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Math Help - [SOLVED] Gamma function simple help needed

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Gamma function simple help needed

    I must show, using integration by parts that  \Gamma(x+1)=x\Gamma (x) for all positive x. I've tried but I don't get the result. The fact that the integral is dt and that the function depends of x is responsible of my fail.
    Can you start me up?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I must show, using integration by parts that  \Gamma(x+1)=x\Gamma (x) for all positive x. I've tried but I don't get the result. The fact that the integral is dt and that the function depends of x is responsible of my fail.
    Can you start me up?
    Let dv=e^{-t}

    and t^x=u

    So v=-e^{-t}

    and du=\color{red}xt^{x-1} \leftarrow\text{ Should look familiar}

    Now notice that your first term dissapears.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Let

    and

    Now notice that your first term dissapears.
    I still don't get it. By the way, do making a substitution would allow me to use integration by parts later? If yes, then can you show me how to get to the integration by parts. Thanks.
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  4. #4
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    {\Gamma(p+1)}=\int_{0}^{\infty}x^{p}e^{-x}dx

    {\Gamma(p)}=\int_{0}^{\infty}x^{p-1}e^{-x}dx

    Let x^{p}=u, \;\ e^{-x}dx=dv, \;\ du=px^{p-1}dx, \;\ v=-e^{-x}

    Then we get:

    {\Gamma(p+1)}=\left[-x^{p}e^{-x}\right]_{0}^{\infty}-\int_{0}^{\infty}\left(-e^{-x}\right)px^{p-1}dx

    =p\underbrace{\int_{0}^{\infty}x^{p-1}e^{-x}dx}_{\text{this is gamma}}=p{\Gamma(p)}
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I still don't get it. By the way, do making a substitution would allow me to use integration by parts later? If yes, then can you show me how to get to the integration by parts. Thanks.
    \Gamma(x+1)=\int_0^{\infty}t^{x}e^{-t}dt

    So letting u=t^x\Rightarrow{du=xt^{x-1}}

    and dv=e^{-t}\Rightarrow{v=-e^{-t}}

    We have

    \Gamma(x+1)=\bigg[-t^{x}e^{-t}\bigg]\bigg|_{0}^{\infty}+\int_0^{\infty}xt^{x-1}e^{-t}dt

    Now for the first term, substitutin zero gives zero, and seeing that \lim_{t\to\infty}\frac{-t^x}{e^t}=0

    We have

    \Gamma(x+1)=0+\int_0^{\infty}xt^{x-1}e^{-t}dt=x\int_0^{\infty}t^{x-1}e^{-t}dt=x\Gamma(x)\quad\blacksquare

    I have never done this before, but I assume that is how its done.
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  6. #6
    MHF Contributor arbolis's Avatar
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    Oh, thanks to both. Now I got it!! But I don't understand one thing.
    With u=t^x, du is the derivative of u, right? u equals to e^{xln(t)}, so the derivative equals ln(t)e^{xln(t)}=ln(t)*t^x. How do you get to du=xt^{x-1}?
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  7. #7
    o_O
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    You're integrating with respect to t so treat x as a real number: \frac{d}{dt} t^{n} = nt^{n-1} \: \: n \in \mathbb{R}

    Also, using your method:
    \frac{d}{dt} t^{x} = \frac{d}{dt} e^{x \ln (t)} = e^{x \ln (t)} \cdot (x \ln (t) )' = e^{x \ln (t)} \cdot \frac{x}{t} = t^{x} \cdot \frac{x}{t} = xt^{x-1}

    -- Totally would've answered this thread but too slow .. Gamma ftw!
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  8. #8
    Moo
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    Hello !

    Quote Originally Posted by arbolis View Post
    Oh, thanks to both. Now I got it!! But I don't understand one thing.
    With u=t^x, du is the derivative of u, right? u equals to e^{xln(t)}, so the derivative equals ln(t)e^{xln(t)}=ln(t)*t^x. How do you get to du=xt^{x-1}?
    You have to be very careful

    The notation that is used is not that right.

    You should say that u is a function of t.

    So this is \frac{du}{dt}, not \frac{du}{dx}




    I think the better way not to be confused is to take n instead of x. So that we're almost sure not to take n as a variable, since we barely use it as a variable
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  9. #9
    MHF Contributor arbolis's Avatar
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    You're integrating with respect to t so treat x as a real number:

    Also, using your method:
    ,
    You have to be very careful

    The notation that is used is not that right.

    You should say that u is a function of t.

    So this is , not




    I think the better way not to be confused is to take n instead of x. So that we're almost sure not to take n as a variable.
    Oops!! I got confused! I thought I had to do it with respect to x... how could I understand then, that if dv=e^{-t}, then v=-e^{-t}!!!
    My exam of Calculus II is coming up next 21st of July, so I have all the time to fix such problems of mine. A big thanks to you.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    , Oops!! I got confused! I thought I had to do it with respect to x... how could I understand then, that if dv=e^{-t}, then v=-e^{-t}!!!
    My exam of Calculus II is coming up next 21st of July, so I have all the time to fix such problems of mine. A big thanks to you.
    Although t is a dummy variable (a place holder) in this context as moo said since x is a fixed value and t is the variable you must differentiate in respect to t
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