the Theory of Dirichlet Series, we deduce that

(1) Landau, Rendiconti del Circolo Matematico di Palermo 24 (1907),81-160.

(11)

212 S. D. CHOWLA:

since

A similar result is obtained by taking k(>0) tobe any odd

number•ß3 (mod 4) and not having any square. factors.

We will now prove (2.42), which is specially interesting as its

proof involves ideas drawn from the theory of probability.

Lemma, if p1, p2, .... pr, are distinct odd primes, then the num-

ber of solutions of the congruence

subject to.

If however p1 = 2, and p2, p3, .... ,pr are oddprimes, the number of such solutions is 2r-1-1.

In other words, the number of values of n for which

and

All these results are easily proved by mathematical. induction.

Now consider the following two sets of numbers,A and B. A con- tains the integers 1, 2, 3, ...., n; and B contains the first n trian- gular numbers; 1, 3, 6, 10, .... ,. The probability that two numbers, chosen at random, one from A and one from B, should be relatively prime, is plainly

where [x] denotes the greatest integer contained in x, and (n, r) denotes the number of positive integers of theform, which are.

If r has no repeated factors, it is clear fromour lemma that if r is odd,

and if r is even,

K being a. positive constant. whose upper bound is independent of n.

(12)

SOME APPLICATIONS OF THE RIEMANN ZETA. FUNCTION. 213

Hence we easily find that Pn, the required probability, is. given by

where p runs through all odd primes, and ƒÃ is an arbitrary positive number.

Now let ƒÓ(r, n) denote the number of positive integers that are less than or equal to n, and prime to r [ƒÓ(1, n)= ƒÓ(n)]; then it is clear that

Again if where p, q, r .... are distinct primes, then

where „ Kr„ < 1.

Hence

where „ K„ <1.

Now

But it is easy to prove that

since, when R(s)>1

(13)

214 S.D.CHOWLA:

where p runs through all the primes.

Hence

We have obtained two expressions for Pn. Comparing these we got

where p runs through all odd primes, and s is an arbitrarily small

positive number.

The rest of the formulae in Section 2. are proved in a way too

similar to the examples given, to require special attention.

4. Let k•ß1 (mod. 4), have no square factors (k>0). Then, we

have, the well-known result(1):

Now let

and its continuations; then, we have

Starting from Mellin's Integral, when. R(y)>0

where the path of integration is a straight line ƒÐ=k (whore k>0), we obtain when k > 1

Hence

(1) See, for example, Hurw1tz, Zeitschrift Math. Phys: 27 (1882), 86-101.

(14)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION. 215

It follows that when k>1,

Making use of the functional equations of Lk(s)and ā(s), noted

above, we get

The difference of the last two integrals is zero by Cauchy' s theorem. Hence(1) :

(4.1)

With the usual notation in the Theory of Elliptic Functions:

where

In this result put (where p is a positive integer of the form 4n+1, and has no square factors), multiplyboth sides by the

Jacobi-Legendre symbol (), and add the results for r=1,2,3,....

(p-1). Making use of a well-known result in the'Cyclotomic

Theory ' we get: ..

where y=ƒÎK'/K.

Combining this with (4.1) we get

(1) This result was suggested by a paper by S. L. Malurkar. See Jour.

Ind. Math. Soc. Dec . 1925, " On the Applicationof Herr Mellin's Integrals to some Series."

(15)

216 S. D.CHOWLA:

if

From the modular equation of the fifth order (in elliptic func- tions) we find that if

Hence, as a particular case of (4.2) we find that if and L'/L=5 then

Using Kronecker's generalization of the Legend re-Jacobi symbol, (4.1) may be written as follows:

5. In this section we will find by rigorous ways the probability that three integers chosen at random should be relatively prime to each other, and allied problems :

Let A, B and C be three equal sets of numbers, each containing the integers 1, 2, 3, ...., n. Let three numbers be chosen at random, one from each of the sets. To find the probability that the number chosen from A should be prime to both of the other numbers.

Let Pn be the probability, and let ƒÓ(r, n) denote the number of positive integers less than n and prime to r. Clearly,

Remembering that when „ Kr„ < 1,

we get by (2.40)

(16)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION. 217

Hence

(5.1)

where p runs through all primes.

Similarly, if three integers are chosen at random, the probability that they be relatively prime to each other, is

(5.2)

where p runs through all primes.

The formula (2.42) can be generalized to the case when triangular numbers are replaced by numbers of the type

In this we require the following lemma:

(5.3) The number of positive integral values ofn for which

and besides

where all the p's are primes greater than r, isrt.

This result is easily proved by mathematical induction. By the method used (at the end of Section 3.) in proving 2.42, we can now, get:

(5.4)

where p runs through all primes like 5, 7, 11, ....that are greater than 3. Incidentally we also find that A he probability that two positive integers, each of the form n(n + 1)(it+ 2)/6, chosen at random, should be relatively prime, is:

(5.5)

where p runs through all primes greater than 3.Similarly we prove that

(17)

218 S. D. CHOWLA:

(5.6)

where p runs through all the odd primes, and 1,3, 15, 21,.... are

the odd triangular numbers.

In this connection it is interesting to consider a few properties

of the following arithmetical function.

If n is odd, let ĵ(n) denote the number of positive triangular

numbers •…(n+1)/2.

The properties of ĵ(n) are very similar to Eu ler's Function

ƒÓ

(n); for example if where p, q, r .... are distinct

odd primes, then

(5.7)

Again it is known that the set of points(1) ƒÓ(n)/n is every-

where dense in the interval 0 to 1. It is also true that the set of

points ĵ(2n -1) /(2n -1) is everywhere dense in the interval 0 to 1.

The proof is exactly similar to that given by Mr. Vijayaraghavan,

except for a few minor changes that have to be made.

6. It is well-known that:

where k is any positive integer and Ek is the k'th Eu1erian number.

Both of these results are capable of being generalized. Sup- pose(2 ), k <0 is•ß1 (mod. 4) then we know that

whence

(1 ) J. Vijayaraghavan, " On the set of points ", Jour. Ind. Math.

Soc. (1922).

(2) We suppose that k has no square factors.

(18)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION. 219

and for ƒÐ> m+1, m>0, when y is small,

It follows that (6.1)

(6.2)

where m is a positive integer.

Similarly we prove the following results:

(6.3)

(6.4)

where a is any real number.

If k(> 0) is •ß1 (mod. 4) and has no square factors, we know that

Using this fact, we find that, when k > 1

(6.5)

(6.6)

Let an,p denote the number of decompositions ofn into p factors, so that when R(s)> l,

Then (6.7)

(19)

220 S. D. CHOWLA:

(6.8)

(6.9)

(6.10)

In the last two formulae k(<0) is •ß1 (mod. 4) and has no square factors.

By using Me11in's integral we also obtain asymptotic expansions such as the following:

(6.11)

as x1-0.

Here p is a positive integer and En is the nth Eulerian number. The right-hand side of this result diverges very rapidly when p is at all large. However, the expansion is useful for cal- culating the series on the left-hand side when x is very near to unity.

Let ƒÓm(n) denote the suin of the mth powers ofthe positive in- tegers less than n and prime to n, so that ƒÓ0(n)=ƒÓ(n). We will prove Lionville's result(1):

where ƒÓm(x)=0 if x is not an integer.

Let, where the p's are distinct primes.

It is easy to verify that

Also it is known that

(1) See Ency. des Sc. Math., Theorie des Nombres, 232-268.

(20)

221 SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION.

where B1, B2, .... are Bernotilli's numbers.

It follows that

the apparently extra term inside the bracket isadded to make the coefficient of (on the right-hand side) equal to unity.

It is obvious that

is equal to the expression within the curvilinear brackets of the equation just written. Hence when R(s)>m+2,

Equating the coefficients of N-3 on both sides,we get

which is Liouville's result.

Let D(<O) be •ß1(mod. 4); it is known that

(i)

where D is supposed to have no square factors.

Also if

and its analytical continuations, then

(ii) Again (iii)

where the path of integration is the straight line ƒÐ=k.

(21)

222 S. D. CHOWLA :

From (iii)

Hence

(iv)

Also from (iv) (v)

Again from (i) and (ii):

(vi)

Froin (v) and (vi) : (vii)

Since the function „C(s) y-8F(s)L-D(s) has no poles between the

lines ƒÐ=k and ƒÐ=1-k [where k>1], it follows from Cauchy's

theorem that the right-hand sides of (iv) and (vii) are equal.

Equating the left-hand sides of (iv) and (vii),we have:

(8.1)

where R(y)>0; D(<0) is •ß1 (mod. 4) and has no square factors.

Similarly we prove the following result:

If D(>O) is •ß1 (mod. 4) and has no repeated factors, then

(8.2)

where () is the Legondre-Jacobi symbol, generalized by

Kronecker. [R(y)>0].

[Note. When D(>0) is •ß1(mod. 4) and has no repeated factors, then

(22)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION. 223

where h is any positive integer, and by convention

are zero if h and D have a common factor.]

9. In connection with the Theory of Binary Quadratic Forms(1),

it is well-known that if D(<0) is •ß1 (mod. 4) and hag no repeated

factors, then

(i)

In this section, I will establish a similar result deduced from Mellin's Integral.

If R(y)>0, it is easy to prove that (ii)

In this put, Multiply both side by () and add

together the results obtained for n=1, 2, 3, ....,(-D-1).

Since

we get

(iii)

Divide both sides of (iii) by y, and then make y tend to zero.

We get

(iv)

By the methods of Section 6, we find quite easily that the right- hand side of (iv) is equal to

4•ã-D L-n(-2).

Now, from Mellin's Integral it follows that

(1) Sae Dickson's History of the Theory of Numbers, Vol. 3. Chap. 6.

(23)

224 S. D.CHOWLA:

(v)

If „ y„ <2ƒÎ,it is easy to prove that left-handside of (v) is

equal to

(vi)

and that the right-hand side is equal to

(vii)

where Bn is the n-th Bernoulliaci numbor.

Equating the coefficients of y2 in (vi) and (vii) we get:

(viii) Since

(i), (iv) and (viii) give (9.1)

In particular, if we put D=-7, we get

which is easily proved by elementary ways.

Similarly we prove the following result:

(9.2) Let p(>0) be •ß3 (mod. 4), and let ƒÊ(p) •‚0. Then

where the sum on the left extends over all positive integers n, that are prime to 6. [x] denotes the greatest integer contained in x,

(24)

SOME APPLICATIONS OF THE RIEMANN ZETA FUNCTION. 225

and (9.3)

From (9.3) it is easy to deduce that (9.4) For primes p =8k + 3,

For primes p=8k + 7,

As an example of (9.2) we have

March, 1927.

References

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