Orthogonal Polynomials and Special Functions

October 1996 Orthogonal Polynomials and Special Functions Newsletter 14
Remark: When this problem was set we thought it was
a one-off. Since then general solutions in closed form have
been found for
π/4
I = ln cos m/n θ ± sin m/n
θ dθ
0
where m and n are coprime integers and a paper detailing
this has been submitted [2].
References
[1] Lewin, L.: Dilogarithms and associated functions. Macdonald,
London, 1958.
[2] Zucker, I.J., Joyce, G.S. and Delves, R.T.: On the evaluation
of the integrals
π/4
ln cos 0
m/n θ ± sin m/n
θ dθ.
Submitted for publication to the Ramanujan Journal.
The following solution using Mathematica was sent by
Victor Adamchik. The calculations require Mathematica
Version 3.
Second Solution of Problem 14
A Trigonometric Integral
by Victor Adamchik
(victor@wolfram.com)
In several steps the proposed integral is brought into a form
such that Mathematica can evaluate it. This will result in
the representation
π/4
ln(sin
0
3/2 (x) + cos 3/2 (x)) dx =
G 5 π
−
12 16
ln 2 + π
8 ln(2 − √ 2)−
π
8 ln(2 + √ 2) − π
3 ln(√3 − 1) + π
3 ln(1 + √ 3) .
First of all we observe that
π/4
ln(sin
0
3/2 (x) + cos 3/2 (x)) dx =
π/2
1
ln(sin
2 0
3/2 (x) + cos 3/2 (x)) dx
which is easy to prove by changing the variable x→π/2−y.
Then we need to transform that integral to the algebraic
form
1
2
π/2
ln(sin
0
3/2 (x) + cos 3/2 (x)) dx
1
ln((1 − z) 3/4 + z3/4 )
√ √ dz .
z 1 − z
= 1
4
0
This can be done by x → arcsin √ z. On the next step we
divide the integrand into two parts:
ln((1 − z) 3/4 + z 3/4 )
√ z √ 1 − z
=
3 ln z
4 √ z √ 1 − z +ln(1 + (1/z − 1)3/4 )
√ √
z 1 − z
so that
1
1 ln((1 − z)
4 0
3/4 + z3/4 )
√ √ dz =
z 1 − z
1
3 ln z
√ √ dz +
16 0 z 1 − z 1
1
ln(1 + (1/z − 1)
4 0
3/4 )
√ √ dz =
z 1 − z
1
3 ln z
√ √ dz +
16 0 z 1 − z 1
∞
ln(1 + y
4 0
3/4 )
√ dy .
y (1 + y)
In the second integral we changed z to 1/(1 + y). Now we
can use Mathematica Version 3 to evaluate both integrals:
In[1]:= Integrate[Log[z]/(Sqrt[z] Sqrt[1-z]),{z,0,1}]
Out[1]= -2 Pi Log[2]
In[2]:= Integrate[Log[1 + y^(3/4)]/(Sqrt[y] (1 + y)),
{y, 0, Infinity}]/.
{r1_ Sum[w1_, {e__}] + r2_ Sum[w2_, {e__}] :>
Sum[r1 w1 + r2 w2 // FullSimplify, {e}]} //
FullSimplify
Out[2]=(4 Catalan + Pi Log[8] + 6 Pi Log[2 - Sqrt[2]] -
> 6 Pi Log[2 + Sqrt[2]] - 16 Pi Log[-1 + Sqrt[3]] +
> 16 Pi Log[1 + Sqrt[3]]) / 12
Finally
In[3]:= 3/16 %1 + 1/4 %2 // Simplify
Out[3]= (4 Catalan - 15 Pi Log[2] +
> 6 Pi Log[2 - Sqrt[2]] - 6 Pi Log[2 + Sqrt[2]] -
> 16 Pi Log[-1 + Sqrt[3]] +
> 16 Pi Log[1 + Sqrt[3]]) / 48
Editor’s Remark: Victor has a WWW Page on the Catalan
Constant, and he searches for different representations
for this constant, see p. 16.
Miscellaneous
1. Publicity about OP-SF in SIAM News
I had a conversation with Gail Corbett, Editor of SIAM News,
during the SIAM meeting in Kansas City. Gail welcomes material
from us as from all the Activity Groups.
But not all items from our own media are suitable for SIAM
News. In particular, announcements or reports of our own minisymposia
are not suitable unless there is something very special
about them. These would include important advances in the
field, profiles of people, etc. Events surrounding a birthday
or anniversary might be suitable especially if they involved a
personality or event of interest to several groups.