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A Gamma function, Bessel function

  1. Dec 5, 2017 #1
    I have question regarding gamma function. It is concerning ##\Gamma## function of negative integer arguments.
    Is it ##\Gamma(-1)=\infty## or ##\displaystyle \lim_{x \to -1}\Gamma(x)=\infty##? So is it ##\Gamma(-1)## defined or it is ##\infty##? This question is mainly because of definition of Bessel function
    [tex] J_p(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+p+1)}(\frac{x}{2})^{2n+p}[/tex]
    For example there is relation
    [tex]J_{-p}(x)=(-1)^pJ_p(x) [/tex]
    What is happening with ##J_{-2}##
    [tex] J_{-2}(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n+1)\Gamma(n-1)}(\frac{x}{2})^{2n-2}[/tex]
    for term ##n=0##?
     
  2. jcsd
  3. Dec 5, 2017 #2

    fresh_42

    Staff: Mentor

    The former is a sloppy notation of the latter, which is as well sloppy, because it doesn't tell from which side and the two limits aren't equal.
    No.
    Yes. And ##-\infty##!
    One may only conclude, that the expression as a series in terms of the gamma function isn't possible in this case. However, the relation ##(*)## is the way out.
     
  4. Dec 5, 2017 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    https://en.wikipedia.org/wiki/Gamma_function
    The above shows a graph of the gamma function, particularly for negative real arguments. At integers ≤ 0, the values are ±∞, depending on which direction x approaches the integer.
     
  5. Dec 6, 2017 #4
    In few books I found, for example
    [tex]J_{-2}(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n-1)\Gamma(n+1)} (\frac{x}{2})^{2n-2}=\sum^{\infty}_{n=2}\frac{(-1)^n}{\Gamma(n-1)\Gamma(n+1)} (\frac{x}{2})^{2n-2}[/tex]
    and this is like that I took that gama values of negative arguments are ##\infty##. Could you give me explanation for this?
     
  6. Dec 6, 2017 #5

    fresh_42

    Staff: Mentor

    I agree with your impression. It is a simple way to treat the first to summands as if ##\Gamma(-1)=\Gamma(0)=\infty## making them zero. As long as there will be no limit ##x \to 0## or ##x=0## later in the text, this is doable. Otherwise one would have to be more careful, that the two limiting processes don't contradict each other. One could as well define ##J_{-2}## by starting at ##n=2##. It all comes down to what is meant by the first equation, the representation of the series itself. What's first: the hen or the egg? All in all it is probably only a matter of how to memorize the definitions or what they are expected to represent. These are the locations, where errors can creep in, because one forgets what has been assumed to justify the notation. Is it sloppy? Probably yes. But is it false, too? Probably not, because there are alternative ways to define the series.
     
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