EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 4, 2021, 1457-1466
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Fourier Expansion, Integral Representation and Explicit
Formula at Rational Arguments of the Tangent

Polynomials of Higher-Order

Cristina B. Corcino1,2, Roberto B. Corcino1,2,∗, Jeremar S. Casquejo1

1 Research Institute for Computational Mathematics and Physics, Cebu Normal University,
6000 Cebu City, Philippines
2 Mathematics Department, Cebu Normal University, 6000 Cebu City, Philippines

Abstract. In this paper, Fourier series expansion of Tangent polynomials are derived and the
integral representation and explicit formula at rational arguments of these polynomials are estab-
lished.
2020 Mathematics Subject Classifications: 11B68, 42A16, 11M35
Key Words and Phrases: Genocchi polynomials, Tangent polynomials, Bernoulli polynomials,
Euler polynomials, Genocchi polynomials, generating functions, Fourier series, integral represen-
tation

1. Introduction

For r ∈ N, the higher-order tangent polynomials, T r
n(x) (n ≥ 0), are defined by the

following generating function (see [1])(
2

e2t + 1

)r

ext =
∞∑
n=0

T r
n(x)

tn

n!
, |2t| < π. (1)

When r = 1, the above equation gives the generating function for the classical tangent
polynomials (see [2]).

The study of tangent polynomials has become an interesting area for many mathemati-
cians for they possess significant properties that can be found in the field of mathematics
and physics (see [3], [4]). Analogues, explicit identities and symmetric properties for tan-
gent polynomials are derived in (see [5], [6], [7]).

In this paper, the researchers derive the Fourier expansion and integral representation
of the tangent polynomials of order r, r ∈ Z+ and present an explicit formula of these
polynomials at rational arguments using the method of Luo [8].

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4152

Email addresses: corcinoc@cnu.edu.ph (C. Corcino), rcorcino@yahoo.com (R. Corcino),
casquejoj@cnu.edu.ph (J. Casquejo)

http://www.ejpam.com 1457 © 2021 EJPAM All rights reserved.



C. Corcino, R. Corcino, J. Casquejo / Eur. J. Pure Appl. Math, 14 (4) (2021), 1457-1466 1458

2. Fourier Expansions for Tangent polynomials of Higher-order

In this section, we give Fourier expansion for tangent polynomials of higher order.

Theorem 2.1. For 0 ≤ x ≤ 1,

T r
n(x) = 2 · n!

(
2

π

)r+n ∞∑
k=0

r−1∑
j=0

(−1)j
(
r + n− j − 1

r − j − 1

)
πj

j!
Br

j

(x
2

)
× cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j
, (2)

where Br
j

(
x
2

)
denotes the Bernoulli polynomials of order r defined by(

w

ew − 1

)r

exw =
∞∑
j=0

Br
j (x)

wn

n!
.

Proof. For r ≥ 2,

Res (f(t), t = tk) =
1

(r − 1)!
lim
t→tk

dr−1

dtr−1
(t− tk)

r

(
2

e2t + 1

)r ext

tn+1
.

Consider the function

(t− tk)
r

(
2

e2t + 1

)r ext

tn+1
= 2r

(t− tk)
r

(e2t + 1)r
ext

tn+1
.

Writing
(
e2t + 1

)r
as (

e2t + 1
)r

= (−1)r
(
e2t(−1)− 1

)r
= (−1)r

(
e2t · e−2tk − 1

)r
= (−1)r

(
e2(t−tk) − 1

)r
,

and since e−2tk = e−2(2k+1)π
2
i = e−(2k+1)πi = −1, we have

(t− tk)
r

(
2

e2t + 1

)r ext

tn+1
=

2r(t− tk)
r

(−1)r
(
e2(t−tk) − 1

)r · ext

tn+1

= (−1)r
(2(t− tk))

r

(e2(t−tk) − 1)r
· ext

tn+1

= (−1)r

( ∞∑
n=0

Br
n

(2(t− tk))
n

n!

)
ext t−(n+1),

where Br
n denotes the Bernoulli numbers of order r defined by the generating function



C. Corcino, R. Corcino, J. Casquejo / Eur. J. Pure Appl. Math, 14 (4) (2021), 1457-1466 1459

(
w

ew − 1

)r

=
∞∑
n=0

Br
n

wn

n!
.

To get the derivative, applying the Leibniz Rule yields

dr−1

dtr−1

{
(t− tk)

r

(
2

e2t + 1

)r ext

tn+1

}
=

dr−1

dtr−1

{
(−1)r

( ∞∑
n=0

Br
n

(2(t− tk))
n

n!

)
ext t−(n+1)

}

= (−1)r
dr−1

dtr−1

{(
ext

∞∑
n=0

Br
n

(2(t− tk))
n

n!

)
t−(n+1)

}

= (−1)r
r−1∑
j=0

(
r − 1

j

)
dr−1−j

dtr−1−j
t−(n+1) · dj

dtj

(
ext

∞∑
n=0

Br
n

(2(t− tk))
n

n!

)
,

dj

dtj

(
ext

∞∑
n=0

Br
n

2n(t− tk)
n)

n!

)

=

j∑
l=0

(
j

l

)
xj−lext

∞∑
n=l

Br
n

2n

n!
(n)l (t− tk)

n−l

= ext
j∑

l=0

(
j

l

)
xj−l

∞∑
n=l

2nBr
n

(t− tk)
n−l

(n− l)!
,

dr−1

dtr−1

(
(t− tk)

r

(
2

e2t + 1

)r ext

tn+1

)
= (−1)r

r−1∑
j=0

(
r − 1

j

)
dr−1−j

dtr−1−j
t−(n+1)

× ext
j∑

l=0

(
j

l

)
xj−l

∞∑
n=l

2nBr
n

(t− tk)
n−l

(n− l)!
.

Thus,

Res (f(t), t = tk) =
1

(r − 1)!
lim
t→tk

(−1)r
r−1∑
j=0

(
r − 1

j

)
dr−1−j

dtr−1−j
t−(n+1)

× lim
t→tk

ext
j∑

l=0

(
j

l

)
xj−l

∞∑
n=l

2nBr
n

(t− tk)
n−l

(n− l)!
.



C. Corcino, R. Corcino, J. Casquejo / Eur. J. Pure Appl. Math, 14 (4) (2021), 1457-1466 1460

Note that Br
n

(t−tk)
n−l

(n−l)! → 0 as t → tk except when n = l. This gives

Res (f(t), t = tk) =
1

(r − 1)!
(−1)r

r−1∑
j=0

(
r − 1

j

)
(−1)r−1−j(n+ r − 1− j)r−1−j t

−(n+r−j)
k

× extk
j∑

l=0

(
j

l

)
xj−l2lBr

l

=
(−1)r

(r − 1)!

r−1∑
j=0

(r − 1)!

j!(r − 1− j)!
(−1)r−1−j(n+ r − 1− j)r−1−j t

−(n+r−j)
k

× extk
j∑

l=0

(
j

l

)
xj−l2lBr

l

=
r−1∑
j=0

(−1)j−1

(
n+ r − 1− j

r − 1− j

)
tj−n−r
k

j!
extk2j

j∑
l=0

(
j

l

)
xj−l

2j−l
Br

l .

Recall that Br
j

(
x
2

)
=
∑j

l=0

(
j
l

)
Br

l

(
x
2

)j−l
. Thus,

Res (f(t), t = tk) =
r−1∑
j=0

(−1)j−12j
(
n+ r − 1− j

r − 1− j

)
tj−n−r
k

j!
extkBr

j

(x
2

)

=

r−1∑
j=0

(−1)j−12j
(
n+ r − 1− j

r − 1− j

)
Br

j

(
x
2

)
j!

extk

tn+r−j
k

=
r−1∑
j=0

(−1)j−12j
(
r + n− j − 1

r − j − 1

)
Br

j

(
x
2

)
j!

extk

tr+n−j
k

.

Taking tk = 1
2(2k + 1)πi, we get

Res (f(t), t = tk) =

r−1∑
j=0

(−1)j−12j
(
r + n− j − 1

r − j − 1

)
Br

j

(
x
2

)
j!

e
1
2
(2k+1)πix(

1
2(2k + 1)πi

)r+n−j

=
1(

1
2πi
)r+n

r−1∑
j=0

(−1)j−12j
(
r + n− j − 1

r − j − 1

)(1
2πi
)j

j!
Br

j

(x
2

) e
1
2
(2k+1)πix

(2k + 1)r+n−j
.

This gives

T r
n(x) = n!

(
2

πi

)r+n∑
k∈Z

r−1∑
j=0

(−1)j
(
r + n− j − 1

r − j − 1

)
(πi)j

j!
Br

j

(x
2

) e
1
2
(2k+1)πi

2
x

(2k + 1)r+n−j
. (3)



C. Corcino, R. Corcino, J. Casquejo / Eur. J. Pure Appl. Math, 14 (4) (2021), 1457-1466 1461

Now, from (3), we look at

i−(r+n−j)
∑
k∈Z

e(2k+1)πi
2
x

(2k + 1)r+n−j
. (4)

Noting that i−(r+n−j) = e−(r+n−j)πi/2 and (−1)r+n−j = e(r+n−j)πi, we see that

i−(r+n−j)
∑
k∈Z

e(2k+1)πi
2
x

(2k + 1)r+n−j
= i−(r+n−j)

{ ∞∑
k=0

e(2k+1)πi
2
x

(2k + 1)r+n−j
+ (−1)r+n−j

∞∑
k=0

e−(2k+1)πi
2
x

(2k + 1)r+n−j

}

=

∞∑
k=0

e[(2k+1)x/2−(r+n−j)/2]πi + e−[(2k+1)x/2−(r+n−j)/2]πi

(2k + 1)r+n−j

=

∞∑
k=0

2 cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

= 2
∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j
. (5)

Replacing (4) with (5) in (3), we get the desired formula (2).

3. Integral representation for Tangent polynomials of Higher-order

In this section, we establish an integral representation for tangent polynomials of
higher order.

Theorem 3.1. For n ∈ N, r ≥ 2, and 0 ≤ R(x) ≤ 1,

T r
n(x) = 2r+n

r−1∑
j=0

(−1)j

j!
·

Br
j

(
x
2

)
(r − j − 1)!

{∫ ∞

0

eπt cos [πx/2− (r + n− j)π/2]

cosh (2πt)− cos (πx)
tr+n−j−1 dt

−
∫ ∞

0

e−πt cos [πx/2 + (r + n− j)π/2]

cosh (2πt)− cos (πx)
tr+n−j−1 dt

}
. (6)

Proof. From (2), we get

T r
n(x) = 2 · n!

(
2

π

)r+n ∞∑
k=0

r−1∑
j=0

(−1)j
(r + n− j − 1)!

(r − j − 1)! n!
· π

j

j!
Br

j

(x
2

)
× cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

= 2r+n+1
r−1∑
j=0

(−1)j Br
j

(
x
2

)
(r − j − 1)! j!

· (r + n− j − 1)!

πr+n−j

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j
.

(7)



C. Corcino, R. Corcino, J. Casquejo / Eur. J. Pure Appl. Math, 14 (4) (2021), 1457-1466 1462

We look at

(r + n− j − 1)!

πr+n−j

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

=
1

πr+n−j

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]
(r + n− j − 1)!

(2k + 1)r+n−j
. (8)

Applying the integral formula ∫ ∞

0
tne−at dt =

n!

an+1
,

for n ≥ 0 and R(a) > 0, then (8) becomes

(r + n− j − 1)!

πr+n−j

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

=
1

πr+n−j

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

∫ ∞

0
tr+n−j−1e−(2k+1)t dt

=
1

πr+n−j

∫ ∞

0
tr+n−j−1

∞∑
k=0

e−(2k+1)t cos [(2k + 1)πx/2− (r + n− j)π/2] dt

=
1

πr+n−j

∫ ∞

0
tr+n−j−1

∞∑
k=0

e−(2k+1)t

{
cos [(2k + 1)πx/2] cos [(r + n− j)π/2]

+ sin [(2k + 1)πx/2] sin [(r + n− j)π/2]

}
dt

=
1

πr+n−j

∫ ∞

0

{
cos [(r + n− j)π/2]

∞∑
k=0

e−(2k+1)t cos [(2k + 1)πx/2]

+ sin [(r + n− j)π/2]

∞∑
k=0

e−(2k+1)t sin [(2k + 1)πx/2]

}
tr+n−j−1 dt.

(9)

By making use of

∞∑
k=0

e−(2k+1)t sin [(2k + 1)x] =
sinx cosh t

cosh (2t)− cos (2x)
,

and
∞∑
k=0

e−(2k+1)t cos [(2k + 1)x] =
cosx sinh t

cosh (2t)− cos (2x)
,

which may be deduced from

∞∑
k=0

e(xi−t)(2k+1) =
cosx sinh t+ i sinx cosh t

cosh (2t)− cos (2x)
,



C. Corcino, R. Corcino, J. Casquejo / Eur. J. Pure Appl. Math, 14 (4) (2021), 1457-1466 1463

for t > 0, (9) then becomes

(r + n− j − 1)!

πr+n−j

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

=
1

πr+n−j

∫ ∞

0

{
cos [(r + n− j)π/2]

cos πx
2 sinh t

cosh (2t)− cos (πx)

+ sin [(r + n− j)π/2]
sin πx

2 cosh t

cosh (2t)− cos (πx)

}
tr+n−j−1 dt. (10)

Applying the transformation t = πt, (10) becomes

(r + n− j − 1)!

πr+n−j

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

=
1

πr+n−j

∫ ∞

0

{
cos [(r + n− j)π/2]

cos πx
2 sinhπt

cosh (2πt)− cos (πx)

+ sin [(r + n− j)π/2]
sin πx

2 coshπt

cosh (2πt)− cos (πx)

}
πr+n−jtr+n−j−1 dt

=

∫ ∞

0

{
cos [(r + n− j)π/2] cos πx

2 (eπs − e−πs)

2 [cosh (2πt)− cos (πx)]

+
sin [(r + n− j)π/2] sin πx

2 (eπs + e−πs)

2 [cosh (2πt)− cos (πx)]

}
tr+n−j−1 dt

=
1

2

∫ ∞

0

{
eπs
(
cos [(r + n− j)π/2] cos πx

2 + sin [(r + n− j)π/2] sin πx
2

)
cosh (2πt)− cos (πx)

−
e−πs

(
cos [(r + n− j)π/2] cos πx

2 − sin [(r + n− j)π/2] sin πx
2

)
cosh (2πt)− cos (πx)

}
tr+n−j−1 dt

=
1

2

∫ ∞

0

eπs cos [πx/2− (r + n− j)π/2]− e−πs cos [πx/2 + (r + n− j)π/2]

cosh (2πt)− cos (πx)
tr+n−j−1 dt.

(11)

Applying (11) to (7), we get the desired formula (6).

4. Explicit formula for Tangent polynomials of Higher-order at rational
arguments

In this section, we obtain an explicit formula for tangent polynomials of higher
order at rational arguments by applying the Fourier expansion (2). Here let Z−

0 =
{0,−1,−2, · · · } denote the set of nonpositive integers.

Theorem 4.1. For n, q ∈ N and p ∈ Z,

T r
n

(
2p

q

)
=

2 · n!
(qπ)r+n

r−1∑
j=0

(−1)j
(
r + n− j − 1

r − j − 1

)
(2qπ)j

j!
Br

j

(
p

q

)



C. Corcino, R. Corcino, J. Casquejo / Eur. J. Pure Appl. Math, 14 (4) (2021), 1457-1466 1464

×
q∑

l=1

ζ

(
r + n− j,

2l − 1

2q

)
cos

[
(2l − 1)pπ

q
− (r + n− j)π

2

]
, (12)

where

ζ(s, a) =

∞∑
n=0

1

(n+ a)s
, (13)

for R(s) > 1 and a /∈ Z−
0 is Hurwitz zeta function.

Proof. We look at

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j
. (14)

Replacing k with k − 1:

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j
=

∞∑
k=1

cos [(r + n− j)π/2− (2k − 1)πx/2]

(2k − 1)r+n−j
. (15)

Applying the elementary series identity

∞∑
k=1

f(k) =

q∑
l=1

∞∑
k=0

f(qk + l), q ∈ N,

(15) becomes

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

=

q∑
l=1

∞∑
k=0

cos [(r + n− j)π/2− (2qk + 2l − 1)πx/2]

(2qk + 2l − 1)r+n−j

=

q∑
l=1

∞∑
k=0

cos [(r + n− j)π/2− (2l − 1)πx/2− qkπx][
2q
(
k + 2l−1

2q

)]r+n−j

=

q∑
l=1

∞∑
k=0

cos [(r + n− j)π/2− (2l − 1)πx/2− qkπx]

(2q)r+n−j
· 1(

k + 2l−1
2q

)r+n−j
. (16)

Setting x = 2p/q, (16) becomes

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

=

q∑
l=1

∞∑
k=0

cos [(r + n− j)π/2− (2l − 1)pπ/q − 2π(pk)]

(2q)r+n−j
· 1(

k + 2l−1
2q

)r+n−j



REFERENCES 1465

=
1

(2q)r+n−j

q∑
l=1

∞∑
k=0

cos [(r + n− j)π/2− (2l − 1)pπ/q] · 1(
k + 2l−1

2q

)r+n−j

=
1

(2q)r+n−j

q∑
l=1

cos [(r + n− j)π/2− (2l − 1)pπ/q]
∞∑
k=0

1(
k + 2l−1

2q

)r+n−j
. (17)

By (13), (17) becomes

∞∑
k=0

cos [(2k + 1)πx/2− (r + n− j)π/2]

(2k + 1)r+n−j

=
1

(2q)r+n−j

q∑
l=1

cos

[
(2l − 1)pπ

q
− (r + n− j)π

2

]
ζ

(
r + n− j,

2l − 1

2q

)
. (18)

Replacing (14) with (18) in (2), we obtain the desired formula (12).

Acknowledgements

The authors would like to thank the anonymous referees for reviewing the paper thor-
oughly The authors would also like to thank Cebu Normal University (CNU) for funding
this research project through its Research Institute for Computational Mathematics and
Physics (RICMP).

References

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Phys 7(9) (2013), 447-454.

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