Some Fun with Integrals involving Hyperbolic Functions Jean-Yves Pitarakis

A few years ago I came across some rather messy integrals which were not considered in the Soviet Bible (i.e. Gradhsteyn and Ryzhik). Here is a selection I solved analytically. Use at your own risk.

I3(s) =

R∞

0

z s dz (cosh z) 2

Expanding and integrating termwise we can formulate I3 (s) as s

∞ (−1)k Γ( 2s + k) 2 2 +2 X I3 (s) = Γ( 2s ) k=0 k!(4k + s)2

which can be used for values of s < 6 but is not absolutely convergent for s ≥ 6. For s = 1 we have I3 (1) = 5.56286 and we may proceed similarly for s < 6. It is of interest to notice the following particular cases I3 (2) = 2 I3 (4) = 2

∞ X

(−1)k = 2G 2 k=0 (2k + 1) ∞ X (−1)k k=0

(k + 1)

= log 2

where G is Catalan’s constant (G = 0.915965594..). In order to evaluate I3 (s) for s ≥ 6 it is convenient to perform the change of variable u = e−2z leading to I3 (s) = −2

s −2 2

Z 0

1

s

u 4 −1 log u s du (1 + u) 2

For s = 6 we have I3 (6) = −2

Z 0

1

√

u log u 1 du = − + G 3 (1 + u) 2 1

for s = 8, Z

I3 (8) = −4

0

1

u log u −1 + 4 log 2 = (1 + u)4 6

and for s = 12, I3 (12) = −16

Z 0

1

u2 log u −11 + 32 log 2 du = 6 (1 + u) 60

using successive integration by parts and intermediate results from Gradhsteyn & Rhyzhik (1980). For the case s = 6 for instance we may proceed as follows: Integrating by parts I3 (6) once leads to I3 (6) = = = = =

1Z 1 1 log u √ du − du − 2 u(1 + u) 2 0 u(1 + u)2 0 √ Z 1 1 1 1Z 1 1 Z 1 arctan u √ √ − du + du + du u(1 + u)2 2 0 u(1 + u) 2 0 u 0 Z 1 1 π 1 √ − du + + (2G) 2 u(1 + u) 4 2 0 √ Z 1 π 1 u du + + G − −4 3 2 4 0 (1 + u) 1 − +G 2 Z

1

√

where G is Catalan’s constant. The reason I proceeded as above was due to the divergence of the infinite series representation for larger values of s. However it is sometimes possible to obtain the values of the integrals by manipulating the initial non convergent series. To illustrate this point I will focus on the case s = 8. From the infinite series representation we have I3 (8) =

∞ 2X (−1)k (k + 1)(k + 3) . 3 k=0 (k + 2)

We can now write the same series by means of the Lerch Transcendent1 as I3 (8) =

2 (Φ(−1, −1, 2) − Φ(−1, 1, 2)) 3

with Φ(−1, 1, 2) = 12 (ζ(1, 1) − ζ(1, 32 )) and Φ(−1, −1, 2) = 2(ζ(−1, 1) − ζ(−1, 23 )) where ζ(., .) denotes Hurwitz’s zeta function. The result I3 (8) = (4 log 2 − 1)/6 follows immediately from the properties of the Hurwitz zeta function. 1

Φ(z, s, a) =

zk k=0 (k+a)s

P∞

2

I4(s) =

R∞

0

z3 s (cosh z) 2

Expanding the integrand appropriately and integrating termwise leads to s ∞ (−1)k Γ( 2s + k) 96 2 2 X I4 (s) = Γ( 2s ) k=0 k!(s + 4k)4

which can be evaluated directly for s < 10: √ ∞ √ (−1)k Γ( 12 + k) 96 2 X 96 2 I4 (1) = √ = √ 1.77112 = 135.662 π k=0 k!(4k + 1)4 π

I4 (2) = 12

I4 (4) =

∞ X

(−1)k = 12β(4) = 11.8674 4 k=0 (2k + 1)

∞ 96 X (−1)k 9 = ζ(3) = 1.35231 3 3 4 k=0 (k + 1) 8

where β(.) denotes Dirichlet’s beta function and ζ() is Riemann’s zeta. Proceeding similarly, we also obtain I4 (6) = 0.437874 and I4 (8) = 0.208395. The series is not convergent (absolutely) for s ≥ 10. In such cases it is convenient to use the log transformation and integrate by parts. Alternatively, one may express the infinite series in terms of the Lerch Transcendent. Under s = 12 for instance we have ∞ 1X (−1)k (k + 1)(k + 2)(k + 4)(k + 5) 5 k=0 (k + 3)2 1 = (Φ(−1, −1, 3) + 4Φ(−1, 3, 3) − 5Φ(−1, 1, 3)) 5

I4 (12) =

where Φ(−1, −1, 3) = 45 , Φ(−1, 3, 3) = − 87 + 34 ζ(3) and Φ(−1, 1, 3) = − 12 + log 2, leading to I4 (12) =

I5(s) =

R∞

0

1 − 10 log 4 + 12ζ(3) = 0.078087. 20

z 2 sinh z s+2

(cosh z) 2

Expanding the integrand appropriately and integrating termwise leads to s

∞ (−1)k Γ(k + 2 2 +4 X I5 (s) = k! Γ( s+2 ) k=0 2

3

s+2 ) 2

"

1 1 − 3 (4k + s) (4k + s + 4)3

#

Under s = 1 the direct evaluation of the above infinite series leads to I5 (1) = 22.2514. For s = 2 we have ∞ ∞ (−1)k (k + 1) X 1 X (−1)k (k + 1) I5 (2) = − 2 k=0 (k + 21 )3 (k + 32 )3 k=0

"

#

which can also be evaluated directly. Alternatively, we can express the series in terms of the Lerch Transcendent and write 1 1 1 1 3 1 3 1 Φ(−1, 3, ) + Φ(−1, 2, ) + Φ(−1, 3, ) − Φ(−1, 2, ) 4 2 2 2 4 2 2 2 = 4G

I5 (2) =

Similarly, for s = 4 we obtain 1 1 1 1 Φ(−1, 2, 2) + Φ(−1, 2, 1) − Φ(−1, 1, 1) − Φ(−1, 1, 2) 2 2 2 2 = log 2.

I5 (4) =

and I5 (6) = 0.27731, I5 (8) = − 21 +

I6(s) =

log 2 , 3

11 I5 (12) = − 180 +

32 log 2 . 180

R ∞ z(sinh z)2 s+4 0 (cosh z) 2

An infinite series representation for I6 (s) can be written as s

∞ (−1)k Γ( s+4 + k) 4 22 X 1 8 2 + I6 (s) = − s s+4 k! (2k + 2 + 4)2 (4k + s)2 (4k + s + 4)2 Γ( 2 ) k=0

"

#

The series can be used to evaluate the integral for values of s up to 5. For s = 1 and s = 2 for instance we have I6 (1) = 5.04191 and I6 (2) = 0. Manipulating the sum by means of the Lerch Transcendent in order to obtain simpler expressions in terms of Hurwitz Zeta functions leads to the following simplifications I6 (4) =

1 log 2 + 6 3

I6 (8) =

1 + 8 log 2 60

I6 (12) =

32 log 2 − 1 420

4

I7(s) =

R∞

0

sinh z s+2

z(cosh z) 2

Focusing on the case s = 1, I7 (1) can be written as ∞

Z

I7 (1) =

0

tanh z 1 dz. z (cosh z) 12

Expanding the second portion and integrating termwise using Gradhsteyn & Rhyzhik (1980, p. 361) we obtain √ ∞ " # Γ( 4k+1 ) 2 X (−1)k Γ( 12 + k) 4k + 1 8 log( ) + 2 log 4k+5 I7 (1) = √ π k=0 k! 8 Γ( 8 ) leading to I7 (1) = 1.56151 We next focus on the case I7 (2) for which the above technique is not computationally very useful. We initially transform the integral via the change of variable u = e−2z and obtain I7 (2) = −2

1

Z 0

(1 − u) √ du. (1 + u)2 u log u

Expanding appropriately, I7 (2) can be rewritten as I7 (2) = −2

∞ X

k

(−1) (k + 1)

Z 0

k=0

1

1

uk− 2 (1 − u) du. log u

It is now useful to perform an integration by part (recalling that and obtain I7 (2)

= −2

∞ X k=0

R

1/(u log u) = log log( u1 ))

  Z 1 Z 1 1 1 3 1 1 1 (−1)k (k + 1) (k + ) uk+ 2 log log( )du − (k + ) uk− 2 log log( )du . 2 0 u 2 0 u

Next, using the result (Adamchik (1997)) R1 0

uα log log( u1 )du = − log(α+1)+γ (α+1)

where γ is the Euler-Mascheroni constant, our original problem reduces to I7 (2) = −2

∞ X

(−1)k (k + 1) log(

k=0

2k + 1 ) 2k + 3

which is equivalent to I7 (2) = −2

∞ X

(−1)k (2k + 1) log(2k + 1).

k=0

5

In order to evaluate the above expression we rewrite it as follows ∞ X (−1)k ∂ 22−µ I7 (2) = 2 lim 1 µ−2 µ→1 ∂µ k=0 (k + 2 )   ∂ 1 2−µ = 2 lim 2 Φ(−1, µ − 2, ) µ→1 ∂µ 2

"

#

Finally, using the properties of the Hurwitz’s zeta function as well as the following relationship between the Lerch transcendent and Hurwitz’s zeta function Φ(−1, µ − 2, 12 ) = 22−µ (ζ(µ − 2, 41 ) − ζ(µ − 2, 43 )) leads to I7 (2) =

4G . π

Next we treat the case I7 (4): making the change of variable u = e−2z in the initial integral we have Z 1 (1 − u) du I7 (4) = −4 0 (1 + u)3 log u Expanding the denominator leads to I7 (4) = −2

∞ X

(−1)k (k + 1)(k + 2)

Z 0

k=0

1

uk (1 − u) du log u

Integrating by parts, we have I7 (4)

= −2

∞ X

1

 Z (−1)k (k + 1)(k + 2) (k + 2) 0

k=0

1 uk+1 log log( )du − (k + 1) u

Z 0

1

 1 uk log log( )du u

which can be reduced to the following I7 (4) = −2

∞ X

(−1)k (k + 1)(k + 2) log(

k=0

k+1 ) k+2

At this stage it is important to observe that the above series can also be written as I7 (4) = 4

∞ X

(−1)k k 2 log k.

k=1

and proceeding as before we rewrite it as ∞ ∂ X (−1)k+1 I7 (4) = 4 lim ( ) µ−3 µ→1 ∂µ k=1 k = 4η(−2) = 28 log A2 = 0.8525567379 . . .

#

"

where η(.) is Dirichlet’s eta function and A2 is known as a general Glaisher’s constant of order 2. Proceeding similarly for s = 12 we obtain I7 (12) = 0.504767.

6

I8(s) =

(sinh z)2

R∞

0

s+4

z(cosh z) 2

For the case s = 1 we proceed as previously and obtain 3 ∞ (−1)k Γ( 32 + k) Γ( 4k+1 )Γ( 4k+5 + 12 ) 22 X 4k + 1 8 8 I8 (1) = √ log( ) + 2 log 4k+1 π k=0 k! 4k + 5 Γ( 8 + 12 )Γ( 4k+5 ) 8

"

#

leading to I8 (1) = 1.00117 Similarly for I8 (2) we have I8 (2)

∞ X

"

Γ( 2k+1 ) Γ( 2k+3 ) 2k + 1 2k + 3 (−1) (k + 1) log( = 2 ) − log( ) + 2 log 2k+14 1 − 2 log 2k+34 1 4 4 Γ( 4 + 2 ) Γ( 4 + 2 ) k=0

#

k

leading2 to I8 (2) = 0.64348 When s = 4, we express I8 (4) as ∞ (k + 1)(k + 3) 2X (−1)k (k + 1)(k + 2)(k + 3) log( ) I8 (4) = − 3 k=0 (k + 2)2 = 0.384928

"

#

and I8 (12) = 0.15061.

REFERENCES • Adamchik, V. (1997). ”A Class of Logarithmic Integrals,” Wolfram Research, Manuscript. • Gradhshteyn, I.S. and I.M. Ryzhik (1980). Tables of Integrals, Series and Products. New-York: Academic Press.

2

When appropriate, all infinite series have been evaluated using Mathematica

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