Trigonometric integrals evaluated in terms of Riemann zeta and Dirichlet beta functions

Jing Li; Wenchang Chu

doi:10.1515/math-2024-0052

Artikel Open Access

Trigonometric integrals evaluated in terms of Riemann zeta and Dirichlet beta functions

Jing Li und Wenchang Chu

Veröffentlicht/Copyright: 13. September 2024

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Abstract

Three classes of trigonometric integrals involving an integer parameter are evaluated by the contour integration and the residue theorem. The resulting formulae are expressed in terms of Riemann zeta function and Dirichlet beta function. Several remarkable integral identities are presented.

Keywords: Cauchy residue theorem; contour integration; formal power series

MSC 2010: 30E20; 11M32

1 Introduction and outline

Evaluating integrals has been always a fascinating subject in mathematics, physics, and engineering [1–5]. By employing the contour integration and the residue theorem, we shall explicitly evaluate, for m ∈ N , the following integral:

(1) J ( m ) ≔ ∫ 0 π 2 sin m ( 2 x ) cot ( 2 x ) log ( tan x ) d x ,

where N denotes the set of natural numbers with N 0 = { 0 } ∪ N throughout the article.

Perform the change of variables

tan x → e z , which implies that sin ( 2 x ) = 2 e z 1 + e 2 z .

Equivalently, we can write

x → arctan ( e z ) , which implies that cos ( 2 x ) = 1 − e 2 z 1 + e 2 z .

Therefore, we have d x = e z d z 1 + e 2 z and can express the integral in question as

(2) J ( m ) = 2 m − 1 ∫ − ∞ ∞ e m z ( 1 − e 2 z ) z ( 1 + e 2 z ) m + 1 d z .

Observe that the integrand

(3) G m ( z ) ≔ e m z ( 1 − e 2 z ) z ( 1 + e 2 z ) m + 1

is a meromorphic function whose poles are determined by 1 + e 2 z = 0 . In order to evaluate the integral J , consider the anti-clockwise contour ℒ , consisting of two segments [ − π M , − ε ] and [ ε , π M ] along the real axis plus two semi-circles C ( ε ) and C ( π M ) over the upper half-plane centered at the origin with radii ε and π M (where M ∈ N ), respectively. Hence, all the poles of G m ( z ) inside the contour ℒ are of order m + 1 and given explicitly by

z k = k − 1 2 π i , where k ∈ N with 1 ≤ k ≤ M .

According to the residue theorem (see [6] and [7, §3.11] for example), we have the following equality:

∫ ℒ G m ( z ) d z = 2 π i ∑ k = 1 M Res z = z k G m ( z ) = ∫ C ( ε ) G m ( z ) d z + ∫ C ( π M ) G m ( z ) d z + ∫ − π M − ε G m ( z ) d z + ∫ ε π M G m ( z ) d z .

As we shall show in the next section that

lim ε → 0 ∫ C ( ε ) G m ( z ) d z = lim M → ∞ ∫ C ( π M ) G m ( z ) d z = 0 ,

the limiting case of the above equation as ε → 0 and M → ∞ will reduce to the following integral–sum identity.

Lemma 1

The following integral–sum formula holds:

J ( m ) = ∫ 0 π 2 sin m ( 2 x ) cot ( 2 x ) log ( tan x ) d x = 2 m π i ∑ k = 1 ∞ Res z = z k G m ( z ) .

The rest of the article will be organized in the following manner. In Section 2, the two integrals along semi-circles C ( ε ) and C ( π M ) will be shown to tend to zero. Then, in Section 3, we shall explicitly determine the residues of G m ( z ) at z = z k by making use of the Lagrange inversion formula, and alternatively, by power series expansions. Analytic formulae will be established in Section 4 for J ( m ) , as well as for two related integrals Φ ( m ) and Ψ ( m ) . Finally, the article will conclude in Section 5, where Laplace transforms of logarithmic powers will be reviewed by employing the Γ -integrals.

2 Estimating integrals along semi-circles

Observe that G m ( z ) is continuous at z = 0 and

lim z → 0 G m ( z ) = lim z → 0 e m z ( 1 − e 2 z ) z ( 1 + e 2 z ) m + 1 = − 1 2 m .

Therefore, z = 0 is a removable singularity. In fact, for any sufficiently small ε > 0 , the function G m ( z ) is bounded on C ε . Then we can estimate the integral

∫ C ( ε ) G m ( z ) d z = O ( 1 ) × ∫ C ( ε ) d z = O ( ε π ) ,

which implies consequently the following limiting relation:

lim ε → 0 ∫ C ( ε ) G m ( z ) d z = 0 .

Now, we turn to examine the integral along C ( π M ) . By making the change in variables z = π M e i θ on the large semi-circle C ( π M ) , we can reformulate the integral

∫ C ( π M ) G m ( z ) d z = i ∫ 0 π Ω ( θ ) d θ , where Ω ( θ ) ≔ ( e π M e i θ ) m ( 1 − e 2 π M e i θ ) ( 1 + e 2 π M e i θ ) m + 1 .

• Observe that Ω ( θ ) is a continuous function and Ω ( π 2 ) = 0 . Then, for an arbitrarily small ε > 0 , there exists a δ subject to 0 < δ ≪ π 2 such that

∣ Ω ( θ ) ∣ < ε for π 2 − δ < θ < π 2 + δ .

Consequently, we can estimate the integral

∫ π 2 − δ π 2 + δ Ω ( θ ) d θ ≤ ∫ π 2 − δ π 2 + δ ∣ Ω ( θ ) ∣ d θ < 2 ε δ .

• When 0 ≤ θ ≤ π 2 − δ , the following limiting relation holds:

∣ e 2 z ∣ = e 2 π M cos θ → ∞ as M → ∞ .

Assuming 2 < e 2 π M cos θ under the same conditions, we deduce further that

∣ Ω ( θ ) ∣ ≤ ( e π M cos θ ) m ( 1 + e 2 π M cos θ ) ∣ 1 − e 2 π M cos θ ∣ m + 1 < 2 ( e π M cos θ ) m + 2 ( 1 2 e 2 π M cos θ ) m + 1 = 2 m + 2 ( e π M cos θ ) m → 0 as M → ∞ .

Consequently, we can estimate the integral

lim M → ∞ ∫ 0 π 2 − δ Ω ( θ ) d θ ≤ lim M → ∞ ∫ 0 π 2 − δ 2 m + 2 ( e π M cos θ ) m d θ = 0 .

• When π 2 + δ ≤ θ ≤ π , by making the change in variable θ = π − ϑ , we have, for 0 ≤ ϑ ≤ π 2 − δ , two limiting relations

∣ e 2 z ∣ = e − 2 π M cos ϑ → 0 ∣ e − 2 z ∣ = e 2 π M cos ϑ → ∞ as M → ∞ .

Assuming 2 < e 2 π M cos ϑ under the precedent conditions, we deduce as before that

∣ Ω ( π − ϑ ) ∣ ≤ ( e π M cos ( π − ϑ ) ) m ( 1 + e 2 π M cos ( π − ϑ ) ) ∣ 1 − e 2 π M cos ( π − ϑ ) ∣ m + 1 = ( e − π M cos ϑ ) m ( 1 + e − 2 π M cos ϑ ) ∣ 1 − e − 2 π M cos ϑ ∣ m + 1 = ( e π M cos ϑ ) m ( 1 + e 2 π M cos ϑ ) ∣ 1 − e 2 π M cos ϑ ∣ m + 1 < 2 ( e π M cos ϑ ) m + 2 ( 1 2 e 2 π M cos ϑ ) m + 1 = 2 m + 2 ( e π M cos ϑ ) m → 0 as M → ∞ .

Hence, we can also estimate the integral

lim M → ∞ ∫ π 2 + δ π Ω ( θ ) d θ = lim M → ∞ ∫ 0 π 2 − δ Ω ( π − ϑ ) d ϑ ≤ lim M → ∞ ∫ 0 π 2 − δ 2 m + 2 ( e π M cos ϑ ) m d ϑ = 0 .

In conclusion, we have confirmed that

lim M → ∞ ∫ C ( π M ) G m ( z ) d z = lim M → ∞ ∫ 0 π Ω ( θ ) d θ ≤ lim M → ∞ ∫ π 2 − δ π 2 + δ + ∫ 0 π 2 − δ + ∫ π 2 + δ π Ω ( θ ) d θ = 0 .

3 Computing residues

For n ∈ N and two integers j , m , let j ≡ n m stand for that “ j is congruent to m modulo n .” Denote further by [ x n ] f ( x ) the coefficient of x n in the Laurent series f ( x ) . In order to determine the residues of G m ( z ) at z k explicitly, we first express the residue in terms of the following coefficient:

Res z = z k G m ( z ) = [ ( z − z k ) − 1 ] e m z ( 1 − e 2 z ) z ( 1 + e 2 z ) m + 1 z = y + z k , e 2 z k = − 1 = ( − 1 ) m ( k + 1 ) i m [ y − 1 ] e m y ( 1 + e 2 y ) ( y + z k ) ( 1 − e 2 y ) m + 1 e m z k = ( − 1 ) m ( k + 1 ) i m = ∑ 1 ≤ j ≤ m j ≡ 2 m ( − 1 ) m k + j + 1 i m z k j + 1 [ y 0 ] y j + 1 ( e y + e − y ) ( e y − e − y ) m + 1 ,

where ( y + z k ) − 1 has been expanded into the power series.

1 y + z k = ∑ j = 0 ∞ ( − 1 ) j y j z k j + 1 which reduces to ∑ 1 ≤ j ≤ m j ≡ 2 m ( − 1 ) j y j z k j + 1 .

The above replacement is justified as follows:

First, when the summation index j > m , the constant term corresponding to the coefficient [ y 0 ] vanishes due to the fact that y j + 1 will cancel the factor y m + 1 appearing in the denominator ( e y − e − y ) m + 1 .
Second, the constant term determined by [ y 0 ] vanishes too when j and m have different parities, since in this case, the function y j + 1 ( e y + e − y ) ( e y − e − y ) m + 1 is odd.
Third, when j = 0 , the corresponding coefficient
[ y 0 ] y ( e y + e − y ) ( e y − e − y ) m + 1 = 0 .
In general for k ∈ N 0 , there holds
[ y 0 ] y ( e y + e − y ) ( e y − e − y ) k = 0 , k ≠ 1 ; 1 , k = 1 .
To verify this identity, we have to appeal to the Lagrange inversion theorem (cf. Comtet [8, §3.8]). Let ϕ ( y ) be a formal power series subject to ϕ ( 0 ) ≠ 0 . Then, the functional equation x ≔ y ⁄ ϕ ( y ) determines an implicit function y = y ( x ) . Suppose further that F ( y ) is another formal power series, then the composite function F ( y ( x ) ) can be expanded into power series with its coefficients being given explicitly by
k [ x k ] F ( y ) = [ y k − 1 ] { F ′ ( y ) ϕ k ( y ) } .
In the above equality, specifying by
F ( y ) = x = e y − e − y and ϕ ( y ) = y e y − e − y ,
we obtain the corresponding equality
k [ x k ] x = [ y k − 1 ] y k e y + e − y ( e y − e − y ) k = [ y 0 ] y ( e y + e − y ) ( e y − e − y ) k .
Clearly, the left-hand side reduces to 1 for k = 1 and 0 otherwise.

Furthermore, recalling the following variant of the Lagrange inversion theorem:

[ x k ] F ( y ) 1 − y ϕ ′ ( y ) ⁄ ϕ ( y ) = [ y k ] { F ( y ) ϕ k ( y ) } ,

we can also reformulate the coefficient

[ y 0 ] y j + 1 ( e y + e − y ) ( e y − e − y ) m + 1 = [ y m − j ] { ( e y + e − y ) ϕ m + 1 ( y ) } = [ y m − j ] { ( e y + e − y ) ϕ j + 1 ( y ) ϕ m − j ( y ) } = [ x m − j ] ( e y + e − y ) ϕ j + 1 ( y ) 1 − y ϕ ′ ( y ) ⁄ ϕ ( y ) .

Writing explicitly

( e y + e − y ) ϕ j + 1 ( y ) 1 − y ϕ ′ ( y ) ⁄ ϕ ( y ) = ( e y + e − y ) y j + 1 x j + 1 ( 1 − x ϕ ′ ( y ) ) = y j x j ,

we find the following simplified expression:

[ y 0 ] y j + 1 ( e y + e − y ) ( e y − e − y ) m + 1 = [ x m − j ] y j x j = [ x m ] y j .

In view of the functional equations

x = e y − e − y and y = ln x + 4 + x 2 2 = arcsinh x 2 ,

we can further reformulate

[ y 0 ] y j + 1 ( e y + e − y ) ( e y − e − y ) m + 1 = [ x m ] arcsinh j x 2 = 2 − m [ x m ] arcsinh j x .

When j = 0 , this confirms again the previously justified third fact.

By substitution, we have established the following explicit formula:

(4) Res z = z k G ( z ) = ∑ 1 ≤ j ≤ m j ≡ 2 m ( − 1 ) m k + j + 1 i m 2 m z k j + 1 [ x m ] arcsinh j x .

The initial coefficients [ x m ] arcsinh j x are tabulated as follows:

m \ j	1	2	3	4	5	6	7	8	9
1	1
2	0	1
3	− 1 6	0	1
4	0	− 1 3	0	1
5	3 40	0	− 1 2	0	1
6	0	8 45	0	− 2 3	0	1
7	− 5 112	0	37 120	0	− 5 6	0	1
8	0	− 4 35	0	7 15	0	− 1	0	1
9	35 1,152	0	− 3,229 15,120	0	47 72	0	− 7 6	0	1

We remark that there exist multiple sum expressions for coefficients [ x m ] arcsin j x (see [9,10]). In accordance with the fact arcsinh x = − i arcsin ( i x ) , we record the corresponding multiple sum expressions as follows:

[ x 2 m ] arcsin 2 j x = ( − 4 ) m − j m 2 2 m m ∑ 1 ≤ n 1 < n 2 < … < n j − 1 < m ( 2 j ) ! ∏ i = 1 j − 1 k i 2 , [ x 2 m + 1 ] arcsin 2 j + 1 x = ( − 1 ) m − j 2 m m 4 m ( 2 m + 1 ) ∑ 0 ≤ n 1 < n 2 < … < n j < m ( 2 j + 1 ) ! ∏ i = 1 j ( 2 n i + 1 ) 2 .

Instead of the above multiple sum expressions, we are going to derive, by means of power series expansions (cf. [11]), a simplified expression of triple sums. For the sake of brevity, we can rewrite the residue by

(5) Res z = z k G m ( z ) = ∑ 1 ≤ j ≤ m j ≡ 2 m ( − 1 ) m k + j + 1 i m z k j + 1 [ y m ] y j U ( y ) V ( y ) 2 m + 1 ,

where U ( y ) and V ( y ) are two even functions given by U ( y ) = e y + e − y and

V ( y ) = 2 y e y − e − y m + 1 = 1 − 1 − e y − e − y 2 y − m − 1 .

In order to extract the coefficient of y m in (5), it is enough to replace the function V ( y ) by its Maclaurin polynomial of order m . Observing the initial terms

e y − e − y 2 y = 1 + y 2 6 + y 4 120 + O ( y 6 ) ,

we can expand V ( y ) into the following binomial sum:

V ( y ) = O ( y m + 1 ) + ∑ ℓ = 0 m 2 ( − 1 ) ℓ − m − 1 ℓ 1 − e y − e − y 2 y ℓ = O ( y m + 1 ) + ∑ ℓ = 0 m 2 ( − 1 ) ℓ − m − 1 ℓ ∑ μ = 0 ℓ ( − 1 ) μ ℓ μ e y − e − y 2 y μ = O ( y m + 1 ) + ∑ μ = 0 m 2 − m − 1 μ e y − e − y 2 y μ ∑ ℓ = μ m 2 ( − 1 ) ℓ − μ − m − 1 − μ ℓ − μ .

Evaluating the rightmost sum

∑ ℓ = μ m 2 ( − 1 ) ℓ − μ − m − 1 − μ ℓ − μ = ( − 1 ) m 2 − μ − m − μ − 2 m 2 − μ = m + m 2 + 1 m 2 − μ

and then making use of the binomial expansion

e y − e − y 2 y μ = ∑ ν = 0 μ ( − 1 ) ν μ ν e ( μ − 2 ν ) y ( 2 y ) μ ,

we derive another double sum expression

V ( y ) = O ( y m + 1 ) + ∑ 0 ≤ ν ≤ μ ≤ m 2 ( − 1 ) ν m + m 2 + 1 m 2 − μ − m − 1 μ μ ν e ( μ − 2 ν ) y ( 2 y ) μ .

Multiplying the last equation by U ( y ) = e y + e − y , we can further manipulate

U ( y ) V ( y ) = O ( y m + 1 ) + ∑ μ = 0 m 2 ∑ ν = 0 μ ( − 1 ) ν m + m 2 + 1 m 2 − μ − m − 1 μ μ ν e ( μ − 2 ν + 1 ) y ( 2 y ) μ + ∑ μ = 0 m 2 ∑ ν = 0 μ ( − 1 ) ν m + m 2 + 1 m 2 − μ − m − 1 μ μ ν e ( μ − 2 ν − 1 ) y ( 2 y ) μ = O ( y m + 1 ) + ∑ μ = 0 m 2 ∑ ν = 0 μ + 1 ( − 1 ) ν m + m 2 + 1 m 2 − μ − m − 1 μ 1 + μ − 2 ν 1 + μ 1 + μ ν e ( 1 + μ − 2 ν ) y ( 2 y ) μ ,

where we have made the replacement ν → ν − 1 for the second sum and then unified it to the first sum by employing the binomial relation

μ ν − μ ν − 1 = 1 + μ − 2 ν 1 + μ 1 + μ ν .

According to (5), we can compute the residue by

Res z = z k G ( z ) = ∑ 1 ≤ j ≤ m j ≡ 2 m ∑ μ = 0 m 2 ∑ ν = 0 μ + 1 ( − 1 ) m k + j + ν + 1 i m 2 m + μ + 1 z k j + 1 m + m 2 + 1 m 2 − μ − m − 1 μ × 1 + μ − 2 ν 1 + μ 1 + μ ν [ y m + μ − j ] e ( 1 + μ − 2 ν ) y = ∑ 1 ≤ j ≤ m j ≡ 2 m ∑ μ = 0 m 2 ∑ ν = 0 μ + 1 ( − 1 ) m k + j + ν + 1 i m 2 m + μ + 1 z k j + 1 m + m 2 + 1 m 2 − μ − m − 1 μ × 1 + μ ν ( 1 + μ − 2 ν ) 1 + m + μ − j ( 1 + μ ) ( m + μ − j ) ! .

Taking into account the binomial relation

m + m 2 + 1 m 2 − μ − m − 1 μ = ( − 1 ) μ ( m + 1 ) μ μ ! ( m + μ + 2 ) m 2 − μ m 2 − μ ! = ( − 1 ) μ m 2 μ m + m 2 + 1 m + 1 m + 1 m + μ + 1 ,

we finally find the following triple sum expression:

(6) Res z = z k G m ( z ) = m + m 2 + 1 m + 1 ∑ 1 ≤ j ≤ m j ≡ 2 m ∑ μ = 0 m 2 ∑ ν = 0 μ + 1 ( − 1 ) m k + j + μ + ν + 1 i m 2 m + μ + 1 z k j + 1 × m + 1 m + μ + 1 m 2 μ 1 + μ ν ( 1 + μ − 2 ν ) 1 + m + μ − j ( 1 + μ ) ( m + μ − j ) ! .

4 Integral identities

Now, we are in position to explicitly evaluate not only the integral J ( m ) but also two further integrals Φ ( m ) and Ψ ( m ) .

4.1 Integral J ( m )

By combining Lemma 1 with (4), we can express the integral J ( m ) as follows:

J ( m ) = ∫ 0 π 2 sin m ( 2 x ) cot ( 2 x ) log ( tan x ) d x = 2 m π i ∑ k = 1 ∞ Res z = z k G m ( z ) = π ∑ 1 ≤ j ≤ m j ≡ 2 m i m + 1 [ x m ] arcsinh j x ∑ k = 1 ∞ ( − 1 ) m k + j + 1 z k j + 1 = π ∑ 1 ≤ j ≤ m j ≡ 2 m 2 π j + 1 ( − 1 ) 2 + m − j 2 [ x m ] arcsinh j x ∑ k = 1 ∞ ( − 1 ) j ( k − 1 ) ( 2 k − 1 ) j + 1 .

Recall the Riemann zeta function

ζ ( λ ) = ∑ n = 1 ∞ 1 n λ and ∑ k = 1 ∞ 1 ( 2 k − 1 ) λ = ( 1 − 2 − λ ) ζ ( λ ) ,

as well as the Dirichlet beta function (cf. [12])

β ( λ ) = ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) λ

with particular values

β ( 1 ) = π 4 , β ( 3 ) = π 3 32 and β ( 2 ) = G (Catalan constant).

We derive the explicit formulae as in the following theorem.

Theorem 2

m ≡ 2 0 J ( m ) = ∑ 1 ≤ j ≤ m j ≡ 2 0 ( − 1 ) m − j 2 ( 1 − 2 j + 1 ) ζ ( j + 1 ) π j [ x m ] arcsinh j x , m ≡ 2 1 J ( m ) = ∑ 1 ≤ j ≤ m j ≡ 2 1 ( − 1 ) m − j 2 ( 0 − 2 j + 1 ) β ( j + 1 ) π j [ x m ] arcsinh j x .

Instead, by combining Lemma 1 with (6), we come to the expression

J ( m ) = ∫ 0 π 2 sin m ( 2 x ) cot ( 2 x ) log ( tan x ) d x = 2 m π i ∑ k = 1 ∞ Res z = z k G m ( z ) = m + m 2 + 1 m + 1 ∑ 1 ≤ j ≤ m j ≡ 2 m ∑ μ = 0 m 2 ∑ ν = 0 μ + 1 ( − 1 ) j + μ + ν + 1 i m + 1 π 2 μ + 1 × m + 1 m + μ + 1 m 2 μ 1 + μ ν ( 1 + μ − 2 ν ) 1 + m + μ − j ( 1 + μ ) ( m + μ − j ) ! ∑ k = 1 ∞ ( − 1 ) m k z k j + 1 .

Evaluating the sum of residues

∑ k = 1 ∞ ( − 1 ) m k z k j + 1 = 2 j + 1 π j + 1 i j + 1 ∑ k = 1 ∞ ( − 1 ) m k ( 2 k − 1 ) j + 1 = 2 j + 1 − 1 π j + 1 i j + 1 ζ ( j + 1 ) , m ≡ 2 0 ; − 2 j + 1 π j + 1 i j + 1 β ( j + 1 ) , m ≡ 2 1 ;

we conclude, after some simplifications, with the following triple sum formula.

Theorem 3

m ≡ 2 0 J ( m ) = m + m 2 + 1 m + 1 ∑ 1 ≤ j ≤ m j ≡ 2 0 1 − 2 j + 1 π j ∑ μ = 0 m 2 ∑ ν = 0 μ + 1 ( − 1 ) μ + ν + m − j 2 2 μ + 1 × m + 1 m + μ + 1 m 2 μ 1 + μ ν ( 1 + μ − 2 ν ) 1 + m + μ − j ( 1 + μ ) ( m + μ − j ) ! ζ ( j + 1 ) , m ≡ 2 1 J ( m ) = m + m 2 + 1 m + 1 ∑ 1 ≤ j ≤ m j ≡ 2 1 ( − 2 j + 1 ) π j ∑ μ = 0 m 2 ∑ ν = 0 μ + 1 ( − 1 ) μ + ν + m − j 2 2 μ + 1 × m + 1 m + μ + 1 m 2 μ 1 + μ ν ( 1 + μ − 2 ν ) 1 + m + μ − j ( 1 + μ ) ( m + μ − j ) ! β ( j + 1 ) .

For 1 ≤ m ≤ 10 , the corresponding formulae are displayed below as examples.

J ( 1 ) = − 4 β ( 2 ) π , J ( 2 ) = − 7 ζ ( 3 ) π 2 , J ( 3 ) = − 2 β ( 2 ) 3 π − 16 β ( 4 ) π 3 , J ( 4 ) = − 7 ζ ( 3 ) 3 π 2 − 31 ζ ( 5 ) π 4 , J ( 5 ) = − 3 β ( 2 ) 10 π − 8 β ( 4 ) π 3 − 64 β ( 6 ) π 5 , J ( 6 ) = − 56 ζ ( 3 ) 45 π 2 − 62 ζ ( 5 ) 3 π 4 − 127 ζ ( 7 ) π 6 , J ( 7 ) = − 5 β ( 2 ) 28 π − 74 β ( 4 ) 15 π 3 − 160 β ( 6 ) 3 π 5 − 256 β ( 8 ) π 7 , J ( 8 ) = − 4 ζ ( 3 ) 5 π 2 − 217 ζ ( 5 ) 15 π 4 − 127 ζ ( 7 ) π 6 − 511 ζ ( 9 ) π 8 , J ( 9 ) = − 35 β ( 2 ) 288 π − 3,229 β ( 4 ) 945 π 3 − 376 β ( 6 ) 9 π 5 − 896 β ( 8 ) 3 π 7 − 1,024 β ( 10 ) π 9 , J ( 10 ) = − 128 ζ ( 3 ) 225 π 2 − 10,168 ζ ( 5 ) 945 π 4 − 1,651 ζ ( 7 ) 15 π 6 − 2,044 ζ ( 9 ) 3 π 8 − 2,047 ζ ( 11 ) π 10 .

In addition, the following binomial formula is widely known

(7) u n + v n = ∑ k = 0 n 2 ( − 1 ) k n n − k n − k k ( u v ) k ( u + v ) n − 2 k .

Albert Girard published it in Amsterdam in 1629 and similar results were subsequently published by Edward Waring in Cambridge during 1762–1782. They can be derived from the earlier work of Sir Isaac Newton (see Gould [13]). There is a companion formula due to Carlitz [14]

(8) u n + 1 − v n + 1 = ( u − v ) ∑ k = 0 n 2 ( − 1 ) k n − k k ( u v ) k ( u + v ) n − 2 k .

They appeared also in more accessible references by Chu and Li [15] and Comtet [8, §4.9].

By making use of the binomial formula (8), we can evaluate, in the sequel, two further classes of related trigonometric integrals.

4.2 Integral Φ ( m )

For m ∈ N , we examine first the integral

Φ ( m ) ≔ ∫ 0 π 2 sin ( 4 m x ) log ( tan x ) d x .

The special case m = 1 was proposed by Stewart [16], which has been the primary inspiration for the authors to work on this topic. This is the only reference about integrals with log ( tan x ) in the denominator of the integrand, that the author can locate in the literature.

By specifying u → e 2 x i , v → − e − 2 x i and n → 2 m − 1 in the binomial formula (8), we can express

sin ( 4 m x ) = e 4 m x i − e − 4 m x i 2 i = u 2 m − v 2 m 2 i = u − v 2 i ∑ k = 0 m − 1 ( − 1 ) k 2 m − k − 1 k ( u v ) k ( u + v ) 2 m − 2 k − 1 = u − v 2 i ∑ k = 1 m ( − 1 ) m − k m + k − 1 2 k − 1 ( u v ) m − k ( u + v ) 2 k − 1 ,

where the last line is justified by inverting the summation index k → m − k . Therefore, we have derived the following expansion formula:

sin ( 4 m x ) = 2 cot ( 2 x ) × ∑ k = 1 m ( − 4 ) k − 1 m + k − 1 2 k − 1 sin 2 k ( 2 x ) .

This implies the integral expression below:

Φ ( m ) = 2 ∑ k = 1 m ( − 4 ) k − 1 m + k − 1 2 k − 1 J ( 2 k ) .

Replacing J ( 2 k ) by the explicit formula given in Theorem 2 and then making some routine simplifications, we find the formula as in the proposition below.

Proposition 4

Φ ( m ) = ∑ j = 1 m ( − 1 ) j − 1 ( 1 − 2 2 j + 1 ) ζ ( 2 j + 1 ) π 2 j ∑ k = j m 2 2 k − 1 m + k − 1 2 k − 1 [ x 2 k ] arcsinh 2 j x .

Furthermore, by applying Theorem 3, we have alternatively another expression

Φ ( m ) = 2 ∑ k = 1 m ( − 4 ) k − 1 m + k − 1 2 k − 1 3 k + 1 k × ∑ j = 1 k 1 − 2 2 j + 1 π 2 j ζ ( 2 j + 1 ) ∑ μ = 0 k ∑ ν = 0 μ + 1 ( − 1 ) μ + ν + k − j 2 μ + 1 × 2 k + 1 2 k + μ + 1 k μ 1 + μ ν ( 1 + μ − 2 ν ) 1 + 2 k + μ − 2 j ( 1 + μ ) ( 2 k + μ − 2 j ) ! .

By interchanging the summation order and then making some routine simplifications, we establish the following formula in terms of Riemann zeta function.

Theorem 5

Φ ( m ) = ∑ j = 1 m 2 2 j + 1 − 1 π 2 j ζ ( 2 j + 1 ) ∑ k = j m m + k − 1 2 k − 1 3 k + 1 k × ∑ μ = 0 k ∑ ν = 0 μ + 1 ( − 1 ) j + μ + ν 2 2 + μ − 2 k k μ 1 + μ ν ( 2 k + 1 ) ( 1 + μ − 2 ν ) 1 + 2 k − 2 j + μ ( 1 + μ ) ( 2 k + μ + 1 ) ( 2 k − 2 j + μ ) ! .

The first five formulae are given as follows:

Φ ( 1 ) = − 14 ζ ( 3 ) π 2 , Φ ( 2 ) = − 28 ζ ( 3 ) 3 π 2 + 248 ζ ( 5 ) π 4 , Φ ( 3 ) = − 322 ζ ( 3 ) 45 π 2 + 992 ζ ( 5 ) 3 π 4 − 4,064 ζ ( 7 ) π 6 , Φ ( 4 ) = − 88 ζ ( 3 ) 15 π 2 + 5,456 ζ ( 5 ) 15 π 4 − 8,128 ζ ( 7 ) π 6 + 65,408 ζ ( 9 ) π 8 , Φ ( 5 ) = − 1,126 ζ ( 3 ) 225 π 2 + 356,128 ζ ( 5 ) 945 π 4 − 174,752 ζ ( 7 ) 15 π 6 + 523,264 ζ ( 9 ) 3 π 8 − 1,048,064 ζ ( 11 ) π 10 .

4.3 Integral Ψ ( m )

For m ∈ N 0 , we can also evaluate the integral

Ψ ( m ) ≔ ∫ 0 π 2 cos ( 2 x + 4 m x ) log ( tan x ) d x .

Analogously by specifying u → e 2 x i , v → − e − 2 x i , and n → 2 m in the binomial formula (8), we can express

cos ( 2 x + 4 m x ) = e 2 x i + 4 m x i + e − 2 x i − 4 m x i 2 = u 2 m + 1 − v 2 m + 1 2 = u − v 2 ∑ k = 0 m ( − 1 ) k 2 m − k k ( u v ) k ( u + v ) 2 m − 2 k = u − v 2 ∑ k = 0 m ( − 1 ) m − k m + k 2 k ( u v ) m − k ( u + v ) 2 k .

This leads us to another expansion formula

cos ( 2 x + 4 m x ) = cot ( 2 x ) × ∑ k = 0 m ( − 4 ) k m + k 2 k sin 2 k + 1 ( 2 x ) .

Consequently, we obtain the following integral formula:

Ψ ( m ) = ∑ k = 0 m ( − 4 ) k m + k 2 k J ( 2 k + 1 ) .

According to Theorem 2, this can be stated as in the following proposition:

Proposition 6

Ψ ( m ) = ∑ j = 0 m ( − 4 ) j + 1 β ( 2 j + 2 ) π 2 j + 1 ∑ k = j m 4 k m + k 2 k [ x 2 k + 1 ] arcsinh 2 j + 1 x .

In addition, by applying Theorem 3, we can deduce a more explicit expression

Ψ ( m ) = ∑ k = 0 m ( − 4 ) k m + k 2 k 3 k + 2 k × ∑ j = 0 k ( − 2 2 j + 2 ) π 2 j + 1 β ( 2 j + 2 ) ∑ μ = 0 k ∑ ν = 0 μ + 1 ( − 1 ) μ + ν + k − j 2 μ + 1 × 2 k + 2 2 k + μ + 2 k μ 1 + μ ν ( 1 + μ − 2 ν ) 1 + 2 k − 2 j + μ ( 1 + μ ) ( 2 k − 2 j + μ ) ! .

By exchanging the order of summations with respect to k and j , we can further simplify it into the formula as in the following theorem.

Theorem 7

Ψ ( m ) = ∑ j = 0 m 4 j + 1 π 2 j + 1 β ( 2 j + 2 ) ∑ k = j m m + k 2 k 3 k + 2 k × ∑ μ = 0 k ∑ ν = 0 μ + 1 ( − 1 ) 1 + j + μ + ν 2 1 + μ − 2 k k μ 1 + μ ν ( 2 k + 2 ) ( 1 + μ − 2 ν ) 1 + 2 k − 2 j + μ ( 1 + μ ) ( 2 k + μ + 2 ) ( 2 k − 2 j + μ ) ! .

The initial formulae are recorded below as consequences:

Ψ ( 0 ) = − 4 β ( 2 ) π , Ψ ( 1 ) = − 4 β ( 2 ) 3 π + 64 β ( 4 ) π 3 , Ψ ( 2 ) = − 4 β ( 2 ) 5 π + 64 β ( 4 ) π 3 − 1,024 β ( 6 ) π 5 , Ψ ( 3 ) = − 4 β ( 2 ) 7 π + 896 β ( 4 ) 15 π 3 − 5,120 β ( 6 ) 3 π 5 + 16,384 β ( 8 ) π 7 , Ψ ( 4 ) = − 4 β ( 2 ) 9 π + 52,352 β ( 4 ) 945 π 3 − 19,456 β ( 6 ) 9 π 5 + 114,688 β ( 8 ) 3 π 7 − 262,144 β ( 10 ) π 9 , Ψ ( 5 ) = − 4 β ( 2 ) 11 π + 9,024 β ( 4 ) 175 π 3 − 467,968 β ( 6 ) 189 π 5 + 917,504 β ( 8 ) 15 π 7 − 786,432 β ( 10 ) π 9 + 4,194,304 β ( 12 ) π 11 .

5 Concluding remarks

In this article, we have explicitly evaluated, in closed form, three classes of definite integrals involving ln ( tan x ) in the denominator of the integrands. All the results do not seem to have appeared previously, except for Φ ( 1 ) , which was proposed as a Monthly problem by Stewart [16]. However, according to the test made by an anonymous referee, Mathematica can evaluate the reformulated integral (2) for a specified small integer m , even though it fails to do that for the original integral (1). This fact may suggest the existence of some articles about (1) scattered in the literature, that the authors have not succeeded in locating them.

Recently, McKee [17] evaluated the Laplace transforms of { ln n x } 1 ≤ n ≤ 4 . Thanks to another reviewer for drawing our attention, who demanded for alternative proofs of McKee’s formulae. In general, consider the Laplace transform

L ( x ε − 1 ln n x ) ≔ ∫ 0 ∞ x ε − 1 e − x y ln n x d x , where ε > 0 and n ∈ N 0 .

We are going to derive analytic formulae with two parameters n and ε . Taking into account the importance of Laplace transforms, the explicit formulae produced in the sequel may have potential applications in computations of physics and engineering.

Denote the Euler constant by γ = lim n → ∞ ( H n − ln n ) . Then, the logarithmic differentiation of the Γ -function results in the digamma function (cf. Rainville [18, §9])

ψ ( z ) = d d z ln Γ ( z ) = Γ ′ ( z ) Γ ( z ) = − γ + ∑ n = 0 ∞ z − 1 ( n + 1 ) ( n + z ) .

For a real number ε ∉ Z \ N , the following expansion formula (cf. [19,20]) holds

(9) Γ ( ε + x ) Γ ( ε ) = exp x ψ ( ε ) + ∑ k = 2 ∞ ( − x ) k k ζ ( k , ε ) ,

where the Hurwitz zeta function is defined by

ζ ( m , z ) = ( − 1 ) m ( m − 1 ) ! D z m − 1 ψ ( z ) = ∑ n = 0 ∞ 1 ( n + z ) m .

We are ready now to show the following general formula.

Theorem 8

For ε > 0 and n ∈ N 0 , we have the Laplace transform formula

L ( x ε − 1 ln n x ) = n ! ( − 1 ) n Γ ( ε ) y − ε ∑ σ ( n ) { ln y − ψ ( ε ) } j 1 j 1 ! ∏ k = 2 n ζ j k ( k , ε ) j k ! k j k ,

where the summation domain σ ( n ) consists of all the solutions ( j 1 , j 2 , … , j n ) (in non-negative integers) of the Diophantine equation j 1 + 2 j 2 + … + n j n = n .

Proof of Theorem 8

By making use of the Γ -integral, we can express

L ( x ε − 1 ln n x ) = L ( x ε + τ − 1 ln n x ) ∣ τ = 0 = ∫ 0 ∞ x ε + τ − 1 e − x y ln n x d x τ = 0 = d n d τ n ∫ 0 ∞ x ε + τ − 1 e − x y d x τ = 0 x → T ⁄ y = d n d τ n y − ε − τ ∫ 0 ∞ T ε + τ − 1 e − T d T τ = 0 = d n d τ n { y − ε − τ Γ ( ε + τ ) } τ = 0 .

By substitution, we can further reformulate

L ( x ε − 1 ln n x ) = d n d τ n { y − ε − τ Γ ( ε + τ ) } τ = 0 = Γ ( ε ) y − ε d n d τ n exp τ ψ ( ε ) − τ ln y + ∑ k = 2 ∞ ( − τ ) k k ζ ( k , ε ) τ = 0 .

Then, the formula in Theorem 8 follows by extracting the coefficient of τ n from the above expression inside the braces “ { … } .”□

As applications, we record the initial formulae in the following corollary.

Corollary 9

( ε > 0 : Θ ( y , ε ) ≔ ln y − ψ ( ε ) )

L ( x ε − 1 ) = Γ ( ε ) y ε , L ( x ε − 1 ln x ) = Γ ( ε ) − y ε Θ ( y , ε ) , L ( x ε − 1 ln 2 x ) = Γ ( ε ) y ε { ζ ( 2 , ε ) + Θ 2 ( y , ε ) } , L ( x ε − 1 ln 3 x ) = Γ ( ε ) − y ε { 2 ζ ( 3 , ε ) + 3 ζ ( 2 , ε ) Θ ( y , ε ) + Θ 3 ( y , ε ) } , L ( x ε − 1 ln 4 x ) = Γ ( ε ) y ε { 6 ζ ( 4 , ε ) + 3 ζ ( 2 , ε ) 2 + 8 ζ ( 3 , ε ) Θ ( y , ε ) + 6 ζ ( 2 , ε ) Θ 2 ( y , ε ) + Θ 4 ( y , ε ) } , L ( x ε − 1 ln 5 x ) = Γ ( ε ) − y ε { 24 ζ ( 5 , ε ) + 20 ζ ( 2 , ε ) ζ ( 3 , ε ) + 30 ζ ( 4 , ε ) Θ ( y , ε ) + 15 ζ 2 ( 2 , ε ) Θ ( y , ε ) + 20 ζ ( 3 , ε ) Θ 2 ( y , ε ) + 10 ζ ( 2 , ε ) Θ 3 ( y , ε ) + Θ 5 ( y , ε ) } .

When ε = 1 , we recover the formulae due to McKee [17], where Δ ( y ) ≔ ( γ + ln y ) for brevity.

L ( 1 ) = 1 y , L ( ln x ) = − 1 y Δ ( y ) , L ( ln 2 x ) = 1 y { ζ ( 2 ) + Δ ( y ) 2 } , L ( ln 3 x ) = − 1 y { 2 ζ ( 3 ) + 3 ζ ( 2 ) Δ ( y ) + Δ ( y ) 3 } , L ( ln 4 x ) = 1 y { 6 ζ ( 4 ) + 3 ζ 2 ( 2 ) + 8 ζ ( 3 ) Δ ( y ) + 6 ζ ( 2 ) Δ ( y ) 2 + Δ ( y ) 4 } , L ( ln 5 x ) = − 1 y { 24 ζ ( 5 ) + 20 ζ ( 2 ) ζ ( 3 ) + 30 ζ ( 4 ) Δ ( y ) + 15 ζ 2 ( 2 ) Δ ( y ) + 20 ζ ( 3 ) Δ ( y ) 2 + 10 ζ ( 2 ) Δ ( y ) 3 + Δ ( y ) 5 } .

Recall the Dirichlet lambda function

λ ( z ) ≔ ∑ k = 0 ∞ 1 ( 2 k + 1 ) z for ℜ ( z ) > 1 .

For ε = 1 2 , we find the following formulae, where ∇ ( y ) ≔ ( γ + ln ( 4 y ) ) for simplicity.

L ( x − 1 ⁄ 2 ) = π y , L ( x − 1 ⁄ 2 ln x ) = − π y ∇ ( y ) , L ( x − 1 ⁄ 2 ln 2 x ) = π y { { 4 λ ( 2 ) + ∇ ( y ) } 2 } , L ( x − 1 ⁄ 2 ln 3 x ) = − π y { 16 λ ( 3 ) + 12 λ ( 2 ) ∇ ( y ) + ∇ 3 ( y ) } , L ( x − 1 ⁄ 2 ln 4 x ) = π y { 96 λ ( 4 ) + 24 λ ( 2 ) ∇ 2 ( y ) + 64 λ ( 3 ) ∇ ( y ) + 48 λ ( 2 ) 2 + ∇ 4 ( y ) } , L ( x − 1 ⁄ 2 ln 5 x ) = − π y { 768 λ ( 5 ) + 40 λ ( 2 ) ∇ 3 ( y ) + 240 λ ( 2 ) 2 ∇ ( y ) + ∇ 5 ( y ) + 640 λ ( 2 ) λ ( 3 ) + 160 λ ( 3 ) ∇ 2 ( y ) + 480 λ ( 4 ) ∇ ( y ) } .

Acknowledgments

The authors express their sincere gratitude to the two reviewers for the careful reading, critical comments, and valuable suggestions that contributed significantly to improving the manuscript during the revision.

Funding information: This research received no external funding.
Author contributions: J. Li: Text editing and computation; W. Chu: Original draft and supervision.
Conflict of interest: The authors declare no conflict of interests.
Data availability statement: Data sharing is not applicable to this article.

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Received: 2023-12-06

Revised: 2024-07-29

Accepted: 2024-07-29

Published Online: 2024-09-13

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https://doi.org/10.1515/math-2024-0052

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Cauchy residue theorem; contour integration; formal power series

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