Interesting integral...

[cosine]

Could anyone help me find this integral? I've been going back to it occasionally on and off for a year and a half. I once had access to mathematica, which gave me a strange string for the integral. At the time I couldn't understand it, but I may be able to understand it now. Here is the integral I'm trying to find:

 

[math]\int^{1}_{-1}\left[(-x)^{k}(1-x^{k})^{1-k} + 1\right]^{\frac{1}{k}}dx[/math]

 

If anyone could help I would be very much obliged, it is an interesting integral.

[TD]

I don't think it exists when x = 0, since you then get 1^(1/0).

Mathematica tells me it doesn't converge on [-1,1]...

[cosine]

I don't think it exists when x = 0' date=' since you then get 1^(1/0).

Mathematica tells me it doesn't converge on [-1,1']...

 

Do you mean when k=0?

[TD]

Eh, I misread: it says a k there, my bad

 

Mathematica still says it doesn't converge though.

[NeonBlack]

Assuming k is an integer > 0, here are some midpoint approximations:

k=1; 2.0

k=2; 3.129495

k=3;

k=4; 3.380247

k=5;

k=6; 3.292757

k=7;

k=8; 3.145137

k=9;

 

So it looks like it's undefined for odd k > 1. Who knows how accurate these are. Maybe someone will give the FTo'C a shot. I'm not even sure if that's possible though.

 

EDIT: as not to confuse anyone, I meant to say that the function itself is defined for odd k, but it has no area on [-1,1]

[cosine]

Assuming k is an integer > 0' date=' here are some midpoint approximations:

k=1; 2.0

k=2; 3.129495

k=3;

k=4; 3.380247

k=5;

k=6; 3.292757

k=7;

k=8; 3.145137

k=9;

 

So it looks like it's undefined for odd k > 1. Who knows how accurate these are. Maybe someone will give the FTo'C a shot. I'm not even sure if that's possible though.

 

EDIT: as not to confuse anyone, I meant to say that the function itself is defined for odd k, but it has no area on [-1,1']

 

Cool thanks, those are very interesting values. How did you get at them exactly? What does FT o'C mean? and I can tell you that for k = 2 the answer is pi.

[NeonBlack]

I got these using the midpoint rule for approximations. FTo'C is the Fundamental Theorem o' Calculus.

 

edit: the values I gave you were computed using n=10 000.

I just tried it with n= 100 million k=2 and came up with 3.141461, it took my computer like 4 full minutes to compute. So we are getting closer to pi.

[cosine]

I got these using the midpoint rule for approximations. FTo'C is the Fundamental Theorem o' Calculus.

 

edit: the values I gave you were computed using n=10 000.

I just tried it with n= 100 million k=2 and came up with 3.141461' date=' it took my computer like 4 full minutes to compute. So we are getting closer to pi.[/quote']

 

Awesome. Wow thats alot of iterations though. <s>What are you taking the midpoint of though? Perhaps I just haven't heard of it referred to by that term.</s>

 

Edit: LOL wow I just realized what you meant. Man such a log time since Reimann sums. Thanks though!

[cosine]

By the way, I feel like you guys may like to know just what it is this integral means. For various distance formulas of the form [math]x^{k} + y^{k} = z^{k}[/math] what is the value of pi? Thus in the unit circle of that k is:

[math]y = \left(1 - x^{k}\right)^{\frac{1}{k}}[/math]

So the pi is going to be the arclength of the unit circle from -1 to 1.

Though since your distance formula changed, the arclength formula changed.

[math]\left(dx^{2} + dy^{2}\right)^{\frac{1}{2}}[/math]

becomes:

[math]\left(dx^{k} + dy^{k}\right)^{\frac{1}{k}}[/math]

Which leads to the general integral of this discussion. It should give the corresponding pi for every k.