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how to choose the sabstitution for Euler's integrals?
I currently have, and have to calculate the Gamma function:
$$int_2^4 sqrt[4]{(x-2)(4-x)^3},mathbb{d}x$$
As per definition gamma function is:
$$int_0^1t^{z-1}e^{-t},mathbb{d}t$$
Do I understand correctly that first of all we have to reduce the lower and upper boundaries so they must be $0$ and $1$ respectivly?
To make lower boundory zero we have to substract $2$ and to make upper boundary 1 we have to also devide substitution by $2$ (as we have already had substrected $2$ and $2$ is left). As a result:
$$begin{array}{|c|} t = frac{x-2}{2} \ dt = frac{1}{2} dx \
x = 4t +2 end{array}$$
Substituting this into the integral will give us the Gamma function, and relationship between Gamma and Beta functions are clear, but I spent quite lot amount of time to understand what substitution is needed (and I am not sure if this one I used was right), so my question is:
What (if any) is the way to find out what substitution should be used in Euler's integral?
calculus definite-integrals gamma-function substitution
asked Nov 22 '18 at 13:46
M.MassM.Mass
1,8643923
add a comment |
I currently have, and have to calculate the Gamma function:
$$int_2^4 sqrt[4]{(x-2)(4-x)^3},mathbb{d}x$$
As per definition gamma function is:
$$int_0^1t^{z-1}e^{-t},mathbb{d}t$$
Do I understand correctly that first of all we have to reduce the lower and upper boundaries so they must be $0$ and $1$ respectivly?
To make lower boundory zero we have to substract $2$ and to make upper boundary 1 we have to also devide substitution by $2$ (as we have already had substrected $2$ and $2$ is left). As a result:
$$begin{array}{|c|} t = frac{x-2}{2} \ dt = frac{1}{2} dx \
x = 4t +2 end{array}$$
Substituting this into the integral will give us the Gamma function, and relationship between Gamma and Beta functions are clear, but I spent quite lot amount of time to understand what substitution is needed (and I am not sure if this one I used was right), so my question is:
What (if any) is the way to find out what substitution should be used in Euler's integral?
calculus definite-integrals gamma-function substitution
asked Nov 22 '18 at 13:46
M.MassM.Mass
1,8643923
add a comment |
I currently have, and have to calculate the Gamma function:
$$int_2^4 sqrt[4]{(x-2)(4-x)^3},mathbb{d}x$$
As per definition gamma function is:
$$int_0^1t^{z-1}e^{-t},mathbb{d}t$$
Do I understand correctly that first of all we have to reduce the lower and upper boundaries so they must be $0$ and $1$ respectivly?
To make lower boundory zero we have to substract $2$ and to make upper boundary 1 we have to also devide substitution by $2$ (as we have already had substrected $2$ and $2$ is left). As a result:
$$begin{array}{|c|} t = frac{x-2}{2} \ dt = frac{1}{2} dx \
x = 4t +2 end{array}$$
Substituting this into the integral will give us the Gamma function, and relationship between Gamma and Beta functions are clear, but I spent quite lot amount of time to understand what substitution is needed (and I am not sure if this one I used was right), so my question is:
What (if any) is the way to find out what substitution should be used in Euler's integral?
calculus definite-integrals gamma-function substitution
asked Nov 22 '18 at 13:46
M.MassM.Mass
1,8643923
I currently have, and have to calculate the Gamma function:
$$int_2^4 sqrt[4]{(x-2)(4-x)^3},mathbb{d}x$$
As per definition gamma function is:
$$int_0^1t^{z-1}e^{-t},mathbb{d}t$$
Do I understand correctly that first of all we have to reduce the lower and upper boundaries so they must be $0$ and $1$ respectivly?
To make lower boundory zero we have to substract $2$ and to make upper boundary 1 we have to also devide substitution by $2$ (as we have already had substrected $2$ and $2$ is left). As a result:
$$begin{array}{|c|} t = frac{x-2}{2} \ dt = frac{1}{2} dx \
x = 4t +2 end{array}$$
Substituting this into the integral will give us the Gamma function, and relationship between Gamma and Beta functions are clear, but I spent quite lot amount of time to understand what substitution is needed (and I am not sure if this one I used was right), so my question is:
What (if any) is the way to find out what substitution should be used in Euler's integral?
calculus definite-integrals gamma-function substitution
calculus definite-integrals gamma-function substitution
asked Nov 22 '18 at 13:46
M.MassM.Mass
1,8643923
asked Nov 22 '18 at 13:46
M.MassM.Mass
1,8643923
edited Nov 22 '18 at 13:51
M.Mass
asked Nov 22 '18 at 13:46
M.MassM.Mass
1,8643923
asked Nov 22 '18 at 13:46
M.MassM.Mass
1,8643923
asked Nov 22 '18 at 13:46
M.MassM.Mass
1,8643923
1,8643923
add a comment |
add a comment |
1 Answer
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As a side note, the upper bound of your second integral should be $infty$ to get the $Gamma$ function.
A hint. Fubini's theorem is useful here. You can writebegin{align}
Gamma(x)Gamma(y) &= int_{0}^infty e^{-u} u^{x-1},du cdotint_{0}^infty e^{-v} v^{y-1},dv \[6pt]
&=int_{0}^inftyint_{0}^infty e^{-u-v} u^{x-1}v^{y-1},du ,dv,
end{align}
then by the change of variable $u=zt$, $v=z(1-t)$ one obtains
begin{align}
Gamma(x)Gamma(y) &= int_{0}^inftyint_{0}^1 e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z,dt ,dz \[6pt]
&= int_{0}^infty e^{-z}z^{x+y-1} ,dzcdotbbox[5px,border:1px solid blue]{int_{0}^1t^{x-1}(1-t)^{y-1},dt}\
&=Gamma(x+y),{rm{B}}(x,y),
end{align}
as expected.
answered Nov 22 '18 at 14:01
Dan KentDan Kent
237
But how to apply this for my specific case then?
– M.Mass
Nov 22 '18 at 16:28
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
As a side note, the upper bound of your second integral should be $infty$ to get the $Gamma$ function.
A hint. Fubini's theorem is useful here. You can writebegin{align}
Gamma(x)Gamma(y) &= int_{0}^infty e^{-u} u^{x-1},du cdotint_{0}^infty e^{-v} v^{y-1},dv \[6pt]
&=int_{0}^inftyint_{0}^infty e^{-u-v} u^{x-1}v^{y-1},du ,dv,
end{align}
then by the change of variable $u=zt$, $v=z(1-t)$ one obtains
begin{align}
Gamma(x)Gamma(y) &= int_{0}^inftyint_{0}^1 e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z,dt ,dz \[6pt]
&= int_{0}^infty e^{-z}z^{x+y-1} ,dzcdotbbox[5px,border:1px solid blue]{int_{0}^1t^{x-1}(1-t)^{y-1},dt}\
&=Gamma(x+y),{rm{B}}(x,y),
end{align}
as expected.
answered Nov 22 '18 at 14:01
Dan KentDan Kent
237
But how to apply this for my specific case then?
– M.Mass
Nov 22 '18 at 16:28
add a comment |
As a side note, the upper bound of your second integral should be $infty$ to get the $Gamma$ function.
A hint. Fubini's theorem is useful here. You can writebegin{align}
Gamma(x)Gamma(y) &= int_{0}^infty e^{-u} u^{x-1},du cdotint_{0}^infty e^{-v} v^{y-1},dv \[6pt]
&=int_{0}^inftyint_{0}^infty e^{-u-v} u^{x-1}v^{y-1},du ,dv,
end{align}
then by the change of variable $u=zt$, $v=z(1-t)$ one obtains
begin{align}
Gamma(x)Gamma(y) &= int_{0}^inftyint_{0}^1 e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z,dt ,dz \[6pt]
&= int_{0}^infty e^{-z}z^{x+y-1} ,dzcdotbbox[5px,border:1px solid blue]{int_{0}^1t^{x-1}(1-t)^{y-1},dt}\
&=Gamma(x+y),{rm{B}}(x,y),
end{align}
as expected.
answered Nov 22 '18 at 14:01
Dan KentDan Kent
237
But how to apply this for my specific case then?
– M.Mass
Nov 22 '18 at 16:28
add a comment |
As a side note, the upper bound of your second integral should be $infty$ to get the $Gamma$ function.
A hint. Fubini's theorem is useful here. You can writebegin{align}
Gamma(x)Gamma(y) &= int_{0}^infty e^{-u} u^{x-1},du cdotint_{0}^infty e^{-v} v^{y-1},dv \[6pt]
&=int_{0}^inftyint_{0}^infty e^{-u-v} u^{x-1}v^{y-1},du ,dv,
end{align}
then by the change of variable $u=zt$, $v=z(1-t)$ one obtains
begin{align}
Gamma(x)Gamma(y) &= int_{0}^inftyint_{0}^1 e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z,dt ,dz \[6pt]
&= int_{0}^infty e^{-z}z^{x+y-1} ,dzcdotbbox[5px,border:1px solid blue]{int_{0}^1t^{x-1}(1-t)^{y-1},dt}\
&=Gamma(x+y),{rm{B}}(x,y),
end{align}
as expected.
answered Nov 22 '18 at 14:01
Dan KentDan Kent
237
As a side note, the upper bound of your second integral should be $infty$ to get the $Gamma$ function.
A hint. Fubini's theorem is useful here. You can writebegin{align}
Gamma(x)Gamma(y) &= int_{0}^infty e^{-u} u^{x-1},du cdotint_{0}^infty e^{-v} v^{y-1},dv \[6pt]
&=int_{0}^inftyint_{0}^infty e^{-u-v} u^{x-1}v^{y-1},du ,dv,
end{align}
then by the change of variable $u=zt$, $v=z(1-t)$ one obtains
begin{align}
Gamma(x)Gamma(y) &= int_{0}^inftyint_{0}^1 e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z,dt ,dz \[6pt]
&= int_{0}^infty e^{-z}z^{x+y-1} ,dzcdotbbox[5px,border:1px solid blue]{int_{0}^1t^{x-1}(1-t)^{y-1},dt}\
&=Gamma(x+y),{rm{B}}(x,y),
end{align}
as expected.
answered Nov 22 '18 at 14:01
Dan KentDan Kent
237
edited Nov 22 '18 at 14:12
answered Nov 22 '18 at 14:01
Dan KentDan Kent
237
answered Nov 22 '18 at 14:01
Dan KentDan Kent
237
answered Nov 22 '18 at 14:01
Dan KentDan Kent
237
237
But how to apply this for my specific case then?
– M.Mass
Nov 22 '18 at 16:28
add a comment |
But how to apply this for my specific case then?
– M.Mass
Nov 22 '18 at 16:28
But how to apply this for my specific case then?
– M.Mass
Nov 22 '18 at 16:28
But how to apply this for my specific case then?
– M.Mass
Nov 22 '18 at 16:28
add a comment |
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