Nice integral
Han de Bruijn
Pi ln(2)
Yes, that is what MAPLE says. But why?
http://hdebruijn.soo.dto.tudelft.nl/jaar2010/nice.pdf
Han de Bruijn
Robert Israel
OK, now do
int((arctan(1/r)^3,r=0..infinity) = -21/8*Zeta(3)+3/4*Pi^2*ln(2)
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Valeri Astanoff
Good day,
Let me post what purists will probably not consider as a proof :
This function :
f[r_] = r*ArcTan[1/r]^2 - 2*(ArcTan[1/r] *
Log[1 - E^(2*I*ArcTan[1/r])] - (1/2)*I*(ArcTan[1/r]^2 +
PolyLog[2, E^(2*I*ArcTan[1/r])]));
is an antiderivative of arccot(r)^2 :
f'[r] // FullSimplify
ArcCot[r]^2
now :
ArcTan[1/r]^2 == ArcCot[r]^2 // FullSimplify
True
hence the integral :
Simplify[Limit[f[r], r -> Infinity] - Limit[f[r], r -> 0]]
Pi Log[2]
--
Valeri
Valeri Astanoff
wrote:
Good day,
I just do it the same way as arctan(1/r)^2 :
This function :
f[r_] = (I*Pi^3)/8 + (-I + r)*ArcCot[r]^3 -
3*ArcCot[r]^2*Log[(2*I)/(I + r)] -
3*I*ArcCot[r]*PolyLog[2, (-I + r)/(I + r)] -
(3/2)*PolyLog[3, (-I + r)/(I + r)];
is an antiderivative of arccot(r)^3 :
f'[r] // FullSimplify
ArcCot[r]^3
now :
ArcTan[1/r]^3==ArcCot[r]^3 // FullSimplify
True
Simplify[Limit[f[r], r -> Infinity] - Limit[f[r], r -> 0]]
(3/8)*(Pi^2*Log[4] - 7*Zeta[3])
check :
% == -21/8*Zeta[3]+3/4*Pi^2*Log[2] // Simplify
True
--
Valeri, awaiting your objections...
Ray Vickson
> > int((arctan(1/r))^2,r=0..infinity);
>
> Pi ln(2)
>
> Yes, that is what MAPLE says. But why?
Well, Maple can do the indefinite integral F(r) = int(arctan(1/
r)^2,r):
F(r) = 2*I*arctan(1/r)^2/((1+1/r*I)^2/(1+1/(r^2))-1)+2*I*arctan(1/
r)^2-2*arctan(1/r)*ln(1-(1+1/r*I)/(1+1/(r^2))^(1/2))+2*I*polylog(2,
(1+1/r*I)/(1+1/(r^2))^(1/2))-2*arctan(1/r)*ln(1+(1+1/r*I)/(1+1/
(r^2))^(1/2))+2*I*polylog(2,-(1+1/r*I)/(1+1/(r^2))^(1/2)),
so you can evaluate limit(F(r),r=infinity) - limit(F(r),r=0,right) =
Pi*ln(2). Of course, that transfers the question to how Maple gets
F(r).
R.G. Vickson
Robert Israel
> On 22 d=E9c, 06:54, Robert Israel <isr...@math.MyUniversitysInitials.ca>
> wrote:
> > Han de Bruijn <umum...@gmail.com> writes:
> >
> > > > int((arctan(1/r))^2,r=3D0..infinity);
> >
> > > Pi ln(2)
> >
> > > Yes, that is what MAPLE says. But why?
> >
> > >http://hdebruijn.soo.dto.tudelft.nl/jaar2010/nice.pdf
> >
> > OK, now do
> >
> > =A0 int((arctan(1/r)^3,r=3D0..infinity) =3D -21/8*Zeta(3)+3/4*Pi^2*ln(2)
> > --
> > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0
> > =A0isr...@math.MyUniversitysInitial=
> s.ca
> > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel
> > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC,
> > Cana=
> da
>
> Good day,
>
> I just do it the same way as arctan(1/r)^2 :
>
> This function :
>
> f[r_] =3D (I*Pi^3)/8 + (-I + r)*ArcCot[r]^3 -
> 3*ArcCot[r]^2*Log[(2*I)/(I + r)] -
> 3*I*ArcCot[r]*PolyLog[2, (-I + r)/(I + r)] -
> (3/2)*PolyLog[3, (-I + r)/(I + r)];
>
> is an antiderivative of arccot(r)^3 :
>
> f'[r] // FullSimplify
> ArcCot[r]^3
>
> now :
> ArcTan[1/r]^3=3D=3DArcCot[r]^3 // FullSimplify
> True
>
> Simplify[Limit[f[r], r -> Infinity] - Limit[f[r], r -> 0]]
> (3/8)*(Pi^2*Log[4] - 7*Zeta[3])
>
> check :
> % =3D=3D -21/8*Zeta[3]+3/4*Pi^2*Log[2] // Simplify
> True
>
> --
> Valeri, awaiting your objections...
No objections. Now, what about int(arctan(1/r)^n, r=0..infinity)
in general for positive integers n?
Valeri Astanoff
wrote:
>
> - Afficher le texte des messages précédents -
Good day,
Let me try to meet the challenge (even if I'm not
at all specialist of the topic).
Given the form of the computed integral from n=2 to 5,
my guess is that int(arctan(1/r)^n, r=0..infinity)
is of this form :
2^(-n-1)(4 n Pi^(n-1) Log[2] +
Sum[(-1)^k m[n, 2k+1] Pi^(n-2k-1) Zeta[2k+1],
{k, 1, Floor[(n-1)/2]}])
I did it this way (hope you're not allergic to square brackets) :
In[1]:= F[n_] := Integrate[ArcTan[1/r]^n, {r, 0, Infinity}]
In[2]:= G[n_] :=2^(-n-1)(4*n*Pi^(n-1)*Log[2] +
Sum[(-1)^k m[n, 2k+1] Pi^(n-2k-1) Zeta[2k+1],
{k, 1, Floor[(n-1)/2]}])
In[3]:= Do[SolveAlways[F[k]==G[k] /. Zeta[n_] -> z^n, z] /.
Rule[m[n_,k_],m0_] :> (m[n,k]=m0), {k,2,10}]
check n=5 :
In[4]:= F[5]
Out[4]= (5/32)*(Pi^4*Log[4] - 18*Pi^2*Zeta[3] + 93*Zeta[5])
In[5]:= G[5]
Out[5]= (1/64)*(20*Pi^4*Log[2] - 180*Pi^2*Zeta[3] + 930*Zeta[5])
In[6]:= F[5] == G[5] // Simplify
Out[6]= True
check n=6 :
In[7]:= F[6]
Out[7]= (3/32)*(Pi^5*Log[4] - 30*Pi^3*Zeta[3] + 225*Pi*Zeta[5])
In[8]:= G[6]
Out[8]= (1/128)*(24*Pi^5*Log[2] - 360*Pi^3*Zeta[3] + 2700*Pi*Zeta[5])
In[9]:= F[6] == G[6] // Simplify
Out[9]= True
values of unknown integer function 'm' :
In[10]:= ?m
m[3,3]=42
m[4,3]=72
m[5,3]=180 m[5,5]=930
m[6,3]=360 m[6,5]=2700
m[7,3]=630 m[7,5]=9450 m[7,7]=40005
m[8,3]=1008 m[8,5]=25200 m[8,7]=158760
m[9,3]=1512 m[9,5]=56700 m[9,7]=714420 m[9,9]=2897370
m[10,3]=2160 m[10,5]=113400 m[10,7]=2381400 m[10,9]=14458500
...
Identifying these mysterious numbers is beyond my skill !
--
Valeri
Han de Bruijn
<isr...@math.MyUniversitysInitials.ca> wrote:
> Han de Bruijn <umum...@gmail.com> writes:
>
> > > int((arctan(1/r))^2,r=0..infinity);
>
> > Pi ln(2)
>
> > Yes, that is what MAPLE says. But why?
>
> >http://hdebruijn.soo.dto.tudelft.nl/jaar2010/nice.pdf
>
> OK, now do
>
> int((arctan(1/r)^3,r=0..infinity) = -21/8*Zeta(3)+3/4*Pi^2*ln(2)
OK, now let's see what (at least my version of) MAPLE can _not_ do:
> int((Pi/2 - arctan(r))^2,r=0..infinity);
int((1/2*Pi-arctan(r))^2,r = 0 .. infinity) Dduuhh !!
So it seems that MAPLE doesn't know that:
arctan(r) + arctan(1/r) = Pi/2 ?
> evalb(arctan(r) + arctan(1/r) = Pi/2);
false
Han de Bruijn
Axel Vogt
For r = - Pi the identity is false.
However: arctan(r) + arctan(1/r) = Pi/2; simplify(%) assuming 0 < r;
Pi Pi
---- = ----
2 2
Robert Israel
> On 23 d=E9c, 00:12, Robert Israel <isr...@math.MyUniversitysInitials.ca>
> wrote:
> > Valeri Astanoff <astan...@gmail.com> writes:
> > > On 22 d=3DE9c, 06:54, Robert Israel
> > > <isr...@math.MyUniversitysInitials.=
> ca>
> > > wrote:
> > > > Han de Bruijn <umum...@gmail.com> writes:
> >
> > > > > > int((arctan(1/r))^2,r=3D3D0..infinity);
> >
> > > > > Pi ln(2)
> >
> > > > > Yes, that is what MAPLE says. But why?
> >
> > > > >http://hdebruijn.soo.dto.tudelft.nl/jaar2010/nice.pdf
> >
> > > > OK, now do
> >
> > > > =3DA0 int((arctan(1/r)^3,r=3D3D0..infinity) =3D3D
> > > > =-21/8*Zeta(3)+3/4*P=
> i^2*ln(2)
> > > > --
> > > > Robert Israel =3DA0 =3DA0 =3DA0 =3DA0 =3DA0 =3DA0
> > > > =3DA0isr...@math.MyUniversitysInitial=3D
> > > s.ca
> > > > Department of Mathematics =3DA0 =3DA0 =3DA0
> > > > =3DA0http://www.math.ubc.=
> ca/~israel
> > > > University of British Columbia =3DA0 =3DA0 =3DA0 =3DA0 =3DA0
> > > > =3DA0Van=
> couver, BC,
> > > > Cana=3D
> > > da
> >
> > > Good day,
> >
> > > I just do it the same way as arctan(1/r)^2 :
> >
> > > This function :
> >
> > > f[r_] =3D3D (I*Pi^3)/8 + (-I + r)*ArcCot[r]^3 -
> > > =A03*ArcCot[r]^2*Log[(2*I)/(I + r)] -
> > > =A03*I*ArcCot[r]*PolyLog[2, (-I + r)/(I + r)] -
> > > =A0(3/2)*PolyLog[3, (-I + r)/(I + r)];
> >
> > > is an antiderivative of arccot(r)^3 :
> >
> > > f'[r] // FullSimplify
> > > ArcCot[r]^3
> >
> > > now :
> > > ArcTan[1/r]^3=3D3D=3D3DArcCot[r]^3 // FullSimplify
> > > True
> >
> > > Simplify[Limit[f[r], r -> Infinity] - Limit[f[r], r -> 0]]
> > > (3/8)*(Pi^2*Log[4] - 7*Zeta[3])
> >
> > > check :
> > > % =3D3D=3D3D -21/8*Zeta[3]+3/4*Pi^2*Log[2] // Simplify
> > > True
> >
> > > --
> > > Valeri, awaiting your objections...
> >
> > No objections. =A0Now, what about int(arctan(1/r)^n, r=3D0..infinity)
> > in general for positive integers n?
> > --
> > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0
> > =A0isr...@math.MyUniversitysInitial=
> s.ca
> > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel
> > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC,
> > Cana=
> da- Masquer le texte des messages pr=E9c=E9dents -
> >
> > - Afficher le texte des messages pr=E9c=E9dents -
>
> Good day,
>
> Let me try to meet the challenge (even if I'm not
> at all specialist of the topic).
> Given the form of the computed integral from n=3D2 to 5,
> my guess is that int(arctan(1/r)^n, r=3D0..infinity)
> is of this form :
>
> 2^(-n-1)(4 n Pi^(n-1) Log[2] +
> Sum[(-1)^k m[n, 2k+1] Pi^(n-2k-1) Zeta[2k+1],
> {k, 1, Floor[(n-1)/2]}])
>
> I did it this way (hope you're not allergic to square brackets) :
>
> In[1]:=3D F[n_] :=3D Integrate[ArcTan[1/r]^n, {r, 0, Infinity}]
>
> In[2]:=3D G[n_] :=3D2^(-n-1)(4*n*Pi^(n-1)*Log[2] +
> Sum[(-1)^k m[n, 2k+1] Pi^(n-2k-1) Zeta[2k+1],
> {k, 1, Floor[(n-1)/2]}])
>
> In[3]:=3D Do[SolveAlways[F[k]=3D=3DG[k] /. Zeta[n_] -> z^n, z] /.
> Rule[m[n_,k_],m0_] :> (m[n,k]=3Dm0), {k,2,10}]
>
> check n=3D5 :
> In[4]:=3D F[5]
> Out[4]=3D (5/32)*(Pi^4*Log[4] - 18*Pi^2*Zeta[3] + 93*Zeta[5])
>
> In[5]:=3D G[5]
> Out[5]=3D (1/64)*(20*Pi^4*Log[2] - 180*Pi^2*Zeta[3] + 930*Zeta[5])
>
> In[6]:=3D F[5] =3D=3D G[5] // Simplify
> Out[6]=3D True
>
> check n=3D6 :
> In[7]:=3D F[6]
> Out[7]=3D (3/32)*(Pi^5*Log[4] - 30*Pi^3*Zeta[3] + 225*Pi*Zeta[5])
>
> In[8]:=3D G[6]
> Out[8]=3D (1/128)*(24*Pi^5*Log[2] - 360*Pi^3*Zeta[3] + 2700*Pi*Zeta[5])
>
> In[9]:=3D F[6] =3D=3D G[6] // Simplify
> Out[9]=3D True
>
> values of unknown integer function 'm' :
> In[10]:=3D ?m
> m[3,3]=3D42
> m[4,3]=3D72
> m[5,3]=3D180 m[5,5]=3D930
> m[6,3]=3D360 m[6,5]=3D2700
> m[7,3]=3D630 m[7,5]=3D9450 m[7,7]=3D40005
> m[8,3]=3D1008 m[8,5]=3D25200 m[8,7]=3D158760
> m[9,3]=3D1512 m[9,5]=3D56700 m[9,7]=3D714420 m[9,9]=3D2897370
> m[10,3]=3D2160 m[10,5]=3D113400 m[10,7]=3D2381400 m[10,9]=3D14458500
> ...
>
> Identifying these mysterious numbers is beyond my skill !
>
> --
> Valeri
What I get is:
int(arctan(1/r)^n, r=0..infinity) =
n*(Pi/2)^(n-1)*ln(2) - sum((-1)^k/2^(n-1)*Pi^(n-2*k-1)*n!/(n-2*k-1)!
*(2^(-2*k)-1)*Zeta(2*k+1), k=1..floor((n-1)/2))
+ (if n is odd) (-1)^((n-1)/2)/2^(n-1)*n!*Zeta(n)
Valeri Astanoff
wrote:
[...]
> What I get is:
>
> int(arctan(1/r)^n, r=0..infinity) =
> n*(Pi/2)^(n-1)*ln(2) - sum((-1)^k/2^(n-1)*Pi^(n-2*k-1)*n!/(n-2*k-1)!
> *(2^(-2*k)-1)*Zeta(2*k+1), k=1..floor((n-1)/2))
> + (if n is odd) (-1)^((n-1)/2)/2^(n-1)*n!*Zeta(n)
> --
> Robert Israel isr...@math.MyUniversitysInitials.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
>
> - Afficher le texte des messages précédents -
Good day,
Just to check with my own tool :
In[1]:= ri[n_] := n*(Pi/2)^(n - 1)*Log[2] -
Sum[((-1)^k/2^(n - 1))*Pi^(n - 2*k - 1)*(n!/(n - 2*k - 1)!)*
(2^(-2*k) - 1)*Zeta[2*k + 1], {k , 1, Floor[(n - 1)/2]}] +
If[OddQ[n], ((-1)^((n - 1)/2)/2^(n - 1))*n!*Zeta[n], 0]
In[2]:= ri[3]==Integrate[ArcTan[1/r]^3, {r, 0, Infinity}]
Out[2]= (3/4)*Pi^2*Log[2] - (21*Zeta[3])/8 == (3/8)*(Pi^2*Log[4] -
7*Zeta[3])
In[3]:= % // Simplify
Out[3]= True
In[4]:= ri[4]==Integrate[ArcTan[1/r]^4, {r, 0, Infinity}]
Out[4]= (1/2)*Pi^3*Log[2] - (9/4)*Pi*Zeta[3] == (1/4)*(Pi^3*Log[4] -
9*Pi*Zeta[3])
In[5]:= % // Simplify
Out[5]= True
Seems to be just fine.
My conclusion : for someone like me, some numbers
may seem mysterious; for an eminent professor like you,
the mystery doesn't last very long !
With my best regards,
--
Valeri Astanoff
Han de Bruijn
> Han de Bruijn wrote:
> > On Dec 22, 6:54 am, Robert Israel
> > <isr...@math.MyUniversitysInitials.ca> wrote:
> >> Han de Bruijn <umum...@gmail.com> writes:
>
> >>>> int((arctan(1/r))^2,r=0..infinity);
> >>> Pi ln(2)
> >>> Yes, that is what MAPLE says. But why?
> >>>http://hdebruijn.soo.dto.tudelft.nl/jaar2010/nice.pdf
> >> OK, now do
>
> >> int((arctan(1/r)^3,r=0..infinity) = -21/8*Zeta(3)+3/4*Pi^2*ln(2)
>
> > OK, now let's see what (at least my version of) MAPLE can _not_ do:
>
> >> int((Pi/2 - arctan(r))^2,r=0..infinity);
>
> > int((1/2*Pi-arctan(r))^2,r = 0 .. infinity) Dduuhh !!
>
> > So it seems that MAPLE doesn't know that:
>
> > arctan(r) + arctan(1/r) = Pi/2 ?
>
> >> evalb(arctan(r) + arctan(1/r) = Pi/2);
>
> > false
>
You mean for r < 0 ?
> However: arctan(r) + arctan(1/r) = Pi/2; simplify(%) assuming 0 < r;
>
> Pi Pi
> ---- = ----
> 2 2
Ah! Thanks.
> simplify(arctan(r) + arctan(1/r) = Pi/2, assume = positive);
Han de Bruijn