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From: Vladimir Bondarenko on 1 Aug 2010 01:10
Hello,

int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2);

?
Cheers,

Vladimir Bondarenko

Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester Ltd.

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"...... the challenges imply that a solution is built within the
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From: Axel Vogt on 1 Aug 2010 05:42
Vladimir Bondarenko wrote:
> Hello,
>
> int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2);
>
> ?

z=arcsin(t) gives an algebraic integral, which Maple
solves in lengthy terms of elliptic functions.
From: clicliclic on 1 Aug 2010 09:25

Axel Vogt schrieb:
>
> Vladimir Bondarenko wrote:
> >
> > int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2);
> >
> > ?
>
> z=arcsin(t) gives an algebraic integral, which Maple
> solves in lengthy terms of elliptic functions.

Does this

1/2*INT(SQRT(t^(3/2)-1)/(t^(5/4)*(t-1)),t,1,inf)

or this

INT(SQRT(u^3-1)/(u^(3/2)*(u^2-1)),u,1,inf)

lead to a simpler result, perhaps via Meijer-G convolution?

Martin.
From: Axel Vogt on 1 Aug 2010 11:16
clicliclic(a)freenet.de wrote:
> Axel Vogt schrieb:
>> Vladimir Bondarenko wrote:
>>> int(sqrt(1 - sin(z)^3)/cos(z), z= 0..Pi/2);
>>>
>>> ?
>> z=arcsin(t) gives an algebraic integral, which Maple
>> solves in lengthy terms of elliptic functions.
>
> Does this
>
> 1/2*INT(SQRT(t^(3/2)-1)/(t^(5/4)*(t-1)),t,1,inf)
>
> or this
>
> INT(SQRT(u^3-1)/(u^(3/2)*(u^2-1)),u,1,inf)
>
> lead to a simpler result, perhaps via Meijer-G convolution?
>
> Martin.

Directly that does not help, I have not tried with MeijerG.

t=1-x^2 makes it better over 0 ... 1: no singularity and it
becomes symmetric w.r.t. 0 over -1 ... +1 (but still would
result in elliptic functions).
From: clicliclic on 2 Aug 2010 16:25

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.

Martin.
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