how to choose the sabstitution for Euler's integrals?











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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







share|cite|improve this question




























    up vote
    1
    down vote

    favorite












    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







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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







      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 22 at 13:51

























      asked Nov 22 at 13:46









      M.Mass

      1,8493923




      1,8493923






















          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.






          share|cite|improve this answer























          • But how to apply this for my specific case then?
            – M.Mass
            Nov 22 at 16:28











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          1 Answer
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          active

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          1 Answer
          1






          active

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          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          0
          down vote













          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.






          share|cite|improve this answer























          • But how to apply this for my specific case then?
            – M.Mass
            Nov 22 at 16:28















          up vote
          0
          down vote













          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.






          share|cite|improve this answer























          • But how to apply this for my specific case then?
            – M.Mass
            Nov 22 at 16:28













          up vote
          0
          down vote










          up vote
          0
          down vote









          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 at 14:12

























          answered Nov 22 at 14:01









          Dan Kent

          237




          237












          • But how to apply this for my specific case then?
            – M.Mass
            Nov 22 at 16:28


















          • But how to apply this for my specific case then?
            – M.Mass
            Nov 22 at 16:28
















          But how to apply this for my specific case then?
          – M.Mass
          Nov 22 at 16:28




          But how to apply this for my specific case then?
          – M.Mass
          Nov 22 at 16:28


















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