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Calculate these integrals
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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Compare the integrals $int_0^{frac{pi}{2}}sin(cos x)dx$ and $int_0^{frac{pi}{2}}cos(sin x)dx$Duo Fresnel-like integrals $(??)$Calculating values of integrals using Fourier series and uniform convergenceIntegral of Bessel function multiplied with sine $int_0^infty J_0(bx) sin(ax) dx$.How to evaluate these two integrals about hyperbolic functions?How can I calculate the following trigonometric integrals with the methods of contour integrals?Limit of these integralsCalculate integrals 2How to calculate this triple integral?Calculate a definite integral involving sin and exp
$begingroup$
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.
- $$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt;$$
- $$int_0^{pi/2}(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2)^{1/p} dt.$$
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
edited Mar 22 at 22:37
Eric Wofsey
193k14221352
asked Mar 22 at 21:17
Pedro G. MattosPedro G. Mattos
307
$endgroup$
|
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.
- $$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt;$$
- $$int_0^{pi/2}(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2)^{1/p} dt.$$
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
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.
- $$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt;$$
- $$int_0^{pi/2}(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2)^{1/p} dt.$$
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
edited Mar 22 at 22:37
Eric Wofsey
193k14221352
asked Mar 22 at 21:17
Pedro G. MattosPedro G. Mattos
307
$endgroup$
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.
- $$int_0^1(1+p^pt^{p(p-1)})^{1/p} dt;$$
- $$int_0^{pi/2}(cos(t)^{2-p}sin(t)^2+sin(t)^{2-p}cos(t)^2)^{1/p} dt.$$
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
integration definite-integrals
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
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
edited Mar 22 at 22:37
Eric Wofsey
193k14221352
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
asked Mar 22 at 21:17
Pedro G. MattosPedro G. Mattos
307
307
$begingroup$
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 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
$begingroup$
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
$begingroup$
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
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
1 Answer
1
active
oldest
votes
$begingroup$
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 $K_p sim p-frac 12$
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
Your Answer
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $K_p sim p-frac 12$
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 $K_p sim p-frac 12$
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 $K_p sim p-frac 12$
answered Mar 23 at 5:42
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
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 $K_p sim p-frac 12$
answered Mar 23 at 5:42
Claude LeiboviciClaude Leibovici
125k1158135
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
125k1158135
$begingroup$
Thanks a lot. That hypergeometric function is really interesting. It's such a coincidence that I was reading about it just a week ago.
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– Pedro G. Mattos
Mar 24 at 5:09
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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
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@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.
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– 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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
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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$begingroup$
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