NOTES ON RIEMANN S ZETA FUNCTION DRAGAN MILIČIĆ. Gamma function.. Definition of the Gamma function. The integral Γz = t z e t dt is well-defined and defines a holomorphic function in the right half-plane {z C Rez > }. This function is Euler s Gamma function. First, by integration by parts Γz + = t z e t dt = t z e t +z t z e t dt = zγz for any z in the right half-plane. In particular, for any positive integer n, we have On the other hand, Γn = n Γn = n!γ. Γ = and we have the following result. e t dt = e t = ;... Lemma. for any n Z. Γn = n! Therefore, we can view the Gamma function as a extension of the factorial... Meromorphic continuation. Now we want to show that Γ extends to a meromorphic function in C. We start with a technical lemma.... Lemma. Let c n, n Z +, be complex numbers such such that n= c n converges. Let S = { n n Z + and c n }. Then fz = n= c n z +n converges absolutely for z C S and uniformly on bounded subsets of C S. The function f is a meromorphic function on C with simple poles at the points in S and Resf, n = c n for any n S.
D. MILIČIĆ Proof. Clearly, if z < R, we have z +n n R for all n R. Therefore, we have z+n n R for z < R and n R. It follows that for n > R, we have c n z +n c n z +n c n n R c n. n n=n n=n R n=n n=n Hence, the series n>r cn z+n converges absolutely and uniformly on the disk {z z < R} and defines there a holomorphic function. It follows that c n n= z+n is a meromorphic function on that disk with simple poles at the points of S in {z z < R}. Therefore, c n n= z+n is a meromorphic function with simple poles at the points in S. Therefore, for any n S we have fz = c n z +n + m S {n} c m z +m = c n z +n +gz where g is holomorphic at n. This implies that Resf, n = c n. Going back to Γ, we have Γz = t z e t dt = t z e t dt+ t z e t dt. Clearly, the second integral converges for any complex z and represents an entire function. On the other hand, since the exponential function is entire, its Taylor series converges uniformly on compact sets in C, and we have t z e t dt = t z p t p dt p! for any z C. Therefore, Γz = p= = p p= p! t z e t dt+ t p+z dt = p p! z +p p= p p! z +p for any z in the right half-plane. By.., the right side of this equation defines a meromorphicfunctiononthecomplexplanewithsimplepolesat,,, 3,. Hence, we have the following result.... Theorem. The function Γ extends to a meromorphic function on the complex plane. It has simple poles at,,, 3,. The residues of Γ are p are given by for any p Z +. ResΓ, p = p p! This result combined with the above calculation immediately implies the following functional equation...3. Proposition. For any z C we have Γz + = zγz. p=
NOTES ON RIEMANN S ZETA FUNCTION 3.3. Another functional equation. Let Re z >. Then Γz = t z e t dt. Let Rep > and Req >. Then, by change of variable t = u, we get Analogously we have Hence, it follows that Γp = ΓpΓq = 4 t p e t dt = Γq = e v v q dv. e u u p du. e u +v u p v q dudv, and by passing to the polar coordinates by u = rcosϕ, v = rsinϕ, we have ΓpΓq = 4 = π e r r p+q cos p ϕsin q ϕdrdϕ π e r r p+q dr cos p ϕ sin q ϕdϕ = Γp+q We put s = sin ϕ in the integral. Then we have If we define π cos p ϕ sin q ϕdϕ = Bp,q = for Rep >, Req >, we get the identity Here B is Euler s Beta function. Let x,. Then we have ΓxΓ x = ΓxΓ x Γ π s p s q ds Bp,q = Bq,p = ΓpΓq Γp+q. Moreover, if we change the variable s = u u+ s x s x ds = u x u+ x.3.. Lemma. For y, we have = Bx, x = cos p ϕ sin q ϕdϕ. s q s p ds., the integral becomes u u+ u y +u du = π sinπy. s x s x ds. x du u+ = u x +u du.
4 D. MILIČIĆ Proof. We cut the complex plane along the positive real axis. On this region we define the function z z y +z with the argument of z y equal to on the upper side of the cut. This function has a first order pole at z = with the residue e πiy. C R - C ε If we integrate this function along a path which goes along the upper side of the cut from ǫ > to R, then along the circle C R of radius R centered at the origin, then along the lower side of the cut from R to ǫ and finally around the origin along the circle C ǫ of radius ǫ, by the residue theorem we get R ǫ u y +u CR du+ z y dz e πiy +z First we remark that for z we have R ǫ u y +u Cǫ du z y +z dz = πie πiy. z y = e ylogz = e yrelogz = e ylog z = z y. Hence, the integrand in the second and fourth integral satisfies z y +z z y +z z y z. Hence, for small ǫ we have and for large R we have +z dz Cǫ z y +z dz CR z y π ǫ y ǫ ; π R y R ; Clearly, this implies that C R as R and C ǫ as ǫ. This implies that e πiy u y +u du = πie πiy.
NOTES ON RIEMANN S ZETA FUNCTION 5 It follows that and finally This lemma implies that e πiy e πiy ΓxΓ x = u y du = πi +u u y +u du = π sinπy. π sinπ x = π sinπx. for x,. Since both sides are meromorphic, it follows that the following result holds..3.. Proposition. For all z C, we have ΓzΓ z = π sinπz. Since z sinπz is an entire function, the right side obviously has no zeros. So, Γz = is possible only at points where z Γ z has poles. Since the poles of Γ are at,,,, it follows that poles of z Γ z are at,,3,. At these points Γn+ = n!. Therefore, we proved the following result..3.3. Theorem. The function Γ has no zeros.. Zeta function.. Meromorphic continuation. Riemann s zeta function ζ is defined by ζz = n z for Rez >. In that region the series converges uniformly on compact sets and represents a holomorphic function. If we consider the expression for the Gamma function Γz = n= t z e t dt for Rez >, and change the variable t into t = ns for n N, we get Γz = n z s z e ns ds, i.e., we have Γz n z = t z e nt dt for any n N and Rez >. This implies that, for Rez >, we have Γzζz = t z e nt dt n= = t z e dt nt = n= t z e t e tdt = t z e t dt. This establishes the following integral representation for the zeta function.
6 D. MILIČIĆ... Lemma. For Rez > we have Γzζz = This integral can be split in to two parts Γzζz = t z e t dt = t z e t dt. t z e t dt+ t z e t dt. Clearly, the second integral converges for all z C and therefore represents an entire function which we denote by F. On the other hand, the function z e z has a simple pole at with residue. Therefore, e z = z +Gz where G is an meromorphic function with first order poles at mπi with integer m. It follows that e z = z + c n z n for any z < π. By Cauchy s estimates, if we fix < r < π, it follows that c n M r n for some M >. In particular, there exists M > such that c n M n for all n Z +. This implies that, for Rez >, since the above series converges uniformly on [,], we have t z e t dt = Therefore, we have t z dt+ n= c n t dt z+n n= = z + c n n= Γzζz = Fz+ z + c n z +n. t z+n dt = z + n= n= c n z +n. By.., the right side is a meromorphic function, holomorphic for any complex z different form z =,,,,... Hence, z Γzζz extends to a meromorphic function in the complex plane with simple pole at and either simple poles or removable singularities at z =,,,... depending if c n or c n =. Since Γ has no zeros, z Γz is an entire function. This implies that the zeta function ζ extends to a meromorphic function in the complex plane. Since Γ =, ζ has a simple pole at with residue equal to. Moreover, since Γ has simple poles at z =,,,... by.., the function Γ has simple zeros there. It follows that ζ has removable singularities at z =,,,... Therefore, we established the following result.... Theorem. The zeta function ζ is a meromorphic function with simple pole at. The residue at this pole is.
NOTES ON RIEMANN S ZETA FUNCTION 7.. The functional equation. In this section we prove the following result.... Theorem. For any z C, we have πz ζz = π z z sin Γ zζ z. To establish this functional equation we first establish a variant of... Let C be a path sketched in the following figure. πi C ε -πi We cut the complex plane along positive real axis and consider the integral w z e w dw. C Here we fix the argument of w to be on the top side of the cut and π on the bottom. Clearly, the integrand is holomorphic on the complement of the positive real axes except at zeros of the denominator of the integrand, i.e., at the points mπi for m Z. If the radius ǫ of the arc C ǫ is less than π, it follows immediately from the Cauchy theorem that this integral doesn t depend on ǫ and the distance of the horizontal lines from the positive real axis. Therefore, to evaluate it we can let ǫ and that distance tend to. First we remark that for w we have w z = e zlogw = e Rezlogw = e Rezlog w Imzargw = w Rez e Imzargw. Since the argument is in [,π], for a fixed z, we see that w z < M w Rez for some positive constant M. We can estimate the integral over C ǫ as Cǫ w z e w dw π MǫRez e ǫeiφ dφ. Since w e w has a simple zero at the origin, e w = wgw where g is holomorphic near the origin and g =. It follows that e w w
8 D. MILIČIĆ for small w. Hence, we have w Cǫ z e w dw 4πMǫRez for small ǫ. In particular, if Rez >, we see that the integral over C ǫ tends to as ǫ. Hence, by taking the limit as ǫ goes to zero, and the horizontal lines tend to the real axis, we get C w z e w dw = = e πiz t z e t dt eπz i t z e t dt t z e t dt = ieiπz sinπz by... This implies that ie iπz sinπzγzζz = C t z e t dt = ie iπz sinπzγzζz w z e w dw for any path C we considered above and Rez >. Clearly, the right integral makes sense for arbitrary complex z, hence this equality is an equality of meromorphic functions on C. This yields the following result, which is another integral representation of ζ as a meromorphic function.... Lemma. For any z C, we have sinπzγzζz = i e iπz C w z e w dw. On the other hand, we can calculate using the residue theorem the integral along the path γ drawn in the following figure. There the outside square passes thru points m+πi and m+πi. m+πi mπi πi γ -πi -mπi -m+πi
NOTES ON RIEMANN S ZETA FUNCTION 9 We have γ w z m e w dw = πi n= πn z e iπz +πn z e 3iπz = z π z ie iπz m e iπz +e iπz n= n= n z πz m = z+ π z ie iπz cos n z = z+ π z ie iπz sin πz m n z. n= Now we want to estimate the integral along the sides of the square for Rez <. For a fixed z, we see as before that there exists M >, such that on the right vertical side of the square the integrand satisfies the estimate w z e w M w Rez e w. Moreover, we have e w e w = e Rew for any w, hence we have w z e w M w Rez M Rew Rez e Rew e Rew and the right side of the square. This expression clearly tends to zero faster than Rew as Rew +. Hence, the integral over the right side tends to zero as the square grows. On the left side of the square we have the same estimate w z e w M Rew Rez e Rew. Since Rew is negative in this case, for large Rew we have w z e w M Rew Rez. Since Rez <, this bound goes to zero faster that Rew. On the other hand, the length of the side of the square is Rew. Hence, the integral along the left side also tends to zero as the square grows. It remains to treat the top and bottom side. As before, we see that on these sides we have w z e w M w Rez e w M Imw Rez e w. On the other hand, the function w e w on the horizontal lines tends to as Rew + and to as Rew. Hence it is bounded from below. Moreover, since it is periodic with period πi, shifting the line up and down by πi doesn t change that bound. Hence, in our estimates we can assume that e w C
D. MILIČIĆ on top and bottom sides of all squares. This implies that w z e w MC Imw Rez on top and bottom sides of all squares. Since Rez < and the length of these sides is Imw, we see that the integrals over these sides tend to zero as the squares grow. Hence, for Rez <, as the squares grow, i.e., as m tends to infinity, the integral over the sides of the square tends to and the integral along γ converges to the integral along the path C from... In addition, the sum on the right side of the above expression converges to the series for ζ z, i.e., we have Hence, we have This yields and ie πiz sinπzγzζz = z+ π z ie iπz sin sin sinπzγzζz = z π z sin πz cos cos πz Γzζz = z π z sin πz πz ζ z. πz Γzζz = z π z ζ z. ζz. πz ζ z. for all z C. By substituting z for z we get π z z π z ζz = cos Γ zζ z and finally ζz = z π z sin what completes the proof of the theorem. πz Γ zζ z.3. Euler product formula. Let P be the setofallprimenumbersin N. Assume that P = {p,p,...} written in the naturalordering. Foranym N, let S m be the subsetofnconsistingofall integerswhich arenot divisibleby primesp,p,...,p m. Then we claim that, for Rez >, we have m ζz p z = k= k n z. n S m Clearly, p =, and we have z ζz = So, the statement holds for m =. n= n z n z = n z. n S n=
NOTES ON RIEMANN S ZETA FUNCTION Assume that the statement holds for m. Then, by the induction assumption, we have p z ζz m+ m k= p z = k p z m+ n z n S m = n S m n z n S m p m+ n z = n S m+ n z. This establishes the above claim. Therefore, we see that as m tends to we get the formula i.e., the following result holds. ζz k= p z =, k.3.. Theorem Euler product. For Rez > we have ζz = p P p z. The factors in Euler products are nonzero and holomorphic for Rez >. The above observation lead Euler to a proof that the set P is infinite. The finiteness of P would imply that the Euler product is holomorphic for Rez >, contradicting the fact that ζ has a pole at. The Euler product formula implies that ζ has no zeros for Rez >..4. Riemann hypothesis. Since ζ has no zeros for Rez >, and Γ has no zeros at all by.3.3, we see from the functional equation.. that the only zeros of ζ for Rez < come from zeros of the function z sin zπ at points, 4,.... These are called the trivial zeros of ζ. It follows that all other zeros of ζ have to lie in the strip Rez. It is called the critical strip. Actually, Euler didn t use complex variables. One can show that finiteness of primes would contradict the fact that harmonic series diverge.
D. MILIČIĆ Im z Critical Strip - - Re z zero Simple pole Critical line Riemann conjectured that all zeros of ζ in the critical strip lie on the critical line Rez =. This is the Riemann hypothesis. Hadamard and de la Valée-Poussin proved that there are no zeros of ζ on the boundary of the critical strip i.e., for Rez = and Rez =. This implies the Prime Number Theorem conjectured by Gauss.