An exact 1-D integration challenge - 21
Vladimir Bondarenko
Alas, this time it is not me who came up with this nice stuff,
Is there a person who can show how using any computer algebra
system to calculate the exact value of the following integral
int(ln(Pi^2+ln(z)^2)/(1+z^2), z= 0..1);
?
(Are you going to use identify() here? ;)
Best wishes,
Vladimir Bondarenko
VM and GEMM architect
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
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Vladimir Bondarenko
Pi*ln(sqrt(..........
;-)
C W
Vladimir Bondarenko wrote:
>
> Hello the computer algebra fans,
>
> Alas, this time it is not me who came up with this nice stuff,
>
> Is there a person who can show how using any computer algebra
> system to calculate the exact value of the following integral
>
> int(ln(Pi^2+ln(z)^2)/(1+z^2), z= 0..1);
>
If You believe You are able to identify this kind of integrals readily then You
should challenge Yourself with the following :
Int(exp(x*Pi)*x/(1+exp(x*Pi)^2)/(1+x^2),x = 0 .. infinity);
Chris
Vladimir Bondarenko
http://groups.google.com/group/sci.math.symbolic/msg/d0734942a2286095
CW> If You believe You are able to identify this kind of
CW> integrals readily
Readily? I did not claim that this task is simple. My words
were "nice stuff"...
CW> Int(exp(x*Pi)*x/(1+exp(x*Pi)^2)/(1+x^2),x = 0 .. infinity);
As for your integral, obviously, it is equal to
sum((-1)^n*Ci(Pi*(2*n+1)), n= 0..infinity);
but I have a vague feeling that it could be simplified further...
C W
Oops..
I forgot to include the ln :
Int(ln(x)*exp(x*Pi)*x/(1+exp(x*Pi)^2)/(1+x^2),x = 0 .. infinity);
Chris
C6...@shaw.ca
> Vladimir Bondarenko wrote:
> >
> > Hello the computer algebra fans,
> >
> > Alas, this time it is not me who came up with this nice stuff,
> >
> > Is there a person who can show how using any computer algebra
> > system to calculate the exact value of the following integral
> >
> > int(ln(Pi^2+ln(z)^2)/(1+z^2), z= 0..1);
> >
>
> If You believe You are able to identify this kind of integrals readily then You
> should challenge Yourself with the following :
>
> Int(exp(x*Pi)*x/(1+exp(x*Pi)^2)/(1+x^2),x = 0 .. infinity);
Just to clarify: what does your notation
something1/something2/something3 stand for?
R.G. Vickson
Ioannis
news:1147128732.7...@g10g2000cwb.googlegroups.com...
>
> C W wrote:
> > Vladimir Bondarenko wrote:
> > >
> > > Hello the computer algebra fans,
> > >
> > > Alas, this time it is not me who came up with this nice stuff,
> > >
> > > Is there a person who can show how using any computer algebra
> > > system to calculate the exact value of the following integral
> > >
> > > int(ln(Pi^2+ln(z)^2)/(1+z^2), z= 0..1);
> > >
> >
> > If You believe You are able to identify this kind of integrals readily
then You
> > should challenge Yourself with the following :
> >
> > Int(exp(x*Pi)*x/(1+exp(x*Pi)^2)/(1+x^2),x = 0 .. infinity);
>
> Just to clarify: what does your notation
> something1/something2/something3 stand for?
I think Maples translates the expression a/(b*c) as a/b/c when you copy
results as "pretty text".
So in general, a/(b*c*d*...) would be copied/translated as a/b/c/d/...
> R.G. Vickson
--
Ioannis --- http://ioannis.virtualcomposer2000.com/
Igor Khavkine
> Hello the computer algebra fans,
>
> Alas, this time it is not me who came up with this nice stuff,
>
> Is there a person who can show how using any computer algebra
> system to calculate the exact value of the following integral
>
> int(ln(Pi^2+ln(z)^2)/(1+z^2), z= 0..1);
>
> ?
Maple 9.5:
> Digits := 20;
Digits := 20
> Pi*ln(GAMMA(1/4)^2/(4*sqrt(Pi)));
3/2
Pi
Pi ln(1/2 -----------)
2
GAMMA(3/4)
> evalf(%);
1.9395745220275329590
> evalf(Int(ln(Pi^2+ln(z)^2)/(1+z^2), z=0..1));
1.9395745220275329587
Have a nice day.
Igor
igo...@gmail.com
> On 2006-05-01, Vladimir Bondarenko <v...@cybertester.com> wrote:
> > Hello the computer algebra fans,
> >
> > Alas, this time it is not me who came up with this nice stuff,
> >
> > Is there a person who can show how using any computer algebra
> > system to calculate the exact value of the following integral
> >
> > int(ln(Pi^2+ln(z)^2)/(1+z^2), z= 0..1);
> >
> > ?
By popular request, here's a sketch of the evaluation of this integral.
First substitute y = -ln(z), and extend the integration limits from
y=-oo to oo. Then the integral becomes:
1/2 * int(ln(Pi^2+y^2)/cosh(y), y=-infinity..infinity)
Realize that ln(Pi^2+y^2) = 2*Re(ln(Pi+I*y)) and the the Re symbol out
outside the integrand. The integrand now has a branch point at y=I*Pi.
Shift the integration contour, y -> y + I*Pi, so that it passes through
the branch point. Along the way the residue at y = I*Pi/2, du to the
cosh(y) denominator, will contribute to the integral.
Move the Re symbol inside the integral, realize that the integrand is
symmetric, and fold the integration interval in half. Modulo some
constants, the remaining integral will look like:
int(ln(x)/cosh(x), x=0..infinity).
Next, use two tricks. First, write ln(x) = d/ds x^s at s = 0, and take
the derivative outside the integral. Then expand
1/(2*cosh(x)) = exp(-x) - exp(-3*x) + exp(-5*x) - exp(-7*x) + ...
After interchanging summation and integration, each remaining integral
has the form int(x^s*exp(-(2*n+1)*x),x=0..infinity), which can be
expressed as GAMMA(s+1)/(2*n+1)^(s+1).
Realize that the resulting series can be summed using the Dirichlet
Beta function[1], beta(s+1). After taking the derivative with respect
to s, find that the answer is expressible in terms of the beta(1) and
beta'(1), both of which have known values[1].
Put all the pieces together, simplify and get (Maple 9.5):
> > Digits := 20;
> Digits := 20
>
> > Pi*ln(GAMMA(1/4)^2/(4*sqrt(Pi)));
> 3/2
> Pi
> Pi ln(1/2 -----------)
> 2
> GAMMA(3/4)
>
> > evalf(%);
> 1.9395745220275329590
>
>
> > evalf(Int(ln(Pi^2+ln(z)^2)/(1+z^2), z=0..1));
> 1.9395745220275329587
Have a nice day.
Igor