Integrating $x^c (1-x)^d$ from 0 to 1 using the gamma function
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I am trying to solve the following integral:
$$
int_0^1 x^c (1-x)^d dx
$$
for some $c, d in mathbb{R}$
I know I have to use the gamma function, I have tried using the substitutions $u = lnfrac{1}{x}$ as well as $u = lnfrac{1}{1-x}$ to try to make this appear as a gamma function but neither have gotten me closer to an answer.
integration gamma-function
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add a comment |
$begingroup$
I am trying to solve the following integral:
$$
int_0^1 x^c (1-x)^d dx
$$
for some $c, d in mathbb{R}$
I know I have to use the gamma function, I have tried using the substitutions $u = lnfrac{1}{x}$ as well as $u = lnfrac{1}{1-x}$ to try to make this appear as a gamma function but neither have gotten me closer to an answer.
integration gamma-function
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4
$begingroup$
en.wikipedia.org/wiki/Beta_function
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– Lord Shark the Unknown
Jan 2 at 11:17
1
$begingroup$
Try $x=sin^2(t)$
$endgroup$
– Claude Leibovici
Jan 2 at 11:19
add a comment |
$begingroup$
I am trying to solve the following integral:
$$
int_0^1 x^c (1-x)^d dx
$$
for some $c, d in mathbb{R}$
I know I have to use the gamma function, I have tried using the substitutions $u = lnfrac{1}{x}$ as well as $u = lnfrac{1}{1-x}$ to try to make this appear as a gamma function but neither have gotten me closer to an answer.
integration gamma-function
$endgroup$
I am trying to solve the following integral:
$$
int_0^1 x^c (1-x)^d dx
$$
for some $c, d in mathbb{R}$
I know I have to use the gamma function, I have tried using the substitutions $u = lnfrac{1}{x}$ as well as $u = lnfrac{1}{1-x}$ to try to make this appear as a gamma function but neither have gotten me closer to an answer.
integration gamma-function
integration gamma-function
asked Jan 2 at 11:11
Abdelrahman SamehAbdelrahman Sameh
32
32
4
$begingroup$
en.wikipedia.org/wiki/Beta_function
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– Lord Shark the Unknown
Jan 2 at 11:17
1
$begingroup$
Try $x=sin^2(t)$
$endgroup$
– Claude Leibovici
Jan 2 at 11:19
add a comment |
4
$begingroup$
en.wikipedia.org/wiki/Beta_function
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:17
1
$begingroup$
Try $x=sin^2(t)$
$endgroup$
– Claude Leibovici
Jan 2 at 11:19
4
4
$begingroup$
en.wikipedia.org/wiki/Beta_function
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:17
$begingroup$
en.wikipedia.org/wiki/Beta_function
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:17
1
1
$begingroup$
Try $x=sin^2(t)$
$endgroup$
– Claude Leibovici
Jan 2 at 11:19
$begingroup$
Try $x=sin^2(t)$
$endgroup$
– Claude Leibovici
Jan 2 at 11:19
add a comment |
1 Answer
1
active
oldest
votes
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Hint:
$$int_0^1 t^{x-1}(1-t)^{y-1}dt=B(x,y)=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
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$begingroup$
But is there a way to do it using only the gamma function and its definition?
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– Abdelrahman Sameh
Jan 2 at 12:12
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@AbdelrahmanSameh A proof of the Beta Function comes from using the definition of the gamma function. I believe it's on Wikipedia
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– Frank W.
Jan 2 at 19:03
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$$int_0^1 t^{x-1}(1-t)^{y-1}dt=B(x,y)=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
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$begingroup$
But is there a way to do it using only the gamma function and its definition?
$endgroup$
– Abdelrahman Sameh
Jan 2 at 12:12
$begingroup$
@AbdelrahmanSameh A proof of the Beta Function comes from using the definition of the gamma function. I believe it's on Wikipedia
$endgroup$
– Frank W.
Jan 2 at 19:03
add a comment |
$begingroup$
Hint:
$$int_0^1 t^{x-1}(1-t)^{y-1}dt=B(x,y)=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
$endgroup$
$begingroup$
But is there a way to do it using only the gamma function and its definition?
$endgroup$
– Abdelrahman Sameh
Jan 2 at 12:12
$begingroup$
@AbdelrahmanSameh A proof of the Beta Function comes from using the definition of the gamma function. I believe it's on Wikipedia
$endgroup$
– Frank W.
Jan 2 at 19:03
add a comment |
$begingroup$
Hint:
$$int_0^1 t^{x-1}(1-t)^{y-1}dt=B(x,y)=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
$endgroup$
Hint:
$$int_0^1 t^{x-1}(1-t)^{y-1}dt=B(x,y)=frac{Gamma(x)Gamma(y)}{Gamma(x+y)}$$
answered Jan 2 at 11:51
DarkraiDarkrai
6,2971441
6,2971441
$begingroup$
But is there a way to do it using only the gamma function and its definition?
$endgroup$
– Abdelrahman Sameh
Jan 2 at 12:12
$begingroup$
@AbdelrahmanSameh A proof of the Beta Function comes from using the definition of the gamma function. I believe it's on Wikipedia
$endgroup$
– Frank W.
Jan 2 at 19:03
add a comment |
$begingroup$
But is there a way to do it using only the gamma function and its definition?
$endgroup$
– Abdelrahman Sameh
Jan 2 at 12:12
$begingroup$
@AbdelrahmanSameh A proof of the Beta Function comes from using the definition of the gamma function. I believe it's on Wikipedia
$endgroup$
– Frank W.
Jan 2 at 19:03
$begingroup$
But is there a way to do it using only the gamma function and its definition?
$endgroup$
– Abdelrahman Sameh
Jan 2 at 12:12
$begingroup$
But is there a way to do it using only the gamma function and its definition?
$endgroup$
– Abdelrahman Sameh
Jan 2 at 12:12
$begingroup$
@AbdelrahmanSameh A proof of the Beta Function comes from using the definition of the gamma function. I believe it's on Wikipedia
$endgroup$
– Frank W.
Jan 2 at 19:03
$begingroup$
@AbdelrahmanSameh A proof of the Beta Function comes from using the definition of the gamma function. I believe it's on Wikipedia
$endgroup$
– Frank W.
Jan 2 at 19:03
add a comment |
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4
$begingroup$
en.wikipedia.org/wiki/Beta_function
$endgroup$
– Lord Shark the Unknown
Jan 2 at 11:17
1
$begingroup$
Try $x=sin^2(t)$
$endgroup$
– Claude Leibovici
Jan 2 at 11:19