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An Application of the Dirichlet Integrals to the Summation of Series

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An Application of the Dirichlet Integrals to the Summation of Series Some applications of the Dirichlet integrals to the summation of series and the evaluation of integrals involving the Riemann zeta function Donal F. Connon [email protected] 1 December 2012 Abstract Using the Dirichlet integrals, which are employed in the theory of Fourier series, this paper develops a useful method for the summation of series and the evaluation of integrals. CONTENTS Page 1. Introduction 1 2. Application of Dirichlet’s integrals 5 3. Some applications of the Poisson summation formula: 3.1 Digamma function 11 3.2 Log gamma function 13 3.3 Logarithm function 14 3.4 An integral due to Ramanujan 16 3.5 Stieltjes constants 36 3.6 Connections with other summation formulae 40 3.7 Hurwitz zeta function 48 4. Some integrals involving cot x 62 5. Open access to our own work 68 1. Introduction In two earlier papers [19] and [20] we considered the following identity (which is easily verified by multiplying the numerator and the denominator by the complex conjugate (1− e−ix ) ) 1 1 isin x 1 i (1.1) = − = + cot(x / 2) 1− eix 2 2 1− cos x 2 2 Using the basic geometric series identity 1 N y N +1 =∑ yn + 1− y n=0 1− y we obtain N 1 inx (1.1a) ix =∑e + RN () x 1− e n=0 1 iN()+ x iN(1)+ x 2 ei1 iN(1)+ x e where RxN () == e[]1 + i cot( x / 2) = 12− exix 2sin(/2) Separating the real and imaginary parts of (1.1a) produces the following two identities (the first of which is called Lagrange’s trigonometric identity and contains the Dirichlet kernel DxN () which is employed in the theory of Fourier series [55, p.49]) 1 N sin(N+ 1/ 2) x (1.2) = ∑cosnx − 2 n=0 2sin(x / 2) 1 N cos(N+ 1/ 2) x (1.3) cot(x / 2)=∑ sin nx + 2 n=0 2sin(x / 2) (and these equations may obviously be generalised by substituting αx in place of x . We first multiply by p()x and, in this section, p()x is assumed to be twice continuously differentiable on [,ab]. Integration then gives us 1 bbN p(xixdxpxedxR ) 1+= cot( / 2) ( ) inx + ∫∫()∑ N 2 aan=0 and we have bb1iN cos(+−+ 1/2) x sin( N 1/2) x R ==p()xR () xdx p ()xdx NN∫∫ aa2sin(x /2) 1 b ⎧ cos(x / 2) ⎫ =− ∫ p()cosxNx⎨ + sin Nx⎬dx 2sa ⎩⎭in(x /2) ixb ⎧ cos( / 2) ⎫ +−+∫ p()xNx⎨ sin cos Nx⎬dx 2sa ⎩⎭in(x /2) Therefore, provided sin(x / 2) has no zero in [ ab, ], a weak version of the Riemann- Lebesgue lemma (which was previously employed in Eq.(2.17) in [19] where we required p()x to be twice continuously differentiable) tells us that limRN = 0. N →∞ 2 This will also be the case if sin()a / 2= 0 , provided p( a )= 0. From the above we can therefore derive the following trigonometric integral identities: 1 bb∞ (1.4) ∫∫p()xdx= ∑ px ()cos nxdx 2 aan=0 1 bb∞ (1.4a) ∫∫p()cot(/2)x x dx= ∑ p ()sin x nx dx 2 aan=0 and more generally we have (by letting x → α x in (1.2) and (1.3) respectively) 1 bb∞ (1.5) ∫∫p()xdx= ∑ px ()cosα nxdx 2 aan=0 1 bb∞ (1.5a) ∫∫p(x )cot(αα x / 2) dx= ∑ p ( x )sin nx dx 2 aan=0 Equations (1.5) and (1.5a) are valid provided (i) sin(α x / 2)≠ 0 ∀∈ xab[ , ] or, alternatively, (ii) if sin(αa / 2)= 0 then pa()= 0 also. Similarly, using the identity 1 1 isin x 1 i (1.6) = − = − tan(x / 2) 1+ eix 2 2 1+ cos x 2 2 we may easily prove that 1 bb∞ (1.7) ∫∫p()xdx=−∑ px ()(1)cosn nxdx 2 aan=0 1 bb∞ (1.7a) ∫∫p()tan(/2)x x dx=−∑ p ()(1)sin xn+1 nx dx 2 aan=0 and more generally 1 bb∞ (1.8) ∫∫p()xdx=−∑ px ()(1)cosn α nxdx 2 aan=0 1 bb∞ (1.8a) ∫∫p()tan(xαα x /2) dx=−∑ p ()(1)sin x n+1 nx dx 2 aan=0 3 From (1.6) we note that the denominator is cos(x / 2) and hence (1.7) and (1.7a) are only valid provided either (i) cos(x / 2) has no zero in [ ab, ] or (ii) if cos(a / 2) = 0 , then p( a) must also be zero. Equations (1.8) and (1.8a) are valid provided (i) cos(α x / 2)≠ 0 ∀∈ xab[ , ] or, alternatively, (ii) if cos(αa / 2)= 0 then pa() = 0 also. The following simple trigonometric identities are easily proved 2 (1.9) cot(x / 2)+ tan( x / 2) = sin x (1.10) cot(x / 2)− tan( x / 2)= 2cot x Therefore, combining (1.5) and (1.7) produces the following identity bbpx() ∞∞ b (1.11) ∫∫dx=−−∑∑ p( x )sin nx dx ∫ p( x )( 1)n sin nx dx aasin x nn==00 a which simplifies to bbpx() ∞ (1.12) ∫∫dx=+2()sin(21)∑ p x n x dx aasin x n=0 Similarly, using (1.10) we obtain b ∞∞bb (1.13) ∫ p()cotxxdx=+∑∑∫∫p()sinxnxdx (1)()sin−n pxnxdx a nn==00aa which simplifies to bb∞ (1.14) ∫∫p()cotx x dx= 2∑ p ()sin2 x nx dx aan=1 It should be noted that in the above formulae we require either (i) both sin(x / 2) and cos(x / 2) have no zero in [ ab, ] or (ii) if either sin(a / 2) or cos(a / 2) is equal to zero then p( a) must also be zero. Condition (i) is equivalent to the requirement that sin x has no zero in [ ab, ]. Note that (1.14) is equivalent to (1.5a) with α = 2 . More generally we have 4 bbpx() ∞ (1.14a) ∫∫dx=+2()sin(21)∑ p x[] nα x dx aasinα x n=0 bb∞ (1.14b) ∫∫p()cotxαα x dx= 2∑ p ()sin2 x nx dx aan=1 Equations (1.14a) and (1.14b) are valid provided (i) sin(α x)≠ 0 ∀∈ xab[ , ] or, alternatively, (ii) if sin(αa )= 0 then pa( )= 0 also. We may also generalise Poisson’s integral [17, p.250] by applying the above analysis to 1 1 the quotient and useful results may also be obtained by using . 1± reix 1± ireix 2. Application of Dirichlet’s integrals In the previous section we required that p()x was twice continuously differentiable on the interval [,ab]; with the assistance of Dirichlet we now show that this condition may be significantly relaxed. Dirichlet [26] considered the following integrals in his classical treatment of Fourier series in 1829 (when he was then 23 years old) b sin(μt ) (2.1) limf (tdt ) = 0 μ→∞ ∫ a sin t b sin(μπt ) (2.2) limft ( ) dt= f(0+ ) μ→∞ ∫ 0 sint 2 where 0 < a < b < π . Dirichlet showed that these limits are valid if f ()x is continuous on [0,b ] and f ()x only has a finite number of maxima and minima on [0,b] . Jordan [37] subsequently showed in 1881 that the less restrictive condition that f ()x be of bounded variation on [0,]b is sufficient. Proofs may be found in [5, p.314] and [17, p.219]. We recall from [5, p.128] that if f ()x is monotonic on [ab,], then f ()x is of bounded variation on [ab,]. In addition, we also note that if f (x) is continuous on [ab,] and if f ′()x exists and is bounded on the open interval, then f ()x is of bounded variation on [,]ab. Even though Jordan wanted to relax the smoothness requirement for the pointwise convergence of Fourier series, it subsequently turned out that functions of bounded variation are in fact fairly smooth anyway; Lebesgue ([50, p.356 ] and [7, p.192]) proved 5 that any monotonic function on a closed interval is differentiable almost everywhere, and that its derivative is integrable. With the substitution t= α x in (2.1) and (2.2) we see that αb sin(μα x ) (2.3) limfx (α ) dx= 0 μ→∞ ∫ αa sinα x or equivalently b′ sin(μα x ) (2.4) limgx ( ) dx= 0 μ→∞ ∫ a′ sinα x where 0 < a′ < b′ < π . We may write (1.2) as 1sN in(2Nx+1)α (2.5) =−∑cos 2αnx 22n=0 sinα x and, after multiplying this by a function f ()x of bounded variation on [ab,], we easily see that 1sbbN bin(2Nx+1)α (2.6) ∫∫f ()xdx=−∑ f ()cos2 xα nxdx ∫ f () x dx 2 aan=0 a2sinα x Letting N →∞ we obtain 1 bb∞ (2.7) ∫∫f ()xdx= ∑ f ()cos2 xα nxdx 2 aan=0 provided 0 < αa < αb < π . This is a generalised version of (1.4) above and, in particular, it should be noted that we no longer require f ()x to be twice continuously differentiable on [,ab]. With α = π we may write (2.7) as 1 bb∞ (2.8) ∫∫f ()xdx= ∑ f ()cos2 xπ nxdx 2 aan=0 where 0 < a < b < 1. We now consider the case where (2.2) applies. With the substitution t= α x in (2.2) we see that 6 αb sin(μαx ) π (2.9) limααfx ( ) dxf= (0+ ) μ→∞ ∫ 0 sinα x 2 or equivalently b′ sin(μαx ) π (2.10) limgx ( ) dx= g(0+ ) μ→∞ ∫ 0 sinααx 2 where 0 < b′ < π . Letting N →∞ in (2.6) and using (2.10) we then obtain 1 bb∞ π ∫∫f() x dx=−∑ f ()cos2 xα nx dx f (0)+ 2400n=0 α where 0 < αb < π .
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