THE RAMANUJAN JOURNAL, 5, 73–89, 2001
c 2001 Kluwer Academic Publishers. Manufactured in The Netherlands.
°
Some Remarkable Properties of Sinc
and Related Integrals∗
DAVID BORWEIN
dborwein@uwo.ca
Department of Mathematics, University of Western Ontario, London, Ontario, Canada N6A 5B7
JONATHAN M. BORWEIN
jborwein@cecm.sfu.ca
Centre for Experimental and Constructive Mathematics, Department of Mathematics and Statistics,
Simon Fraser University, Burnaby, B.C., Canada V5A 1S6
Received December 3, 1999; Accepted August 14, 2000
Abstract.
Using Fourier transform techniques, we establish inequalities for integrals of the form
Z ∞Y
n
sin(ak x)
d x.
ak x
0 k=0
We then give quite striking closed form evaluations of such integrals and finish by discussing various extensions
and applications.
Key words: sinc integrals, Fourier transforms, convolution, Parseval’s theorem
2000 Mathematics Subject Classification: Primary—42A99; 42A38
1.
Introduction
Motivated by questions about the integral1
Z
µ ¶
x
cos
d x,
k
k=1
∞
∞Y
µ :=
0
(1)
we study the behaviour of integrals of the form
Z
0
n
∞Y
sin(ak x)
d x.
ak x
k=0
In Section 2 we use Fourier transform theory to establish monotonicity properties of these
integrals as functions of n. In Section 3, by direct methods, we give closed forms for these
integrals and for similar integrals also incorporating cosine terms. In Section 4, we provide
∗ Research
1 Through
supported in part by the National Sciences and Engineering Research Council of Canada.
J. Selfridge and R. Crandall we learned that B. Mares discovered that µ < 4z .
74
BORWEIN AND BORWEIN
a very different proof of one of these results following an idea in an 1885 paper of Störmer
[2]. Finally, in Section 5 we return to the study of (1).
2.
Fourier cosine transforms and sinc integrals
Define

 sin x
sinc(x) :=
x

1
if x 6= 0
if x = 0.
and, for a > 0,

1



1
χa (x) :=
2


0
if |x| < a
if |x| = a
if |x| > a.
We first state some standard results about the Fourier cosine transform (FCT) which may
be found in texts such as [4, ch. 13].
The FCT of a function f ∈ L 1 (−∞,∞) is defined to be the function fˆ given by
Z ∞
1
fˆ(t) := √
f (x) cos (xt) d x.
2π −∞
Observe that if f is also even, then so is fˆ and
r Z ∞
2
ˆ
f (x) cos (xt) d x.
f (t) =
π 0
Further, if f is even and f ∈ L 1 (−∞,∞) ∩ L 2 (−∞,∞), then fˆ ∈ L 2 (−∞,∞). If,
in addition, this fˆ ∈ L 1 (−∞,∞), then f is equivalent to the FCT of fˆ, that is
Z ∞
1
f (x) = √
fˆ(t) cos (xt) dt for a.a. x ∈ (−∞, ∞).
2π −∞
Hence, if f is even, f ∈ L 1 (−∞,∞) ∩ L 2 (−∞,∞), fˆ ∈ L 1 (−∞, ∞), and f is continuous
on (−α, α) for some α > 0, then
Z ∞
1
f (x) = √
fˆ(t) cos (xt) dt for x ∈ (−α, α),
2π −∞
since the right-hand term is also continuous
q on (−α,α) by dominated convergence.
q
Note that, for a > 0, the FCT of χa is a π2 sinc(ax), so that the FCT of a π2 sinc(ax) is
equivalent to χa . (In fact it can easily be shown to be identically equal to χa , either directly
75
SINC INTEGRALS
or by appeal to a standard result about inverse Fourier transforms of functions of local
bounded variation.)
Note also that if fˆ1 , fˆ2 are FCTs of even functions f 1 , f 2 ∈ L 1 (−∞, ∞) ∩ L 2 (−∞, ∞),
then f 1 f 2 is the FCT of √12π f 1 ∗ f 2 , where
Z
f 1 ∗ f 2 (x) :=
∞
f 1 (x − t) f 2 (t) dt
−∞
for all real x.
In addition, we have the following version of Parseval’s theorem for such even functions:
Z ∞
Z ∞
f 1 (x) f 2 (x) d x =
fˆ1 (x) fˆ2 (x) d x,
0
0
provided at least one of the functions f 1 , f 2 is real.
We are now in a position to prove:
Theorem 1. Suppose that {an } is a sequence of positive numbers. Let sn :=
Z
n
∞Y
τn :=
0
Pn
k=1
ak and
sinc(ak x) d x.
k=0
(i) Then
0 < τn ≤
1 π
,
a0 2
with equality if n = 0, or if a0 ≥ sn when n ≥ 1.
(ii) If an+1 ≤ a0 < sn with n ≥ 1, then
0 < τn+1 ≤ τn <
(iii) If a0 < sn 0 with n 0 ≥ 1, and
Z
τn ≥
0
∞
∞Y
k=0
P∞
k=0
1 π
.
a0 2
ak2 < ∞, then there is an integer n 1 ≥ n 0 such that
Z
sinc(ak x) d x ≥
0
∞
∞Y
sinc2 (ak x) d x > 0
for all n ≥ n 1 .
k=0
Observe that applying Theorem 1 to different permutations of the parameters will in general
yield different inequalities.
Proof: Part (i). That τ0 = a10 π2 is a standard result (proven e.g., by contour integration
in [1, p. 157] and by Fourier analysis in [3, p. 563]) with the integral in question being
improper (i.e. not absolutely convergent—the integrals in the other cases are absolutely
convergent). Assume therefore that n ≥ 1, and let
r
r
¡√ ¢1−n
1 π
1 π
χa ,
χa .
Fn :=
2π
f 1 ∗ f 2 ∗ · · · ∗ f n , where f n :=
F0 :=
a0 2 0
an 2 n
76
BORWEIN AND BORWEIN
Then it is readily verified by induction that, for n ≥ 1, Fn (x) is an even function which
vanishes on (−∞, −sn ) ∪ (sn , ∞) and is positive on (−sn , sn ). Moreover, Fn+1 = √12π Fn
∗ f n+1 , so that
Z
1
Fn+1 (x) = √
2π
∞
−∞
1
2an+1
Fn (x − t) f n+1 (t) dt =
Z
x+an+1
Fn (u) du.
x−an+1
Hence Fn+1 (x) is absolutely continuous on (−∞, ∞) and, for almost all x ∈ (−∞, ∞),
0
(x) = Fn (x + an+1 ) − Fn (x − an+1 ) = Fn (x + an+1 ) − Fn (an+1 − x).
2an+1 Fn+1
Since (x + an+1 ) ≥ max{(x − an+1 ), (an+1 − x)} ≥ 0 when x > 0, it follows that if Fn (x) is
0
(x) ≤ 0 for a.a. x ∈ (0, ∞), and so Fn+1 (x)
monotone non-increasing on (0, ∞), then Fn+1
is monotone non-increasing on (0, ∞). This monotonicity property of Fn on (0, ∞) is
therefore established by induction for all n ≥ 1. Also
Fn is the FCT of σn (x) :=
n
Y
sinc(ak x),
and
σn is the FCT of Fn .
k=1
Thus, all our functions and transforms are even and are in L 2 (−∞, ∞). Hence, by the
above version of Parseval’s theorem,
Z
∞
τn =
0
1
Fn (x)F0 (x) d x =
a0
r
π
2
Z
min(sn ,a0 )
Fn (x) d x.
(2)
0
p p
When a0 ≥ sn , the final term is equal to a10 π2 π2 σn (0) = a10 π2 since σn (x) is continuous
on (−∞, ∞); and when a0 < sn , the term is positive and less than a10 π2 since Fn (x) is
positive and continuous for 0 < x < sn . This establishes part (i).
Part (ii). Observe again that Fn+1 = √12π Fn ∗ f n+1 , and hence that, for y > 0,
Z
y
0
Z y
Z ∞
1
Fn+1 (x) d x = √
dx
Fn (x − t) f n+1 (t) dt
2π 0
−∞
Z an+1
Z y
Z an+1 Z y
1
1
dx
Fn (x − t) dt =
dt
Fn (x − t) d x
=
2an+1 0
2an+1 −an+1
0
−an+1
Z y
Z an+1 Z y−t
1
1
dt
Fn (u) du =
Fn (u) du +
(I1 + I2 ),
=
2an+1 −an+1
2a
n+1
−t
0
where
Z
I1 :=
Z
an+1
0
dt
−an+1
−t
Z
Fn (u) du
and
I2 :=
Z
an+1
y−t
dt
−an+1
y
Fn (u) du.
77
SINC INTEGRALS
Now I1 = 0 since
R0
−t
Fn (u) du is an odd function of t, and for y ≥ an+1 ,
Z
I2 =
Z
an+1
y−t
dt
0
y
0
y−t
Z
Fn (u) du +
Z
0
y−t
dt
−an+1
Z an+1
Fn (u) du
y
Z y+t
Z an+1 Z y
dt
Fn (u) du +
dt
Fn (u) du
=−
y−t
y
0
0
Z an+1 Z y
dt
(Fn (u + t) − Fn (u)) du ≤ 0
=
since Fn (u) is monotonic non-increasing for u ≥ y − t ≥ y − an+1 ≥ 0. Hence
Z
y
Z
y
Fn+1 (x) d x ≤
0
Fn (x) d x
when an+1 ≤ y < sn .
(3)
0
It follows from (2), and (3) with y = a0 , that 0 < τn+1 ≤ τn if an+1 ≤ a0 < sn , and this
completes part (ii).
Q
2
Part (iii). Let ρ(x) := limn→∞ σn2 (x) = ∞
k=1 sinc (ak x) for x > 0. Observe that the
2
limit exists since 0 ≤ sinc (ak x) < 1, and that there is a set A differing from (0, ∞) by a
countable set such that 0 < sinc2 (ak x) < 1 whenever x ∈ A and k = 1, 2, . . . . Now
sinc(ak x) = 1 − δk ,
where 0 ≤
δk
x2
as k → ∞,
→
3
ak2
P∞
so that
k=1 δk < ∞, and hence, by standard theory of infinite products, σ (x) :=
limn→∞ σn (x) exists and σ 2 (x) = ρ(x) > 0 for x ∈ A. It follows, by part (ii), that
Z
τn ≥
∞
0
Z
σn2 (x) d x
≥
∞
ρ(x) d x > 0
0
for all n ≥ n 1 , where n 1 ≥ n 0 is an integer such that an+1 ≤ a0 for all n ≥ n 1 . In addition,
by dominated convergence,
Z
lim τn =
n→∞
∞
Z
σ (x) d x ≥
0
and this completes the proof of part (iii).
3.
∞
ρ(x) d x,
0
2
Some elementary identities
In this section we prove some identities involving products of sines and cosines by straightforward methods not involving Fourier transform theory. We adopt the usual convention
78
BORWEIN AND BORWEIN
that empty sums have the value 0 and empty products have the value 1, and we define

if x > 0

1
if x = 0
sign(x) := 0


−1 if x < 0.
Theorem 2. Let a0 , a1 , . . . , an be complex numbers with n ≥ 1. For each of the 2n
ordered n-tuples γ := (γ1 , γ2 , . . . , γn ) ∈ {−1, 1}n define
bγ := a0 +
n
X
γk ak ,
²γ :=
k=1
n
Y
γk .
k=1
(i) Then
X
γ ∈{−1,1}n
²γ bγr
=


0
n

 2 n!
n
Y
for r = 1, 2, . . . , n − 1
ak
for r = n,
k=1
and
n
Y
sin(ak x) =
k=0
1
2n
µ
¶
π
²γ cos bγ x − (n + 1) .
2
γ ∈{−1,1}n
X
(ii) If a0 , a1 , . . . , an are real, then
Z
X
π 1
sin(ak x)
dx =
²γ bγn sign(bγ ).
n
x
2
2
n!
k=0
γ ∈{−1,1}n
n
∞Y
0
If, in addition,
a0 ≥
n
X
|ak |,
k=1
then
Z
n
∞Y
0
n
sin(ak x)
πY
ak .
dx =
x
2 k=1
k=0
Proof:
Observe that
e a0 t
n
Y
¡
¢
eak t − e−ak t =
k=1
X
²γ ebγ t .
γ ∈{−1,1}n
Since eak t − e−ak t = 2ak t + O(t 2 ) as t → 0, the first summation formula in part (i) follows
on equating coefficients of t r in the above identity. Note that the formula also holds for
79
SINC INTEGRALS
r = 0 if we define bγ0 = 1 even when bγ = 0. Similarly
n
Y
n
¢Y
¡ iak x
¢
1 ¡ ia0 x
−ia0 x
−
e
− e−iak x
e
e
n+1
(2i)
k=1
X
¡ ibγ x
¢
1
²γ e
− (−1)n e−ibγ x
=
n+1
(2i)
γ ∈{−1,1}n
¶
µ
X
π
1
²γ cos bk x − (n + 1) ,
= n
2 γ ∈{−1,1}n
2
sin(ak x) =
k=0
and this completes the proof of part (i).
To prove part (ii) of the theorem, observe that
Z
Z ∞Y
n
1 ∞ −n−1
sin(ak x)
dx = n
x
Cn (x) d x,
x
2 0
0 k=0
(4)
P
where Cn (x) := γ ∈{−1,1}n ²γ cos(bγ x − π2 (n + 1)). Because Cn (x) is an entire function,
bounded for all real x, with a zero of order n + 1 at x = 0, we can integrate the right-hand
side of (4) by parts n times to get
Z ∞
Z ∞Y
n
1
sin(ak x)
dx X
dx = n
²γ bγn sin(bγ x)
x
2 n! 0 x γ ∈{−1,1}n
0 k=0
Z ∞
X
sin(bγ x)
1
n
dx
²γ bγ
= n
2 n! γ ∈{−1,1}n
x
0
X
π 1
=
²γ bγn sign(bγ ).
2 2n n! γ ∈{−1,1}n
Since the additional hypothesis implies that bγ ≥ 0 for all γ ∈ {−1, 1}n , the final formula
in the theorem follows from part (i).
2
Corollary 1. If 2ak ≥ an > 0 for k = 0, 1, . . . , n − 1 and
n
X
ak > a0 ≥
k=1
n−1
X
ak ,
k=1
then
Z
r
∞Y
0
r
πY
sin(ak x)
ak
dx =
x
2 k=1
k=0
while
Z
n
∞Y
0
π
sin(ak x)
dx =
x
2
k=0
(
for r = 0, 1, . . . , n − 1,
)
(a1 + a2 + · · · + an − a0 )n
.
ak −
2n−1 n!
k=1
n
Y
80
BORWEIN AND BORWEIN
Proof: Let γ 0 := (−1, −1, . . . , −1) ∈ {−1, 1}n , Observe that bγ 0 := a0 − a1 − · · · −an
< 0, that bγ ≥ 0 for every other γ ∈ {−1, 1}n , and that ²γ 0 = (−1)n . It follows that
Z
∞
0
n
Y
sin(ak x)
k=0
x
dx =
X
π 1
²γ bγn sign(bγ )
n
2 2 n! γ ∈{−1,1}n
!
Ã
X
π 1
²γ bγn + ²γ 0 bγn 0 (sign(bγ 0 ) − 1)
=
2 2n n! γ ∈{−1,1}n
(
)
n
2(−bγ 0 )n
π Y
,
ak −
=
2 k=1
2n n!
2
as desired.
Remarks 1.
(a) If a0 , a1 , . . . , an are real and non-zero, then, by Theorem 2(ii),
Z
X
π 1
sin(ak x)
dx =
²γ bγn sign(bγ )
x
2 2n n! γ ∈{−1,1}n
0 k=0
!
Ã
X
X
π 1
n
n
²γ bγ +
²γ bγ (sign(bγ ) − 1)
=
2 2n n! γ ∈{−1,1}n
bγ <0
(
)
X
1
π
n
1 − n−1
²γ bγ .
=
2a0
2 n!a1 a2 · · · an bγ <0
τn : =
n
∞Y
(b) Suppose further that ak > 0 for k = 0, 1, . . . , n. Consider the polyhedra
Pn = Pn (a0 , a1 , . . . , an )
(
:= (x1 , x2 , . . . , xn ) | − a0 ≤
n
X
)
xk ≤ a0 , −ak ≤ xk ≤ ak
for k = 1, 2, . . . , n .
k=1
Q n = Q n (a0 , a1 , . . . , an )
(
:= (x1 , x2 , . . . , xn ) | − a0 ≤
n
X
)
ak xk ≤ a0 , −1 ≤ xk ≤ 1
for k = 1, 2, . . . , n ,
k=1
Hn := {(x1 , x2 , . . . , xn ) | − 1 ≤ xk ≤ 1
for k = 1, 2, . . . , n}.
(i) If we return to Eq. (2) we observe that
Z min(sn ,a0 )
π
1
τn =
χa1 ∗ χa2 ∗ · · · ∗ χan d x
a 0 2n a 1 a 2 · · · a n 0
π Vol(Q n )
Vol(Pn )
π
=
.
=
n
2a0 2 a1 a2 · · · an
2a0 Vol(Hn )
Moreover, we now explain P
the behaviour of τn when we note that the value drops precisely
when the constraint −a0 ≤ nk=1 ak xk ≤ a0 becomes active and bites into the hypercube Hn .
81
SINC INTEGRALS
(ii) We sketch a probabilistic interpretation. From (i) it follows that pn := 2a0 τn /π
may be regarded as the probability that independent
random variables {xk , k = 1, 2, . . .}
P
identically distributed in [−1, 1] satisfy | nk=1 ak xk | ≤ a0 . Correspondingly
p∞
2a0
:=
π
is the probability that the constraint |
decreases monotonically to p∞ .
(c) Consider now the special case
Z
∞
∞Y
0
P∞
k=1
sinc(ak x) d x
k=1
ak xk | ≤ a0 is met. We have also shown that pn
Z
µn := τn−1 =
∞
sincn (x) d x.
0
In this case we have ak = 1 for k = 0, 1, . . . , n − 1, and it is straightforward to verify that
X
γ ∈{−1,1}n−1 ,bγ <0
²γ bγn−1 =
X
1≤r ≤ n2
(−1)r +1
µ
¶
n−1
(n − 2r )n−1 ,
r −1
and hence that
(
)
µ
¶
X
π
2
r +1 n − 1
n−1
(−1)
1 − n−1
(n − 2r )
µn =
2
2 (n − 1)! 1≤r ≤ n
r −1
2
(
)
1 X (−1)r (n − 2r )n−1
π
1 + n−2
.
=
2
2
(r − 1)! (n − 1)!
1≤r ≤ n
2
The following formula for µn appears as an exercise in [5, p. 123]:
µ ¶
X
π
r n
(−1)
(n − 2r )n−1 .
µn = n
2 (n − 1)! 0≤r ≤ n
r
2
To show that this formula for µn is equivalent to the one derived above, it clearly suffices
to prove that
µ ¶
µ
¶
X
n
n−1
n−1
r +1 n − 1
(−1)
= 2 (n − 1)! +
(−1)
(n − 2r )
(n − 2r )n−1 .
r
r
−
1
n
n
0≤r ≤
1≤r ≤
X
r
2
2
Since
µ ¶
µ ¶
µ
¶
n
n−1
1 n
(n − 2r ),
−2
=
n r
r
r −1
82
BORWEIN AND BORWEIN
this is equivalent to proving that
X
(−1)r
0≤r ≤ n2
µ ¶
n
(n − 2r )n = 2n−1 n!,
r
which, by symmetry, is equivalent to proving that
µ ¶
n
n
1 X
(−1)r
(n − 2r )n = 2n .
n! r =0
r
But the left-hand side of this latter identity is the coefficient of t n in
µ ¶
n
X
r n
(−1)
e(n−2r )t = ent (1 − e−2t )n = (2 sinh t)n .
r
r =0
Since 2 sinh t = 2t+O(t 2 ) as t → 0, the coefficient is indeed 2n , and the desired equivalence
of the formulae for µn is proved.
The next theorem extends Theorem 2 by adjoining cosines to the product of sines.
Theorem 3. Let a0 , a1 , . . . , an+m be complex numbers with n ≥ 1 and m ≥ 0. For each
of the 2n+m ordered (n + m)-tuples γ := (γ1 , γ2 , . . . , γn+m ) ∈ {−1, 1}n+m define
bγ := a0 +
n+m
X
γk ak ,
²γ :=
k=1
n
Y
γk .
k=1
(i) Then
X
γ ∈{−1,1}n+m
and
Ã
n
Y
!Ã
sin(ak x)
k=0
²γ bγr
n+m
Y
k=n+1
=


0
n+m
n!

2
n
Y
ak
for r = n,
k=1
!
cos(ak x) =
for r = 1, 2, . . . , n − 1
1
2n+m
µ
¶
π
²γ cos bγ x − (n + 1) .
2
γ ∈{−1,1}n+m
X
(ii) If a0 , a1 , . . . , an+m are real, then
!
!Ã
Z ∞ ÃY
n
n+m
X
Y
π
sin(ak x)
1
cos(ak x) d x =
²γ bγn sign(bγ ).
n+m
x
2
2
n!
0
n+m
k=0
k=n+1
γ ∈{−1,1}
If, in addition,
a0 ≥
n+m
X
k=1
|ak |,
83
SINC INTEGRALS
then
Z
∞
Ã
0
Proof:
n
Y
sin(ak x)
!Ã
x
k=0
n+m
Y
!
cos(ak x) d x =
k=n+1
n
πY
ak .
2 k=1
By Theorem 2 we have that
n+m
Y
sin(ak x) =
k=1
X
1
2n+m
γ ∈{−1,1}n+m
µ
¶
π
cos bγ x − (n + m + 1) ,
2
²γ0
where, for each γ = (γ1 , γ2 , . . . , γn+m ) ∈ {−1, 1}n+m ,
bγ = a 0 +
n+m
X
²γ0 =
γk a k ,
n+m
Y
k=1
and
X
γ ∈{−1,1}n+m
²γ0 bγr
=
γk = ±1,
k=1


0
n+m
(n + m)!

2
n+m
Y
for r = 1, 2, . . . , n + m − 1
ak
for r = n + m.
k=1
Differentiating these expressions partially with respect to an+1 , an+2 , . . . , an+m yields part
(i) of Theorem 3 with
!
Ã
m
n
m
n
Y
Y
Y
Y
0
2
²γ = ²γ
γn+k =
γk
γn+k
=
γk .
k=1
k=1
k=1
k=1
To deal with part (ii) of Theorem 3 we observe that, by Theorem 2, if a0 , a1 , . . . , an+m are
real, then
Z
∞ n+m
Y
0
k=0
X
π
1
sin(ak x)
dx =
² 0 bn+m sign(bγ ).
x
2 2n+m (n + m)! γ ∈{−1,1}n+m γ γ
Differentiating partially with respect to an+1 , an+2 , . . . , an+m , we get
!
!Ã
Z ∞ ÃY
n
n+m
Y
sin(ak x)
cos(ak x) d x
x
0
k=0
k=n+1
X
π
1
=
²γ bγn sign(bγ ).
n+m
22
n! γ ∈{−1,1}n+m
If, in addition,
a1 ≥
n+m
X
k=2
|ak |,
84
BORWEIN AND BORWEIN
then, by Theorem 2,
Z
∞ n+m
Y
0
k=0
Y
π n+m
sin(ak x)
dx =
ak .
x
2 k=2
Differentiating partially with respect to an+1 , an+2 . . . . , an+m , we get
!Ã
!
Z ∞ ÃY
n
n+m
n
Y
πY
sin(ak x)
cos(ak x) d x =
ak .
x
2 k=1
0
k=0
k=n+1
2
Corollary 2. If 2ak ≥ an+m > 0 for k = 0, 1, . . . , n + m − 1 and
n+m
X
ak > a0 ≥
n+m−1
X
k=1
then
Z
∞
0
Ã
r
Y
sin(ak x)
k=0
!Ã
x
rY
+m
!
cos(ak x) d x =
k=r +1
while
Z
∞
Ã
n
Y
sin(ak x)
0
k=0
π
=
2
(
ak ,
k=1
!Ã
x
n+m
Y
n
πY
ak
2 k=1
for r = 1, 2, . . . , n − 1,
!
cos(ak x) d x
k=n+1
)
(a1 + a2 + · · · + an+m − a0 )n
.
ak −
2n+m−1 n!
k=1
n
Y
Proof: The first part follows immediately from Theorem 3, and the second part can be
derived from Corollary 1 with n + m in place of n by differentiating partially with respect
to an+1 , an+2 , . . . , an+m , as above.
2
4.
An alternative proof
The next theorem is a restatement of the last part of Theorem 3 restricted to real numbers.
It appears as an example without proof in [5, p. 122] where it is ascribed to Carl Störmer
[2]. Störmer’s article does not contain the integral in question, but his proof for the series
identity
!
!Ã
Ã
∞
n
m
n
X
Y
Y
1Y
sin(ra
)
k
(−1)r +1
cos(r c j ) =
ak ,
r
2 k=1
r =1
k=1
j=1
provided
n
X
k=1
|ak | +
m
X
j=1
|c j | < π,
85
SINC INTEGRALS
is readily adapted to yield a proof of the theorem which is radically different from the proof
of Theorem 3.
If a, a1 , a2 , . . . , an , c1 , c2 , . . . , cm , are real numbers with a > 0 and
Theorem 4.
a≥
n
X
|ak | +
m
X
k=1
|c j |,
j=1
then
Z
∞
0
Ã
n
Y
sin(ak x)
k=1
!Ã
x
m
Y
!
cos(c j x)
j=1
n
πY
sin(ax)
dx =
ak .
x
2 k=1
(5)
Proof: We prove the theorem by induction. Applying as before the convention that empty
sums have the value 0 and empty products have the value 1, we observe that formula (5)
for the case n = m = 0 reduces to the standard result
Z ∞
sin(ax)
π
dx = .
x
2
0
Formula (5) also holds for the case n = 1, m = 0, by the case n = 1 of Theorem 1 (which
can easily be proved directly).
Assume that the theorem holds for certain integers n ≥ 1 and m ≥ 0. First suppose that
a≥
n
X
|ak | +
m+1
X
k=1
|c j |.
j=1
Then
a ≥ |a1 ± cm+1 | +
n
X
k=2
|ak | +
m
X
|c j |,
j=1
and hence
Z
∞
0
=
!
Ã
!Ã
n
m
Y
sin(ax)
sin(a1 ± cm+1 ) Y
sin(ak x)
cos(c j x)
dx
x
x
x
k=2
j=1
n
Y
π
ak .
(a1 ± cm+1 )
2
k=2
Adding the two identities in (6), we immediately obtain
!Ã
!
Z ∞ÃY
n
m+1
n
Y
πY
sin(ax)
sin(ak x)
dx =
cos(c j x)
ak .
x
x
2 k=1
0
k=1
j=1
(6)
(7)
86
BORWEIN AND BORWEIN
Next suppose that
a≥
n+1
X
|ak | +
k=1
m
X
|c j |,
j=1
and let t lie between 0 and an+1 . Then, by (7), we have
!
!Ã
Z ∞ÃY
n
m
n
Y
sin(ax)
sin(ak x)
πY
cos(c j x) cos(t x)
ak .
dx =
x
x
2 k=1
0
k=1
j=1
Now integrate (8) with respect to t from 0 to an+1 to get
!Ã
!
Z ∞ Ã n+1
m
Y sin(ak x)
Y
Y
π n+1
sin(ax)
dx =
cos(c j x)
ak .
x
x
2 k=1
0
k=1
j=1
(8)
(9)
Identities (7) and (9) show that if the theorem holds for a pair of integers n, m with
n ≥ 1, m ≥ 0, then it also holds for the pairs n, m + 1 and n + 1, m. Since it holds for
n = 1, m = 0, the proof is completed by induction.
2
Remarks 2. Parts of our previous theorems do, of course, overlap with Theorem 4, but
this latter theorem does not deal with cases where the identity in (4) fails, whereas the other
theorems do. Thus, for example,
Z
∞
π
sinc(x) d x = ,
2
µ ¶
Z ∞
x
π
sinc(x) sinc
dx = ,
3
2
0
µ ¶
µ ¶
Z ∞
x
x
π
sinc(x) sinc
· · · sinc
dx = ,
3
13
2
0
0
yet
Z
µ ¶
µ ¶
x
x
sinc(x) sinc
· · · sinc
dx
3
15
0
467807924713440738696537864469
=
π,
935615849440640907310521750000
∞
(10)
and this fraction in (10), in accord with Corollary 1, is approximately equal to
0.499999999992646. When this fact was recently verified by a researcher using a computer
algebra package, he concluded that there must be a “bug” in the software. Not so. In the
1
1
< 1, but with the addition of 15
, the sum exceeds 1 and the
above example, 13 + 15 + · · · + 13
identity no longer holds. This is a somewhat cautionary example for too enthusiastically
inferring patterns from symbolic or numerical computation.
87
SINC INTEGRALS
5.
An infinite product of cosines
We return to the integral, which we denote by µ, in (1). Let
µ ¶
x
.
cos
C(x) :=
n
n=1
∞
Y
2
x
1
This product is absolutely convergent, since cos ( nx ) = 1 − 2n
2 + O( n 4 ) as n → ∞. Here
and elswhere in this section we ignore the countable set of points on which individual terms
of such an infinite product vanish. Recall the absolutely convergent Weierstrass products
[4, p. 144]
sinc(x) =
∞ µ
Y
1−
n=1
¶
x2
,
π 2n2
cos(x) =
∞ µ
Y
1−
k=0
from which it follows that
∞ µ
∞ Y
Y
1−
C(x) =
4x 2
π 2 n 2 (2k + 1)2
n=1 k=0
¶
µ
∞
Y
2x
=
.
sinc
2k + 1
k=1
¶
=
¶
4x 2
,
π 2 (2k + 1)2
∞ µ
∞ Y
Y
k=0 n=1
4x 2
1− 2 2
π n (2k + 1)2
¶
(11)
It is interesting to note that the alternative absolutely convergent product expansion of C(x)
afforded by (11) can also be derived from the Weierstrass expansion of sinc(x) together
with Vieta’s formula [3, p. 419] in the form
sinc(2x) =
∞
Y
µ
cos
n=0
¶
x
,
2n
since every positive integer is uniquely expressible as an odd integer times a power of 2.
Now apply Theorem 1 and (11) to obtain
Z
∞
0<µ=
0
Z
C(x) d x = lim
N →∞ 0
N
∞Y
µ
2x
sinc
2k
−1
k=1
¶
dx <
π
.
4
These sinc integrals areP
essentially those of the
previous Remarks. Note that all parts of
P∞
1
1
<
∞
=
Theorem 1 apply since ∞
k=1 (2k−1)2
k=1 2k−1 .
We observe that Theorem 1 allows for reasonable lower bounds on µ. Indeed, as cos2 x >
1−x 2 > 0 for 0 < x < 1, we see—using the product form for sinc—that C 2 (x) > sinc(π x)
on the same range. Hence, by Theorem 1(iii),
Z ∞
Z
1 π
π
>µ>
C 2 (x) d x >
sinc(x) d x ≈ .5894898722.
4
π 0
0
88
BORWEIN AND BORWEIN
We could produce a better lower bound, and indeed lower bounds for our more general sinc
integrals in the same way.
In fact
Z ∞
C(x) d x ≈ 0.785380557298632873492583011467332524761
0
while π4 ≈ .785398 only differs in the fifth significant place. We note that high precision
numerical evaluation of these highly oscillatory integrals is by no means straightforward.
If C(x) is replaced by
C ∗ (x) := cos(2x)C(x) = cos(2x)
∞
Y
cos
n=1
µ ¶
x
,
n
we similarly obtain
C ∗ (x) = sinc(4x)
∞
Y
n=1
µ
sinc
2x
2n + 1
¶
.
(12)
1
> 2, so that the corresponding integrals
It now takes 55 terms before 13 + 15 + · · · + 2n+1
π
drop below 8 . Indeed, lengthy numerical computation shows that
0<
π
−
8
Z
∞
0
C ∗ (x) d x <
1
.
1041
We finish by recording without details that (11) allows us to obtain the Maclaurin series for
log C(x). It is
log C(x) = −
∞
X
4k − 1 ζ 2 (2k) 2k
x ,
k
π 2k
k=1
with radius of convergence 12 π. This in turn shows that the coefficient of x 2n in the Maclaurin
series for C(x), say cn , is a rational multiple of π 2k and is explicitly given by the recursion
c0 := 1,
cn := −
n
1X
ζ 2 (2k)
(4k − 1) 2k cn−k
n k=1
π
for n > 0.
Thus
C(x) = 1 −
11 4 4
233
1429
1 2 2
π x +
π x −
π6x6 +
π 8 x 8 + O(x 9 ).
12
4320
5443200
3048192000
Incidentally, as pointed out by David Bradley, the Maclaurin series of log C(x) can be
obtained without appeal to (11) via the Weierstrass product for cos(x).
89
SINC INTEGRALS
Acknowledgment
Thanks are due to David Bailey, Richard Crandall, Greg Fee, Richard Lockhart and Frank
Stegner for very useful discussions. Special thanks are due to David Bradley for an extremely
thorough critique of the original draft of the paper and for his many useful and insightful
suggestions which enabled us to greatly improve the presentation. In particular, the present
proof of Theorem 2(i) is his; our original proof was by induction.
Note
1. Through J. Selfridge and R. Crandall we learned that B. Mares discovered that µ <
π
4.
References
1. L.V. Ahlfors, Complex Analysis, McGraw-Hill, New York, 1966.
2. C. Störmer, “Sur un généralisation de la formule φ2 = sin1 φ − sin22φ + sin33φ − · · ·,” Acta Math. 19 (1885)
341–350.
3. K.R. Stromberg, An Introduction to Modern Real Analysis, Wadsworth Inc., Belmont California, 1981.
4. E.C. Titchmarsh, The Theory of Functions, Oxford University Press, London, 1947.
5. E.T. Whittaker and G.N. Watson, A Course of Modern Analysis, 4th edn., Cambridge University Press, London,
1967.