Gamma function, Bessel function

In summary, the gamma function is defined for negative real arguments, but it is not defined for negative imaginary arguments.
  • #1
LagrangeEuler
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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##?
 
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  • #2
LagrangeEuler said:
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##?
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.
So is it ##\Gamma(-1)## defined ...
No.
... or it is ##\infty##?
Yes. And ##-\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) \quad (*)[/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##?
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.
 
  • #3
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.
 
  • #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?
 
  • #5
LagrangeEuler said:
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?
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.
 
  • #6
After all this time I feel that I do not have an answer for the same question. Can someone propose me a book where I can read something about it? Because from the graph of the gamma function, it looks like #\Gamma(-2)## is not defined. Also #\lim_{x \to -2^{+}}=\infty# and #\lim_{x \to -2^{-}}=-\infty#. But in many calculations authors just take. Wolfram Mathematica when I put #\Gamma(-2)# writes just Complex infinity.
 
  • #7
Generally, ##\Gamma(-k)## is undefined when ##k\in\mathbb{Z}##. To show this, we start with the Weierstrass definition of the Gamma function: $$\Gamma(x)=\frac{e^{-\gamma x}}x\prod_{n=1}^\infty\frac{e^{x/n}}{1+\frac xn }$$If ##x## is a negative integer, then for some ##n## the denominator becomes zero.
 
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  • #8
Also, the index of the bessel function can't be a negative integer, see here
 
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  • #9
LagrangeEuler said:
After all this time I feel that I do not have an answer for the same question. Can someone propose me a book where I can read something about it? Because from the graph of the gamma function, it looks like #\Gamma(-2)## is not defined. Also #\lim_{x \to -2^{+}}=\infty# and #\lim_{x \to -2^{-}}=-\infty#. But in many calculations authors just take. Wolfram Mathematica when I put #\Gamma(-2)# writes just Complex infinity.

There is only one complex infinity: It is the single point you must add to the complex plane to make it toplogically conjugate to a sphere.

In the real line we treat [itex]+\infty[/itex] and [itex]-\infty[/itex] as being distinct because the real line is ordered. The complex plane is not.
 
  • #10
You can't put [itex]\alpha = -|m|[/itex] for [itex]m \in \mathbb{Z}[/itex] directly into the definition [tex]
J_\alpha(x) = \sum_{n=0}^\infty \frac{(-1)^n}{n!\Gamma(n + \alpha + 1)}\left(\frac{x}{2}\right)^{2n+\alpha}[/tex] but that doesn't mean that [itex]J_{-m}[/itex] is not defined.

Frobenius's Method for the Bessel equation [tex]
xy'' +xy' +(x^2 - \alpha^2)y = 0[/tex] with [itex]y(x) = \sum_{n=0}^\infty a_nx^{n+r}[/itex] leads to the conditions [tex]
a_n ( (n + r)^2 - \alpha^2) = \begin{cases} 0 & n = 0, 1 \\ -a_{n-2} & n \geq 2 \end{cases}[/tex] Taking [itex]r = \pm |\alpha|[/itex] means that [itex]a_0[/itex] can be chosen arbitrarily; the condition on [itex]n = 1[/itex] then leads to [itex]a_1 = 0[/itex]. The third condition then reduces to [tex]
a_n n(n \pm 2|\alpha|) = -a_{n-2}[/tex] for [itex]r = \pm |\alpha|[/itex]. It follows from this that the coefficients for odd [itex]n[/itex] vanish.

Taking [itex]r = -|\alpha|[/itex] for positive integer [itex]\alpha = m[/itex], we find that [tex]
a_{2m} (2m)0 = -a_{2m - 2}[/tex] is satisfied for any [itex]a_{2m}[/itex], but we must take [itex]a_{2k} = 0[/itex] for [itex]0 \leq k \leq m - 1[/itex], leaving [tex]
y(x) = \sum_{k=m}^\infty a_{2k}x^{2k - m} = \sum_{k=0}^\infty a_{2k + 2m} x^{2k + m}[/tex] where [tex]
a_{2k + 2m}(2k + 2m)(2k + 2m - 2m) = a_{2k + 2m}(2k)(2k + 2m) = -a_{2(k +m - 1)}[/tex] is exactly the recurrence relation for the [itex]r = +m[/itex] case, [tex]
a_{2k}(2k)(2k + 2m) = -a_{2(k - 1)}.[/tex] It follows that [itex]J_{-m}(x) = C_mJ_m(x)[/itex] for some [itex]C_m \neq 0[/itex].

For non-integer [itex]\alpha[/itex] we have the particular linear combination [tex]
Y_\alpha = \frac{ J_\alpha\cos(\pi \alpha) - J_{-\alpha}}{\sin (\pi \alpha)}[/tex] and setting [itex]J_{-m} = (-1)^mJ_m[/itex] allows us to define [tex]
Y_m = \lim_{\alpha \to m} Y_{\alpha}.[/tex]
 

1. What is the Gamma function and what is it used for?

The Gamma function, denoted by Γ(z), is a mathematical function that extends the factorial function to complex and real numbers. It is used in various areas of mathematics, such as number theory, calculus, and statistics, and has applications in physics and engineering as well.

2. How is the Gamma function related to the factorial function?

The Gamma function is closely related to the factorial function, as it generalizes it to non-integer values. In fact, for positive integer values of z, Γ(z) is equal to (z-1)!, making it an extension of the factorial function.

3. What are Bessel functions and where are they used?

Bessel functions, denoted by Jn(z), are a family of solutions to the Bessel differential equation. They are used in many areas of physics and engineering, such as electromagnetism, fluid mechanics, and signal processing, to describe various physical phenomena.

4. How are Bessel functions different from other special functions?

Bessel functions are different from other special functions, such as trigonometric functions and exponential functions, in that they are defined by a differential equation rather than an algebraic equation. They also have unique properties, such as being orthogonal and having zeros at specific values.

5. Can Bessel functions be evaluated for complex or non-real values?

Yes, Bessel functions can be evaluated for complex or non-real values. However, they may not always have a simple closed form solution and may require numerical methods for evaluation. Additionally, for certain values of z, Bessel functions can have complex values, making them useful in describing oscillatory or damped physical systems.

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