Re: Integrate[s^s(1-s)^(1-s)Sin[Pi s],{s,0,1}]
- To: mathgroup at smc.vnet.net
- Subject: [mg76609] Re: Integrate[s^s(1-s)^(1-s)Sin[Pi s],{s,0,1}]
- From: Valeri Astanoff <astanoff at gmail.com>
- Date: Thu, 24 May 2007 06:21:22 -0400 (EDT)
- References: <f2bs3m$ga6$1@smc.vnet.net>
On 15 mai, 10:47, janos <janostothmeis... at gmail.com> wrote:
> Any idea to calculate this integral (symbolically) or reformulate it
> using some special functions?
>
> Thanks, Janos
Good day,
Here is a way of computing that integral (without Plouffe's inverter)
using Stirling's formula and "Recognize" :
In[1]:=<<NumberTheory`
In[2]:=stir = First[Solve[s! == ss*Exp[-s]*Sqrt[2*Pi*s]*(1 + k), ss]]
Out[2]={ss -> (E^s*s!)/((1 + k)*Sqrt[2*Pi]*Sqrt[s])}
In[3]:=fs[s_]=ss/.stir
Out[3]=(E^s*s!)/((1 + k)*Sqrt[2*Pi]*Sqrt[s])
In[4]:=ff=fs[1-s] fs[s] Sin[Pi*s]//FullSimplify[#,0<s<1]&
Out[4]=(E*Sqrt[(-(-1 + s))*s])/(2*(1 + k)^2)
In[5]:=i1 = Integrate[ff, {s, 0, 1}]
Out[5]=(E*Pi)/(16*(1 + k)^2)
In[6]:=i2=NIntegrate[s^s (1-s)^(1-s)Sin[Pi*s],{s,0,1},WorkingPrecision-
>40];
In[7]:=kk=Solve[i1 == i3,k]
Out[7]={{k -> (-4*Sqrt[i3] - Sqrt[E*Pi])/(4*Sqrt[i3])},
{k -> (-4*Sqrt[i3] + Sqrt[E*Pi])/(4*Sqrt[i3])}}
In[8]:=Recognize[k/.kk[[2]]/.i3->i2,4,k]==0
Out[8]=1 - 4*k - 2*k^2 == 0
In[9]:=Solve[%]
Out[9]={{k -> (1/2)*(-2 - Sqrt[6])}, {k -> (1/2)*(-2 + Sqrt[6])}}
In[10]:=i1/.%[[2]]//Simplify
Out[10]=(E*Pi)/24
V.Astanoff