An exact 1-D integration challenge - 63 - (arctan)
Vladimir Bondarenko
Hello,
int(arctan(z)*arctan(2*z)/(1+z)^2, z= 0..infinity);
?
Cheers,
Vladimir Bondarenko
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester Ltd.
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sashap
> Hello,
>
> int(arctan(z)*arctan(2*z)/(1+z)^2, z= 0..infinity);
>
> ?
1..infinity. Change variables z->1/z in the second part
arriving at
int == Integrate[(ArcTan[z] ArcTan[2 z] -
ArcCot[z] ArcCot[z/2])/(1 + z)^2, {z, 0, 1}]
Next, integrate by parts absorbing 1/(1+z)^2 into
the differential:
int == 1/4 Pi (Pi - ArcTan[2]) +
Integrate[1/(1 + z)*
D[(ArcTan[z] ArcTan[2 z] +
ArcCot[z] ArcCot[z/2]), z],
{z, 0, 1}];
Evaluate integrals separately for
each argtrig:
FullSimplify[(Integrate[#, {z, 0, 1}] & /@
Collect[1/(1 + z)
D[(ArcTan[z] ArcTan[2 z] + ArcCot[z] ArcCot[z/2]),
z], _ArcTan | _ArcCot]) + 1/4 Pi (Pi - ArcTan[2])]
After some while, and additional massaging with
ComplexExpand[Re[expr]] I obtained
1/160 (-64 Catalan + (25 - (10 I)/3) Pi^2 - (20 + 20 I) *
ArcTan[4/3]^2 + 4 Pi (5 I ArcTan[4/3] +
ArcTan[3116/237] - 12 Log[2] +
18 Log[3] - 5 Log[5]) + 40 I PolyLog[2, -(3/5) + (4 I)/5] +
10 PolyLog[2, 1/81] - 48 PolyLog[2, 1/9] +
40 PolyLog[2, 1/5 - (4 I)/15] + 40 PolyLog[2, 1/5 + (4 I)/15] -
48 PolyLog[2, 1/3] + 40 I PolyLog[2, 3/5 + (4 I)/5])
In[47]:= N[%, 20]
Out[47]= 0.99966424390838457971 + 0.*10^-21 I
This numerically coincided with NIntegrate:
In[48]:= NIntegrate[ArcTan[z] ArcTan[2 z]/(1 + z)^2, {z, 0, Infinity},
WorkingPrecision -> 20]
Out[48]= 0.99966424390838457972
I hope there is a neat way to reduce those PolyLogs, but I
was not successful at doing so.
Vladimir, if you have a shorter answer, please post it.
Cheers,
Oleksandr
Vladimir Bondarenko
Hi Oleksandr,
Much thanks for your detailed description!
My answer looks a bit shorter
(4*Pi*((9+I)*Pi+3*Log[19683/6250])-96*Catalan+3*(20*
ArcCot[2]*Log[5/4] -(3-4*I)*Log[4/3]*Log[3]-(6-8*I)*
PolyLog[2,-2]+6*PolyLog[2,1/9] - (18+8*I)*PolyLog[2,
1/3]-20*I*PolyLog[2,4/5-2*I/5]+20*I*PolyLog[2,4/5+2*
I/5]))/240
and I am sure there IS still room to compress it...
Cheers,
Vladimir
dimitris
(*To TrigExpand or not to TrigExpand?*)
In[16]:=
symb = Integrate[TrigToExp[ArcTan[z]*ArcTan[2*z]]/(1 + z)^2, {z, 0,
Infinity}]
Out[16]=
(1/40)*(-16*Catalan - 6*Pi^2 - 6*ArcTan[1/3]*ArcTan[1/2] - 6*ArcTan
[1/3]*ArcTan[2] - 6*ArcTan[1/2]*ArcTan[3] -
6*ArcTan[2]*ArcTan[3] - 20*ArcCot[2]*Log[2] - 20*ArcTan[1/2]*Log[2]
- 40*ArcTan[2]*Log[2] - 3*Log[2]^2 +
10*ArcCot[2]*Log[5] - 10*ArcTan[1/2]*Log[5] + 10*ArcTan[2]*Log[5] +
Log[3]*Log[64] +
Pi*(3*ArcTan[1/3] - 13*ArcTan[1/2] + 11*ArcTan[2] + 27*ArcTan[3] +
18*Log[3] - 10*Log[5] + Log[256]) - 6*PolyLog[2, -(1/2)] -
10*I*PolyLog[2, 1/5 - (2*I)/5] + 10*I*PolyLog[2, 1/5 + (2*I)/5])
In[18]:=
Chop[N[symb]]
Out[18]=
0.999664243908386
In[13]:=
NIntegrate[ArcTan[z]*(ArcTan[2*z]/(1 + z)^2), {z, 0, Infinity}]
Out[13]=
0.9996642439083778
(*Also*)
In[21]:=
ff = FullSimplify[symb]
Out[21]=
(1/40)*(-16*Catalan + 6*Pi^2 - 6*Log[2]^2 + 10*ArcTan[2]*Log[5] + Log
[8]*Log[9] - 2*Pi*Log[200000/19683] -
10*I*(PolyLog[2, 1/5 - (2*I)/5] - PolyLog[2, 1/5 + (2*I)/5]) -
3*PolyLog[2, 1/4])
In[22]:=
LeafCount /@ {symb, ff}
Out[22]=
{162, 69}
Best Regards
Dimitris
Axel Vogt
> Hello,
>
> int(arctan(z)*arctan(2*z)/(1+z)^2, z= 0..infinity);
>
> ?
> Cheers,
>
> Vladimir Bondarenko
I see it as Int(f(z)*arctan(2*z)*D(f)(z), z=a..infinity) for f=arctan
and a=0.
Using integration by parts w.r.t. arctan(2*z) finally leads me to
(sorry ...):
-2/5*Catalan-1/4*Im(dilog(1/5-2/5*I)-dilog(1/5+2/5*I))-3/40*ln(2)*ln(3)-3/40*dilog(4)+(-9/4*ln(1/3*5^(1/9)*2^(1/45)*3^(4/5))+11/80*Pi)*Pi-3/40*dilog(2/3)-3/40*dilog(3)+3/40*dilog(4/3)+(-ln(1-1/2*I)+ln(1+1/2*I))*(1/4*I*ln(2)-1/8*I*ln(5))
clicl...@freenet.de
dimitris schrieb:
> On 14 ???, 13:45, Vladimir Bondarenko <v...@cybertester.com> wrote:
> >
> > int(arctan(z)*arctan(2*z)/(1+z)^2, z= 0..infinity);
> >
> > ?
I have polished this a bit:
-(2*pi*LN(400000)-12*LN(3)*(LN(2)+3*pi)-40*LN(2)*ATAN(1/3)-12*LI~
2(-2)+32*CATALAN_CONSTANT-13*pi^2+20*#i*(LI2(-2*#i)-LI2(2*#i)))/~
80
0.99966424390838457970+1.0005761180378663865*10^(-22)*#i
The LN(400000) is crazy.
Martin.