Talk About Network

Google





Education > Math Recreational > Re: An exact 1-...
Latest [ Topics | Posts ] Archive Post A New Topic Post a Reply
<< Topic < Post Post 5 of 5 Topic 2744 of 2947
Post > Topic >>

Re: An exact 1-D integration challenge - 57 - Proud Earthling, raise,

by Vladimir Bondarenko <vb@[EMAIL PROTECTED] > Apr 13, 2008 at 02:17 PM

So... no one can crunch this integral?

On Apr 12, 2:38=A0pm, Vladimir Bondarenko <v...@[EMAIL PROTECTED]
> wrote:
> Please consider, it's trivial to guess the answer in the
> given case. A fro****e can guess it, a math oriented high
> scholer can guess it.
>
> Our challenge is not about guessing but about the concrete
> steps to get the result.
>
> On Apr 12, 1:11=A0pm, Vladimir Bondarenko <v...@[EMAIL PROTECTED]
> wrote:
>
>
>
> > On Apr 12, 1:04 pm, Axel Vogt <&nore...@[EMAIL PROTECTED]
> writes:
>
> > AV> =A0It equals 0.
>
> > Precisely!
>
> > Frankly, I expected (and hoped) that you enter the game
> > this time again.
>
> > Now could you please show the steps to get this result
> > symbolically?
>
> > Cheers from far-off Simferopol,
>
> > Vladimir Bondarenko
>
> > On Apr 12, 1:04=A0pm, Axel Vogt <&nore...@[EMAIL PROTECTED]
> wrote:
>
> > > > int(-sqrt(2)*z*sqrt((sqrt(z^2+2*z+2)-sqrt(z^2+1)+1)/(sqrt(2)*
> > > > sqrt(sqrt(z^2+1)-z)+sqrt(z^2+1)+1))/(z^2+1)^2+sqrt(2)*(z^2-1)
> > > > *sqrt((sqrt(z^2+2*z+2)+sqrt(z^2+1)-1)/(sqrt(2)*sqrt(sqrt(z^2+
> > > > 1)-z)+sqrt(z^2+1)+1))/(2*(z^2+1)^2)+sqrt(2)*z*sqrt((sqrt(z^2-
> > > > 2*z+2)-sqrt(z^2+1)+1)/(sqrt(2)*sqrt(sqrt(z^2+1)+z)+sqrt(z^2+1
> > > > )+1))/(z^2+1)^2+sqrt(2)*(z^2-1)*sqrt((sqrt(z^2-2*z+2)+sqrt(z^
> > > > 2+1)-1)/(sqrt(2)*sqrt(sqrt(z^2+1)+z)+sqrt(z^2+1)+1))/(2*(z^2+
> > > > 1)^2)-(sqrt(2)*(z^2-1)*sqrt(-(sqrt(z^2+2*z+2)-sqrt(z^2+1)+1)/
> > > > (sqrt(2)*sqrt(sqrt(z^2+1)-z)+sqrt(z^2+1)+1))/(2*(z^2+1)^2)+sq
> > > > rt(2)*z*sqrt(-(sqrt(z^2+2*z+2)+sqrt(z^2+1)-1)/(sqrt(2)*sqrt(s
> > > > qrt(z^2+1)-z)+sqrt(z^2+1)+1))/(z^2+1)^2+sqrt(2)*(z^2-1)*sqrt(
> > > > -sqrt(z^2-2*z+2)+sqrt(z^2+1)-1)/(2*(z^2+1)^2*sqrt(sqrt(2)*sqr
> > > > t(sqrt(z^2+1)+z)+sqrt(z^2+1)+1))-sqrt(2)*z*sqrt(-sqrt(z^2-2*z
> > > > +2)-sqrt(z^2+1)+1)/((z^2+1)^2*sqrt(sqrt(2)*sqrt(sqrt(z^2+1)+z
> > > > )+sqrt(z^2+1)+1))), z=3D 0..infinity);
>
> > > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 ?
>
> > > It equals 0.- Hide quoted text -
>
> > > - Show quoted text -- Hide quoted text -
>
> > - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
 




 5 Posts in Topic:
An exact 1-D integration challenge - 57 - Proud Earthling, raise
Vladimir Bondarenko <v  2008-04-12 06:14:59 
Re: An exact 1-D integration challenge - 57 - Proud Earthling, r
Axel Vogt <&norepl  2008-04-12 22:04:03 
Re: An exact 1-D integration challenge - 57 - Proud Earthling, r
Vladimir Bondarenko <v  2008-04-12 13:11:32 
Re: An exact 1-D integration challenge - 57 - Proud Earthling, r
Vladimir Bondarenko <v  2008-04-12 14:38:30 
Re: An exact 1-D integration challenge - 57 - Proud Earthling, r
Vladimir Bondarenko <v  2008-04-13 14:17:14 

Post A Reply:
  Go here to Signup

AddThis Feed Button


About - Advertising - Contact - Frequently Asked Questions - Privacy Policy - Terms of Use - Signup

Contact
localhost-V2008-12-19 Thu Jan 8 11:55:02 PST 2009.