Proof that β-function ∈ C^∞
I need to find correct proof that β-function is smooth on its domain.
Is there some feature of such functions, I guess that we need to prove the continuity of all n-derivatives, or their partial derivatives, but how it will be eventually I hadn't got yet. I will very glad your help.
calculus integration special-functions smooth-functions
add a comment |
I need to find correct proof that β-function is smooth on its domain.
Is there some feature of such functions, I guess that we need to prove the continuity of all n-derivatives, or their partial derivatives, but how it will be eventually I hadn't got yet. I will very glad your help.
calculus integration special-functions smooth-functions
You need to write in latex, click on edit in the other post to see how. That the $beta$ function is $k$-times differentiable in each of its variable is not hard. To see it is analytic you can expand $t^{x-1}$ as a series in powers of $x+c$.
– reuns
Nov 29 at 6:58
Which function do you mean: Dirichlet's $beta$ or Euler's $B$?
– gammatester
Nov 29 at 8:22
Euler's Β function
– user2952487
Nov 29 at 11:39
add a comment |
I need to find correct proof that β-function is smooth on its domain.
Is there some feature of such functions, I guess that we need to prove the continuity of all n-derivatives, or their partial derivatives, but how it will be eventually I hadn't got yet. I will very glad your help.
calculus integration special-functions smooth-functions
I need to find correct proof that β-function is smooth on its domain.
Is there some feature of such functions, I guess that we need to prove the continuity of all n-derivatives, or their partial derivatives, but how it will be eventually I hadn't got yet. I will very glad your help.
calculus integration special-functions smooth-functions
calculus integration special-functions smooth-functions
edited Nov 29 at 6:56
asked Nov 29 at 6:19
user2952487
224
224
You need to write in latex, click on edit in the other post to see how. That the $beta$ function is $k$-times differentiable in each of its variable is not hard. To see it is analytic you can expand $t^{x-1}$ as a series in powers of $x+c$.
– reuns
Nov 29 at 6:58
Which function do you mean: Dirichlet's $beta$ or Euler's $B$?
– gammatester
Nov 29 at 8:22
Euler's Β function
– user2952487
Nov 29 at 11:39
add a comment |
You need to write in latex, click on edit in the other post to see how. That the $beta$ function is $k$-times differentiable in each of its variable is not hard. To see it is analytic you can expand $t^{x-1}$ as a series in powers of $x+c$.
– reuns
Nov 29 at 6:58
Which function do you mean: Dirichlet's $beta$ or Euler's $B$?
– gammatester
Nov 29 at 8:22
Euler's Β function
– user2952487
Nov 29 at 11:39
You need to write in latex, click on edit in the other post to see how. That the $beta$ function is $k$-times differentiable in each of its variable is not hard. To see it is analytic you can expand $t^{x-1}$ as a series in powers of $x+c$.
– reuns
Nov 29 at 6:58
You need to write in latex, click on edit in the other post to see how. That the $beta$ function is $k$-times differentiable in each of its variable is not hard. To see it is analytic you can expand $t^{x-1}$ as a series in powers of $x+c$.
– reuns
Nov 29 at 6:58
Which function do you mean: Dirichlet's $beta$ or Euler's $B$?
– gammatester
Nov 29 at 8:22
Which function do you mean: Dirichlet's $beta$ or Euler's $B$?
– gammatester
Nov 29 at 8:22
Euler's Β function
– user2952487
Nov 29 at 11:39
Euler's Β function
– user2952487
Nov 29 at 11:39
add a comment |
1 Answer
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You have more than enough to differentiate through the integral sign (the Leibniz rule.) For example, thinking of $x,y>0,$ we have
$$frac{d}{dx}int_0^1t^{x-1}(1-t)^{y-1},dt = int_0^1(ln t)t^{x-1}(1-t)^{y-1},dt.$$
You can keep going, piling up factors like $(ln t)^m ln (1-t)^n$ in the integral. None of these factors will destroy integrability.
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You have more than enough to differentiate through the integral sign (the Leibniz rule.) For example, thinking of $x,y>0,$ we have
$$frac{d}{dx}int_0^1t^{x-1}(1-t)^{y-1},dt = int_0^1(ln t)t^{x-1}(1-t)^{y-1},dt.$$
You can keep going, piling up factors like $(ln t)^m ln (1-t)^n$ in the integral. None of these factors will destroy integrability.
add a comment |
You have more than enough to differentiate through the integral sign (the Leibniz rule.) For example, thinking of $x,y>0,$ we have
$$frac{d}{dx}int_0^1t^{x-1}(1-t)^{y-1},dt = int_0^1(ln t)t^{x-1}(1-t)^{y-1},dt.$$
You can keep going, piling up factors like $(ln t)^m ln (1-t)^n$ in the integral. None of these factors will destroy integrability.
add a comment |
You have more than enough to differentiate through the integral sign (the Leibniz rule.) For example, thinking of $x,y>0,$ we have
$$frac{d}{dx}int_0^1t^{x-1}(1-t)^{y-1},dt = int_0^1(ln t)t^{x-1}(1-t)^{y-1},dt.$$
You can keep going, piling up factors like $(ln t)^m ln (1-t)^n$ in the integral. None of these factors will destroy integrability.
You have more than enough to differentiate through the integral sign (the Leibniz rule.) For example, thinking of $x,y>0,$ we have
$$frac{d}{dx}int_0^1t^{x-1}(1-t)^{y-1},dt = int_0^1(ln t)t^{x-1}(1-t)^{y-1},dt.$$
You can keep going, piling up factors like $(ln t)^m ln (1-t)^n$ in the integral. None of these factors will destroy integrability.
answered Nov 29 at 18:02
zhw.
71.5k43075
71.5k43075
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You need to write in latex, click on edit in the other post to see how. That the $beta$ function is $k$-times differentiable in each of its variable is not hard. To see it is analytic you can expand $t^{x-1}$ as a series in powers of $x+c$.
– reuns
Nov 29 at 6:58
Which function do you mean: Dirichlet's $beta$ or Euler's $B$?
– gammatester
Nov 29 at 8:22
Euler's Β function
– user2952487
Nov 29 at 11:39