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Math Help - Convergence of Gamma Function

  1. #1
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    Convergence of Gamma Function

    Check my proof and see if there are any mistakes or needs adjustment.

    Given \Gamma(s)=\int^\infty_{0} e^{-t}t^{s-1}dt
    If s=1 then \Gamma(1)=\int^\infty_{0} e^{-t}dt=1


    If s>1 then consider the function f(t)=e^{-t}t^{s-1} this function is positive and countinous for t\geq0. Also f'(t)=t^{s-2}e^{-t}(s-1-t). By the first derivative test the function is increasing for 0<t<s-1 and decreasing for t>s-1. Thus f(t) is a positive, decreasing and countinous function for t>s-1
    The gamma function can be expressed as \int^\infty_{0} e^{-t}t^{s-1}dt=\int^{s-1}_{0} e^{-t}t^{s-1}dt + \int^\infty_{s-1} e^{-t}t^{s-1}dt
    The first integral \int^{s-1}_{0} e^{-t}t^{s-1}dt definitely converges so the convergence of the gamma function determines on the improper integral \int^\infty_{s-1} e^{-t}t^{s-1}dt. By the integral test for infinite series this improper integral converges if and only if \sum^\infty_{k=s-1} \frac{k^{s-1}}{e^k} converges. But by the ratio test \lim_{k \rightarrow \infty} \frac{(k+1)^{s-1}}{e^{k+1}}\frac{e^k}{k^{s-1}}=\frac{1}{e}<1 Thus, the improper integral \int^\infty_{s-1} e^{-t}t^{s-1}dt converges, thus, \Gamma(s)=\int^\infty_{0} e^{-t}t^{s-1}dt converges for s>1.


    Now finally, if 0<s<1 then 1<s+1<2. But then \Gamma(s+1)=s\Gamma(s) with \Gamma(s+1) converges just as was proven in the last paragraph. Thus, the Gamma function \Gamma(s)=\int^\infty_{0} e^{-t}t^{s-1}dt converges for all s>0

    Note: \Gamma(s+1)=s\Gamma(s) is the fundamental property of the Gamma function and was assumed to have been known.

    Wow, I finally finished, you have no idea how long it took me to write this in Latex form.
    Last edited by ThePerfectHacker; December 30th 2005 at 04:51 PM.
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    Super Member Rebesques's Avatar
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    Dude, I think that's also the standard way to prove convergence It's nice and sharp, but won't get you the Fields Medal

    There's also a shorter proof with the help of complex analysis, but this won't get us the medal (or the money and prestige) either. Dang it, back to Riemann Hypothesis...
    Last edited by ThePerfectHacker; February 17th 2007 at 03:01 PM.
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    Quote Originally Posted by Rebesques View Post
    Dude, I think that's also the standard way to prove convergence It's nice and sharp, but won't get you the Fields Medal
    Thank you! I was waiting for somebody to respond from a long time ago. Did you catch the error in the end, I assumed convergence when integrating by parts. But it no matter, it can be easily fixed, I was just lazy to.

    This is also one of my eariler posts ever. When I was just learning latex.
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    Senior Member bkarpuz's Avatar
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    Talking

    Quote Originally Posted by ThePerfectHacker View Post
    Check my proof and see if there are any mistakes or needs adjustment.

    Given \Gamma(s)=\int^\infty_{0} e^{-t}t^{s-1}dt
    If s=1 then \Gamma(1)=\int^\infty_{0} e^{-t}dt=1


    If s>1 then consider the function f(t)=e^{-t}t^{s-1} this function is positive and countinous for t\geq0. Also f'(t)=t^{s-2}e^{-t}(s-1-t). By the first derivative test the function is increasing for 0<t<s-1 and decreasing for t>s-1. Thus f(t) is a positive, decreasing and countinous function for t>s-1
    The gamma function can be expressed as \int^\infty_{0} e^{-t}t^{s-1}dt=\int^{s-1}_{0} e^{-t}t^{s-1}dt + \int^\infty_{s-1} e^{-t}t^{s-1}dt
    The first integral \int^{s-1}_{0} e^{-t}t^{s-1}dt definitely converges so the convergence of the gamma function determines on the improper integral \int^\infty_{s-1} e^{-t}t^{s-1}dt. By the integral test for infinite series this improper integral converges if and only if \sum^\infty_{k=s-1} \frac{k^{s-1}}{e^k} converges. But by the ratio test \lim_{k \rightarrow \infty} \frac{(k+1)^{s-1}}{e^{k+1}}\frac{e^k}{k^{s-1}}=\frac{1}{e}<1 Thus, the improper integral \int^\infty_{s-1} e^{-t}t^{s-1}dt converges, thus, \Gamma(s)=\int^\infty_{0} e^{-t}t^{s-1}dt converges for s>1.


    Now finally, if 0<s<1 then 1<s+1<2. But then \Gamma(s+1)=s\Gamma(s) with \Gamma(s+1) converges just as was proven in the last paragraph. Thus, the Gamma function \Gamma(s)=\int^\infty_{0} e^{-t}t^{s-1}dt converges for all s>0

    Note: \Gamma(s+1)=s\Gamma(s) is the fundamental property of the Gamma function and was assumed to have been known.

    Wow, I finally finished, you have no idea how long it took me to write this in Latex form.
    Hi TPH!

    I guess I have something simpler.
    Let s\in(0,1], then
    \Gamma(s)=\int_{0}^{\infty}t^{s-1}\mathrm{e}^{-t}\mathrm{d}t
    ....... =\int_{0}^{1}t^{s-1}\mathrm{e}^{-t}\mathrm{d}t+\int_{1}^{\infty}t^{s-1}\mathrm{e}^{-t}\mathrm{d}t
    ....... \leq\int_{0}^{1}t^{s-1}\mathrm{d}t+\int_{1}^{\infty}\mathrm{e}^{-t}\mathrm{d}t
    ....... =\frac{1}{s}+\frac{1}{\mathrm{e}}<\infty.
    Convergence for s\in\mathbb{R}^{+} is an immediate consequence of the functional relation \Gamma(s+1)=s\Gamma(s) for s\in\mathbb{R}^{+}.

    By the way, one may wish to show that \lim\nolimits_{s\to0^{+}}\Gamma(s)=\infty.
    One can reach by using similar steps to those above to
    \Gamma(s)\geq\frac{1}{\mathrm{e}}\int_{0}^{1}t^{s-1}\mathrm{d}t=\frac{1}{\mathrm{e}s} for all s\in(0,1],
    which yields the desired result by letting s\to0^{+}.
    Last edited by bkarpuz; January 13th 2010 at 02:08 PM.
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    Super Member Rebesques's Avatar
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    I think that functional relation will be hard to prove without first proving convergence for general s
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    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Rebesques View Post
    I think that functional relation will be hard to prove without first proving convergence for general s
    If \lambda\geq1, then t^{\lambda-1} is of exponential order any positive number on [0,\infty),
    and therefore the relation
    \mathcal{L}\{t^{\lambda-1}\}(s)=\frac{\Gamma(\lambda)}{s^{\lambda}} for s\in(0,\infty)
    or simply
    \mathcal{L}\{t^{\lambda-1}\}(1)=\Gamma(\lambda)
    implies the convergence of the \Gamma function.
    The problem that \Gamma(\lambda) converges occurs only for \lambda\in(0,1],
    this is another reason that I only considered this case in my previous post.
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    Super Member Rebesques's Avatar
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    So you called in a truck to carry a few bricks? Cool
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    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Rebesques View Post
    So you called in a truck to carry a few bricks? Cool
    Why not if the bricks are golden.
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  9. #9
    Super Member Rebesques's Avatar
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    Sweet But costly. PH just used his bike
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    Senior Member bkarpuz's Avatar
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    Question

    Quote Originally Posted by Rebesques View Post
    Sweet But costly. PH just used his bike
    I have enough golden bricks, who cares about its cost!

    By the way, I have searched the internet too much and I could not find a reference which proves that the \Gamma converges on \mathbb{R}^{+}.
    Is there a book or another reference that proves this?


    Note. The sum in TPH's post must start from an integer (for instance \lfloor s-1\rfloor, where \lfloor\cdot\rfloor is the least integer function) not s-1.
    Last edited by bkarpuz; January 18th 2010 at 11:38 AM.
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  11. #11
    Super Member Rebesques's Avatar
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    If you don't care about expenses (lucky you. spare me some for my rent plz )
    you can check Conway's Functions of a complex variable, pp 176 onwards. Covers many issues regarding the Gamma function, even the (unexpected) Bohr-Mollerup theorem.
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  12. #12
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Rebesques View Post
    If you don't care about expenses (lucky you. spare me some for my rent plz )
    Sure, you may have all I have!

    Quote Originally Posted by Rebesques View Post
    you can check Conway's Functions of a complex variable, pp 176 onwards. Covers many issues regarding the Gamma function, even the (unexpected) Bohr-Mollerup theorem.
    Oh thanks, I will check the book.

    Btw, the Laplace transform directly implies convergence of the \Gamma function on \mathbb{R}^{+}.
    We have
    \Gamma(x)=\mathcal{L}\{t^{x-1}\}(1)
    ......- =\frac{1}{x}\mathcal{L}\bigg\{\frac{\mathrm{d}}{\m  athrm{d}t}t^{x}\bigg\}(1)
    ......- =\frac{1}{x}\bigg(1\times\mathcal{L}\{t^{x}\}(1)-\lim_{t\to0^{+}}t^{x}\bigg)........(Laplace transform of the derivative)
    ......- =\frac{1}{x}\mathcal{L}\{t^{x}\}(1).
    As t^{x} for x\in\mathbb{R}^{+} is of exponential order any positive number, we have convergence on the right-hand side of the equation above (since 1>0), and therefore the left-hand side is convergent too.
    The above result also implies
    \Gamma(x)=\frac{1}{x}\Gamma(x+1) for x\in\mathbb{R}^{+}.

    Note. An interesting feature of the derivative theorem is that we obtain \mathcal{L}\{f^{\prime}(t)\} without requiring that f^{\prime}(t) itself be of exponential order*.

    * See Schiff, The Laplace Transform Theory and Applications, Remark 2.8 on pp. 54.
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