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From: Axel Vogt on 3 Aug 2010 14:54
Vladimir Bondarenko wrote:
> Hello,
>
> int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2);
>
> ?

Here is a variant (n=3 is 'the' question):

Set f:= (z,n) -> sqrt(1-tanh(z)^n).
What is Int( f(z,n), z=0..infinity) ?

Only n=1,2 and 4 are answered by Maple directly.
From: Axel Vogt on 3 Aug 2010 16:18
clicliclic(a)freenet.de wrote:
> clicliclic(a)freenet.de schrieb:
>> Vladimir Bondarenko schrieb:
>>> int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2);
>>>
>>> ?
>> Numerically, this integral evaluates to:
>>
>> 1.7682569591454495299
>>
>> I have transformed it into a sum of three 3F2(1)'s:
>>
>> SQRT(pi)/6*(SUM((2*k-2/3)!/(2*k+5/6)!,k,0,inf)
>> +SUM((2*k)!/(2*k+3/2)!,k,0,inf)
>> +SUM((2*k+2/3)!/(2*k+13/6)!,k,0,inf))
>>
>> - hopefully correctly.
>>
>
> This result can be improved upon as follows. We split the integral as:
>
> 1/2*INT(SQRT(1-t^3)/(1-t),t,0,1)+1/2*INT(SQRT(1-t^3)/(1+t),t,0,1)
>
> Since the Meijer-G functions need only be transformed to G(t^3) now,
> their order will be lower. The first term gives three 2F1(1)'s, which
> can of course be evaluated in closed form:
>
> SQRT(pi)*(1/3)!/(4*(5/6)!)*F21(1,1/3,11/6,1)+3*SQRT(pi)*(2/3)!/(~
> 28*(1/6)!)*F21(1,2/3,13/6,1)+1/9*F21(1,1,5/2,1)
>
> 9*(2/3)!*(5/6)!/(10*SQRT(pi))+3*(1/3)!*(1/6)!/(2*SQRT(pi))+1/3
>
> 1.4656093125343588864
>
> The second term can be transformed into three 2F1(-1)'s, one of which is
> clearly elementary:
>
> SQRT(pi)*(1/3)!/(4*(5/6)!)*F21(1,1/3,11/6,-1)-3*SQRT(pi)*(2/3)!/~
> (28*(1/6)!)*F21(1,2/3,13/6,-1)+1/9*F21(1,1,5/2,-1)
>
> SQRT(pi)/12*SUM((k-2/3)!*COS(pi*k)/(k+5/6)!,k,0,inf)-SQRT(pi)/12~
> *SUM((k-1/3)!*COS(pi*k)/(k+7/6)!,k,0,inf)+SQRT(pi)/12*SUM(k!*COS~
> (pi*k)/(k+3/2)!,k,0,inf)
>
> SQRT(pi)/12*2.484779611412182053-SQRT(pi)/12*0.99195099031185643~
> 773+(SQRT(2)*LN(SQRT(2)+1)-1)/3
>
> 0.30264764661109064355
>
> The numerical values for the other two sums are derived from integral
> representations.
>
> Martin.

Overall I get it as

1/3*2^(1/2)*ln(1+2^(1/2))+ 2^(1/3)*
((1/4-3/28*hypergeom([2/3, 1],[13/6],-1))/Pi*3^(1/2)*GAMMA(2/3)^3+
(1/9+1/15*hypergeom([1/3, 1],[11/6],-1))*Pi^2/GAMMA(2/3)^3*2^(1/3) )

which may be yours (re-typing from Derive is a bit awful for 2F1 ...).
From: clicliclic on 4 Aug 2010 10:58

Axel Vogt schrieb:
> clicliclic(a)freenet.de wrote:
> > clicliclic(a)freenet.de schrieb:
> >> Vladimir Bondarenko schrieb:
> >>>
> >>> int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2);
> >>>
> >>> ?
> >>
> >> Numerically, this integral evaluates to:
> >>
> >> 1.7682569591454495299
> >
> > [...] We split the integral as:
> >
> > 1/2*INT(SQRT(1-t^3)/(1-t),t,0,1)+1/2*INT(SQRT(1-t^3)/(1+t),t,0,1)
> >
> > [...] The first term gives three 2F1(1)'s, which can of course be
> > evaluated in closed form:
> >
> > SQRT(pi)*(1/3)!/(4*(5/6)!)*F21(1,1/3,11/6,1)+3*SQRT(pi)*(2/3)!/(~
> > 28*(1/6)!)*F21(1,2/3,13/6,1)+1/9*F21(1,1,5/2,1)
> >
> > 9*(2/3)!*(5/6)!/(10*SQRT(pi))+3*(1/3)!*(1/6)!/(2*SQRT(pi))+1/3
> >
> > 1.4656093125343588864
> >
> > The second term can be transformed into three 2F1(-1)'s, one of
> > which is clearly elementary:
> >
> > SQRT(pi)*(1/3)!/(4*(5/6)!)*F21(1,1/3,11/6,-1)-3*SQRT(pi)*(2/3)!/~
> > (28*(1/6)!)*F21(1,2/3,13/6,-1)+1/9*F21(1,1,5/2,-1)
> >
> > SQRT(pi)/12*2.484779611412182053-SQRT(pi)/12*0.99195099031185643~
> > 773+(SQRT(2)*LN(SQRT(2)+1)-1)/3
> >
> > 0.30264764661109064355
>
> Overall I get it as
>
> 1/3*2^(1/2)*ln(1+2^(1/2))+ 2^(1/3)*
> ((1/4-3/28*hypergeom([2/3, 1],[13/6],-1))/Pi*3^(1/2)*GAMMA(2/3)^3+
> (1/9+1/15*hypergeom([1/3, 1],[11/6],-1))*Pi^2/GAMMA(2/3)^3*2^(1/3) )
>
> which may be yours (re-typing from Derive is a bit awful for 2F1 ...).

Yours appears to be the same. One might think that one would only have
to look up F21(1,1/3,11/6,-1) and F21(1,2/3,13/6,-1) at

<http://www.planetquantum.com/HyperF/FTable2F1NU6/index.html>

or some other table of special hypergeometrics. Alas, I had no luck
there - as far as I can see, these two 2F1(-1)'s are very hard nuts
indeed!

So, if Vladimir expects an *elementary* solution, it's time for him to
tell us now.

Martin.
From: Axel Vogt on 4 Aug 2010 15:15
clicliclic(a)freenet.de wrote:
> Axel Vogt schrieb:
....
> Yours appears to be the same. One might think that one would only have
> to look up F21(1,1/3,11/6,-1) and F21(1,2/3,13/6,-1) at
>
> <http://www.planetquantum.com/HyperF/FTable2F1NU6/index.html>
>
> or some other table of special hypergeometrics. Alas, I had no luck
> there - as far as I can see, these two 2F1(-1)'s are very hard nuts
> indeed!
>
> So, if Vladimir expects an *elementary* solution, it's time for him to
> tell us now.
>
> Martin.

I have some variants (but nothing essential). Thx for reminding Kelly Roach.
Yes, Vladimir Bondarenko should post his solution.
From: Vladimir Bondarenko on 4 Aug 2010 15:29
On Aug 4, 10:15 pm, Axel Vogt <&nore...(a)axelvogt.de> wrote:
> cliclic...(a)freenet.de wrote:
> > Axel Vogt schrieb:
> ...
> > Yours appears to be the same. One might think that one would only have
> > to look up F21(1,1/3,11/6,-1) and F21(1,2/3,13/6,-1) at
>
> >   <http://www.planetquantum.com/HyperF/FTable2F1NU6/index.html>
>
> > or some other table of special hypergeometrics. Alas, I had no luck
> > there - as far as I can see, these two 2F1(-1)'s are very hard nuts
> > indeed!
>
> > So, if Vladimir expects an *elementary* solution, it's time for him to
> > tell us now.
>
> > Martin.
>
> I have some variants (but nothing essential). Thx for reminding Kelly Roach.
> Yes, Vladimir Bondarenko should post his solution.

Hello,

It boils down to MeijerG... Alas, no elementary solution
is known to me; most probably as pointed out Oleksandr
Pavlyk, it simply does not exist.

Yet another point. Usually, but not always, the examples
are selected in a way to show this or that bug in a CAS.

In this case, with Mathematica 7

f = Sqrt[z^(3/2) - 1]/(z^(5/4) (z - 1));

N[Integrate[f, {z, 1, Infinity}]]
NIntegrate[f, {z, 1, Infinity}]

-1.3668 - 0.292523 I
3.53651

Cheers,

Vladimir Bondarenko
Cyber Tester
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