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Calculate these integrals

2

$begingroup$

I want to know how to calculate any of these integrals, which arise from computing the perimeter of the unit ball in the $p$-norm.

1. $$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt;$$


2. $$int_0^{pi/2}(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2)^{1/p} dt.$$

In the first one, I tried to expand $(1+p^pt^{p(p-1)})^{1/p}$ as
$$sum_{k =0}^infty binom{1/p}{k}(pt^{p(p-1)})^k,$$
where $binom{1/p}{k} = frac{(1/p)cdots((1/p)-k+1)}{k!}$.

Since
$$int_0^1 t^{p(p-1)k} dt = frac{t^{1+p(p-1)k}}{1+p(p-1)k} = frac{1}{1+p(p-1)k},$$

it follows that
$$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt = sum_{0}^infty binom{1/p}{k}frac{p^{pk}}{1+p(p-1)k}.$$

I'm not sure if this converges and how to calculated it. I don't know how to proceed in the second one.

integration definite-integrals

share|cite|improve this question

edited Mar 22 at 22:37

Eric Wofsey

193k14221352

asked Mar 22 at 21:17

Pedro G. MattosPedro G. Mattos

307

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't mean to be rude, but is it really necessary? I just want to calculate these and move on, but I can't find it anywhere, so I figured someone here could give me a hand.
  $endgroup$
  – Pedro G. Mattos
  Mar 22 at 21:22
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  On MSE, we do not answer problem-statement-questions. We are not here to do your homework for you. So, yes it is really necessary if you want an answer. Give the integrals a shot, show us what you did, then you'll get an answer.
  $endgroup$
  – clathratus
  Mar 22 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I want to calculate the perimeter of a the unit ball in the p-norm, that's it. These integrals come from two different ways I found of parameterizing the curves.
  $endgroup$
  – Pedro G. Mattos
  Mar 22 at 21:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  These integrals are already simplified. I got some constants out etc. I thought about expanding the first in series and integrating after that, but I'm not sure if it converges.
  $endgroup$
  – Pedro G. Mattos
  Mar 22 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the second one, I tried to factorize it as $cos^2sin^2(cos^{-p}+sin^{-p})$, but I'm not sure if this is going to help.
  $endgroup$
  – Pedro G. Mattos
  Mar 22 at 21:32
  

  
  
  
  

 | 
show 5 more comments

2

$begingroup$

I want to know how to calculate any of these integrals, which arise from computing the perimeter of the unit ball in the $p$-norm.

1. $$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt;$$


2. $$int_0^{pi/2}(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2)^{1/p} dt.$$

In the first one, I tried to expand $(1+p^pt^{p(p-1)})^{1/p}$ as
$$sum_{k =0}^infty binom{1/p}{k}(pt^{p(p-1)})^k,$$
where $binom{1/p}{k} = frac{(1/p)cdots((1/p)-k+1)}{k!}$.

Since
$$int_0^1 t^{p(p-1)k} dt = frac{t^{1+p(p-1)k}}{1+p(p-1)k} = frac{1}{1+p(p-1)k},$$

it follows that
$$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt = sum_{0}^infty binom{1/p}{k}frac{p^{pk}}{1+p(p-1)k}.$$

I'm not sure if this converges and how to calculated it. I don't know how to proceed in the second one.

integration definite-integrals

share|cite|improve this question

edited Mar 22 at 22:37

Eric Wofsey

193k14221352

asked Mar 22 at 21:17

Pedro G. MattosPedro G. Mattos

307

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't mean to be rude, but is it really necessary? I just want to calculate these and move on, but I can't find it anywhere, so I figured someone here could give me a hand.
  $endgroup$
  – Pedro G. Mattos
  Mar 22 at 21:22
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  On MSE, we do not answer problem-statement-questions. We are not here to do your homework for you. So, yes it is really necessary if you want an answer. Give the integrals a shot, show us what you did, then you'll get an answer.
  $endgroup$
  – clathratus
  Mar 22 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I want to calculate the perimeter of a the unit ball in the p-norm, that's it. These integrals come from two different ways I found of parameterizing the curves.
  $endgroup$
  – Pedro G. Mattos
  Mar 22 at 21:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  These integrals are already simplified. I got some constants out etc. I thought about expanding the first in series and integrating after that, but I'm not sure if it converges.
  $endgroup$
  – Pedro G. Mattos
  Mar 22 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the second one, I tried to factorize it as $cos^2sin^2(cos^{-p}+sin^{-p})$, but I'm not sure if this is going to help.
  $endgroup$
  – Pedro G. Mattos
  Mar 22 at 21:32
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

I want to know how to calculate any of these integrals, which arise from computing the perimeter of the unit ball in the $p$-norm.

1. $$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt;$$


2. $$int_0^{pi/2}(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2)^{1/p} dt.$$

In the first one, I tried to expand $(1+p^pt^{p(p-1)})^{1/p}$ as
$$sum_{k =0}^infty binom{1/p}{k}(pt^{p(p-1)})^k,$$
where $binom{1/p}{k} = frac{(1/p)cdots((1/p)-k+1)}{k!}$.

Since
$$int_0^1 t^{p(p-1)k} dt = frac{t^{1+p(p-1)k}}{1+p(p-1)k} = frac{1}{1+p(p-1)k},$$

it follows that
$$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt = sum_{0}^infty binom{1/p}{k}frac{p^{pk}}{1+p(p-1)k}.$$

I'm not sure if this converges and how to calculated it. I don't know how to proceed in the second one.

integration definite-integrals

share|cite|improve this question

edited Mar 22 at 22:37

Eric Wofsey

193k14221352

asked Mar 22 at 21:17

Pedro G. MattosPedro G. Mattos

307

$endgroup$

share|cite|improve this question

edited Mar 22 at 22:37

Eric Wofsey

193k14221352

asked Mar 22 at 21:17

Pedro G. MattosPedro G. Mattos

307

share|cite|improve this question

edited Mar 22 at 22:37

Eric Wofsey

193k14221352

asked Mar 22 at 21:17

Pedro G. MattosPedro G. Mattos

307

edited Mar 22 at 22:37

Eric Wofsey

193k14221352

edited Mar 22 at 22:37

Eric Wofsey

193k14221352

asked Mar 22 at 21:17

Pedro G. MattosPedro G. Mattos

307

asked Mar 22 at 21:17

Pedro G. MattosPedro G. Mattos

307

1 Answer
1

active

oldest

votes

1

$begingroup$

Concerning the first question (excluding the trivial cases $p=0$ and $p=1$)
$$I_p=intleft(1+p^p t^{(p-1) p}right)^{frac{1}{p}},dt=t,, _2F_1left(-frac{1}{p},frac{1}{(p-1) p};1+frac{1}{(p-1) p};-p^p t^{(p-1) p}right)$$ where appears the Gaussian or ordinary hypergeometric function (have a look here). So,
$$J_p=int_0^1left(1+p^p t^{(p-1) p}right)^{frac{1}{p}},dt=, _2F_1left(-frac{1}{p},frac{1}{(p-1) p};frac{p^2-p+1}{(p-1) p};-p^pright)$$

What is interesting is that $J_p$ goes through a minimum value as shown in the table below
$$left(
begin{array}{cc}
1 & 2.00000 \
2 & 1.47894 \
3 & 1.45510 \
4 & 1.49983 \
5 & 1.54817 \
6 & 1.58972 \
7 & 1.62429 \
8 & 1.65314 \
9 & 1.67752 \
10 & 1.69837
end{array}
right)$$

Remarkable is $J_2=frac{10+sqrt{5} sinh ^{-1}(2)}{4 sqrt{5}}$.

For the second integral
$$K_p=int_0^{pi/2}left(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2right)^{frac 1p}, dt$$ I have not been able to get anything analytical beside $K_1=frac 23$ and $K_2=frac pi 2$. Numerical integration shows that $K_p sim p-frac 12$

share|cite|improve this answer

edited Mar 27 at 5:18

answered Mar 23 at 5:42

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot. That hypergeometric function is really interesting. It's such a coincidence that I was reading about it just a week ago.
  $endgroup$
  – Pedro G. Mattos
  Mar 24 at 5:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is also curious that it is undefined for $p=1$, since the integral is clearly equal to $2$.
  $endgroup$
  – Pedro G. Mattos
  Mar 26 at 22:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PedroG.Mattos. I wanted to mean that the formula cannot be used for such a case. Sorry for the lack of precision of my wording. Cheers.
  $endgroup$
  – Claude Leibovici
  Mar 27 at 5:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's alright, I get it. Which formula did you use to calculate the integral? I checked at the wiki and found only the Euler integral formula. I couldn't see how you applied it, because I think there has to be a change of variables. I don't see how you did it. I changed variables, taking $u=t^{p(p-1)}$, and what I ended up with is not the same as you.
  $endgroup$
  – Pedro G. Mattos
  Mar 27 at 5:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK, I messed up in the process, now I see it.
  $endgroup$
  – Pedro G. Mattos
  Mar 27 at 5:31
  

  
  
  
  

 | 
show 2 more comments

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1

$begingroup$

Concerning the first question (excluding the trivial cases $p=0$ and $p=1$)
$$I_p=intleft(1+p^p t^{(p-1) p}right)^{frac{1}{p}},dt=t,, _2F_1left(-frac{1}{p},frac{1}{(p-1) p};1+frac{1}{(p-1) p};-p^p t^{(p-1) p}right)$$ where appears the Gaussian or ordinary hypergeometric function (have a look here). So,
$$J_p=int_0^1left(1+p^p t^{(p-1) p}right)^{frac{1}{p}},dt=, _2F_1left(-frac{1}{p},frac{1}{(p-1) p};frac{p^2-p+1}{(p-1) p};-p^pright)$$

What is interesting is that $J_p$ goes through a minimum value as shown in the table below
$$left(
begin{array}{cc}
1 & 2.00000 \
2 & 1.47894 \
3 & 1.45510 \
4 & 1.49983 \
5 & 1.54817 \
6 & 1.58972 \
7 & 1.62429 \
8 & 1.65314 \
9 & 1.67752 \
10 & 1.69837
end{array}
right)$$

Remarkable is $J_2=frac{10+sqrt{5} sinh ^{-1}(2)}{4 sqrt{5}}$.

For the second integral
$$K_p=int_0^{pi/2}left(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2right)^{frac 1p}, dt$$ I have not been able to get anything analytical beside $K_1=frac 23$ and $K_2=frac pi 2$. Numerical integration shows that $K_p sim p-frac 12$

share|cite|improve this answer

edited Mar 27 at 5:18

answered Mar 23 at 5:42

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot. That hypergeometric function is really interesting. It's such a coincidence that I was reading about it just a week ago.
  $endgroup$
  – Pedro G. Mattos
  Mar 24 at 5:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is also curious that it is undefined for $p=1$, since the integral is clearly equal to $2$.
  $endgroup$
  – Pedro G. Mattos
  Mar 26 at 22:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PedroG.Mattos. I wanted to mean that the formula cannot be used for such a case. Sorry for the lack of precision of my wording. Cheers.
  $endgroup$
  – Claude Leibovici
  Mar 27 at 5:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's alright, I get it. Which formula did you use to calculate the integral? I checked at the wiki and found only the Euler integral formula. I couldn't see how you applied it, because I think there has to be a change of variables. I don't see how you did it. I changed variables, taking $u=t^{p(p-1)}$, and what I ended up with is not the same as you.
  $endgroup$
  – Pedro G. Mattos
  Mar 27 at 5:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK, I messed up in the process, now I see it.
  $endgroup$
  – Pedro G. Mattos
  Mar 27 at 5:31
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

Concerning the first question (excluding the trivial cases $p=0$ and $p=1$)
$$I_p=intleft(1+p^p t^{(p-1) p}right)^{frac{1}{p}},dt=t,, _2F_1left(-frac{1}{p},frac{1}{(p-1) p};1+frac{1}{(p-1) p};-p^p t^{(p-1) p}right)$$ where appears the Gaussian or ordinary hypergeometric function (have a look here). So,
$$J_p=int_0^1left(1+p^p t^{(p-1) p}right)^{frac{1}{p}},dt=, _2F_1left(-frac{1}{p},frac{1}{(p-1) p};frac{p^2-p+1}{(p-1) p};-p^pright)$$

What is interesting is that $J_p$ goes through a minimum value as shown in the table below
$$left(
begin{array}{cc}
1 & 2.00000 \
2 & 1.47894 \
3 & 1.45510 \
4 & 1.49983 \
5 & 1.54817 \
6 & 1.58972 \
7 & 1.62429 \
8 & 1.65314 \
9 & 1.67752 \
10 & 1.69837
end{array}
right)$$

Remarkable is $J_2=frac{10+sqrt{5} sinh ^{-1}(2)}{4 sqrt{5}}$.

For the second integral
$$K_p=int_0^{pi/2}left(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2right)^{frac 1p}, dt$$ I have not been able to get anything analytical beside $K_1=frac 23$ and $K_2=frac pi 2$. Numerical integration shows that $K_p sim p-frac 12$

share|cite|improve this answer

edited Mar 27 at 5:18

answered Mar 23 at 5:42

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot. That hypergeometric function is really interesting. It's such a coincidence that I was reading about it just a week ago.
  $endgroup$
  – Pedro G. Mattos
  Mar 24 at 5:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is also curious that it is undefined for $p=1$, since the integral is clearly equal to $2$.
  $endgroup$
  – Pedro G. Mattos
  Mar 26 at 22:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PedroG.Mattos. I wanted to mean that the formula cannot be used for such a case. Sorry for the lack of precision of my wording. Cheers.
  $endgroup$
  – Claude Leibovici
  Mar 27 at 5:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's alright, I get it. Which formula did you use to calculate the integral? I checked at the wiki and found only the Euler integral formula. I couldn't see how you applied it, because I think there has to be a change of variables. I don't see how you did it. I changed variables, taking $u=t^{p(p-1)}$, and what I ended up with is not the same as you.
  $endgroup$
  – Pedro G. Mattos
  Mar 27 at 5:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK, I messed up in the process, now I see it.
  $endgroup$
  – Pedro G. Mattos
  Mar 27 at 5:31
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Concerning the first question (excluding the trivial cases $p=0$ and $p=1$)
$$I_p=intleft(1+p^p t^{(p-1) p}right)^{frac{1}{p}},dt=t,, _2F_1left(-frac{1}{p},frac{1}{(p-1) p};1+frac{1}{(p-1) p};-p^p t^{(p-1) p}right)$$ where appears the Gaussian or ordinary hypergeometric function (have a look here). So,
$$J_p=int_0^1left(1+p^p t^{(p-1) p}right)^{frac{1}{p}},dt=, _2F_1left(-frac{1}{p},frac{1}{(p-1) p};frac{p^2-p+1}{(p-1) p};-p^pright)$$

What is interesting is that $J_p$ goes through a minimum value as shown in the table below
$$left(
begin{array}{cc}
1 & 2.00000 \
2 & 1.47894 \
3 & 1.45510 \
4 & 1.49983 \
5 & 1.54817 \
6 & 1.58972 \
7 & 1.62429 \
8 & 1.65314 \
9 & 1.67752 \
10 & 1.69837
end{array}
right)$$

Remarkable is $J_2=frac{10+sqrt{5} sinh ^{-1}(2)}{4 sqrt{5}}$.

For the second integral
$$K_p=int_0^{pi/2}left(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2right)^{frac 1p}, dt$$ I have not been able to get anything analytical beside $K_1=frac 23$ and $K_2=frac pi 2$. Numerical integration shows that $K_p sim p-frac 12$

share|cite|improve this answer

edited Mar 27 at 5:18

answered Mar 23 at 5:42

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

share|cite|improve this answer

edited Mar 27 at 5:18

answered Mar 23 at 5:42

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 23 at 5:42

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 23 at 5:42

Claude LeiboviciClaude Leibovici

125k1158135