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MathGroup Archive 2007

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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



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