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From: Peter Luschny on 28 Jul 2010 17:19
> Vladimir Bondarenko wrote:

Hi Vladimir!

It is always a pleasure to see a master at work!

> Both of you have a subtle sense of humor. I was almost laughing
> looking at how masterly you pretended you do not know what is
> the K, bravo!

> evalf(ln(EllipticK(1/sqrt(2)))/ln(exp(1/Pi)), 50);

> 1.9395745220275329586655005909882008730942339448308

Great! But this is not the end of the story.
We can get rid of the EllipticK.

The way I look at things involve the LemniscateConstant.
http://mathworld.wolfram.com/LemniscateConstant.html

L := GAMMA(1/4)^2/sqrt(8*Pi) = 2.622057556...

So if we replace the constat K in Plouffe's formula by L we get

ln(L(1/sqrt(2)))/ln(exp(1/Pi))

Nice, isn't it? ;-) Summing up:

/
| Integral(ln(Pi^2+log(x)^2)/(1+x^2), x = 0..1)
| = Pi*ln(1/2*Pi^(3/2)/(GAMMA(3/4)^2))
\

evalf(%,50); 1.9395745220275329586655005909882008730942339448307

Cheers Peter
From: Vladimir Bondarenko on 28 Jul 2010 20:30
On Jul 29, 12:19 am, Peter Luschny <peter.lusc...(a)googlemail.com>
wrote:
> > Vladimir Bondarenko wrote:
>
> Hi Vladimir!
>
> It is always a pleasure to see a master at work!
>
> > Both of you have a subtle sense of humor. I was almost laughing
> > looking at how masterly you pretended you do not know what is
> > the K, bravo!
> > evalf(ln(EllipticK(1/sqrt(2)))/ln(exp(1/Pi)), 50);
> > 1.9395745220275329586655005909882008730942339448308
>
> Great! But this is not the end of the story.
> We can get rid of the EllipticK.
>
> The way I look at things involve the LemniscateConstant.http://mathworld.wolfram.com/LemniscateConstant.html
>
>    L := GAMMA(1/4)^2/sqrt(8*Pi) = 2.622057556...
>
> So if we replace the constat K in Plouffe's formula by L we get
>
>    ln(L(1/sqrt(2)))/ln(exp(1/Pi))
>
> Nice, isn't it? ;-) Summing up:
>
>  /
> |  Integral(ln(Pi^2+log(x)^2)/(1+x^2), x = 0..1)
> |        = Pi*ln(1/2*Pi^(3/2)/(GAMMA(3/4)^2))
>  \
>
> evalf(%,50); 1.9395745220275329586655005909882008730942339448307
>
> Cheers Peter

From which it is obvious that

sum((-1)^n*Ci((2*n+1)*Pi)/(2*n+1), n= 0..infinity)

=

Pi/2*ln(Pi^(3/2)/(2*GAMMA(3/4)^2)) - ln(Pi^Pi)/4

heh, quite a nice sum :)

So, you invented a beautiful integral...

I wonder how on planet you guessed it?

Cheers Vladimir
From: Passerby on 28 Jul 2010 23:40
On Wed, 28 Jul 2010 17:30:16 -0700 (PDT), Vladimir Bondarenko
<vb(a)cybertester.com> wrote:

>On Jul 29, 12:19 am, Peter Luschny <peter.lusc...(a)googlemail.com>
>wrote:
>> > Vladimir Bondarenko wrote:
>>
>> Hi Vladimir!
>>
>> It is always a pleasure to see a master at work!
>>
>> > Both of you have a subtle sense of humor. I was almost laughing
>> > looking at how masterly you pretended you do not know what is
>> > the K, bravo!
>> > evalf(ln(EllipticK(1/sqrt(2)))/ln(exp(1/Pi)), 50);
>> > 1.9395745220275329586655005909882008730942339448308
>>
>> Great! But this is not the end of the story.
>> We can get rid of the EllipticK.

[snip]
>>  /
>> |  Integral(ln(Pi^2+log(x)^2)/(1+x^2), x = 0..1)
>> |        = Pi*ln(1/2*Pi^(3/2)/(GAMMA(3/4)^2))
>>  \
>>
>> evalf(%,50); 1.9395745220275329586655005909882008730942339448307
>>
>> Cheers Peter
>
>From which it is obvious that
>
>sum((-1)^n*Ci((2*n+1)*Pi)/(2*n+1), n= 0..infinity)
>
>=
>
>Pi/2*ln(Pi^(3/2)/(2*GAMMA(3/4)^2)) - ln(Pi^Pi)/4
>
>heh, quite a nice sum :)
>
>So, you invented a beautiful integral...
>
>I wonder how on planet you guessed it?
>
>Cheers Vladimir

Speaking only for myself, knowing that K( 1/sqrt(2) ) is "special" is
just one of those things I remember. 8-)

There are a number of articles at MathWorld that provide relevant
information, in particular:

o Elliptic Integral Singular Value ... eqn. (3)
<http://mathworld.wolfram.com/EllipticIntegralSingularValue.html>

o Elliptic Lambda Function ... eqn. (11)
<http://mathworld.wolfram.com/EllipticLambdaFunction.html>

o Gamma Function ... eqn. (64) and the preceeding paragraph
<http://mathworld.wolfram.com/GammaFunction.html>

From: Peter Luschny on 29 Jul 2010 04:15
> Vladimir Bondarenko wrote:
> > Peter Luschny wrote:

> > /
> > | Integral(ln(Pi^2+log(x)^2)/(1+x^2), x = 0..1)
> > | = Pi*ln(1/2*Pi^(3/2)/(GAMMA(3/4)^2))
> > \

> From which it is obvious that
>
> sum((-1)^n*Ci((2*n+1)*Pi)/(2*n+1), n= 0..infinity)
> =
> Pi/2*ln(Pi^(3/2)/(2*GAMMA(3/4)^2)) - ln(Pi^Pi)/4
>
> heh, quite a nice sum :)

I like the way you put it. Especially the Pi^Pi pleases me.

So we have numerically .7071851108147692981e-1
Back to Plouffe's Inverter. It says:

> 7071851023016875 = (p241)
> Your value of 7071851108147692 would be here.
> 7071851108577178 = (m414) Riemannzero2*LemniGauss^2/exp(TravelSale)

This is interesting because LemniGauss appears in the adjacent
formula.
But does it not also confirm Axel's complaint? What is TravelSale?

> So, you invented a beautiful integral...
> I wonder how on planet you guessed it?

It was divine inspiration after five days of abstinence ;)
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