Is the Gamma Function of Negative Integers Defined?

[LagrangeEuler]

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
$$J_p(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+p+1)}(\frac{x}{2})^{2n+p}$$
For example there is relation
$$J_{-p}(x)=(-1)^pJ_p(x)$$
What is happening with $J_{-2}$
$$J_{-2}(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n+1)\Gamma(n-1)}(\frac{x}{2})^{2n-2}$$
for term $n=0$?

 

[fresh_42]

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
$$J_p(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+p+1)}(\frac{x}{2})^{2n+p}$$
For example there is relation
$$J_{-p}(x)=(-1)^pJ_p(x) \quad (*)$$
What is happening with $J_{-2}$
$$J_{-2}(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n+1)\Gamma(n-1)}(\frac{x}{2})^{2n-2}$$
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.

 

[mathman]

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.

 

[LagrangeEuler]

In few books I found, for example
$$J_{-2}(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n-1)\Gamma(n+1)} (\frac{x}{2})^{2n-2}=\sum^{\infty}_{n=2}\frac{(-1)^n}{\Gamma(n-1)\Gamma(n+1)} (\frac{x}{2})^{2n-2}$$
and this is like that I took that gama values of negative arguments are $\infty$. Could you give me explanation for this?

 

[fresh_42]

In few books I found, for example
$$J_{-2}(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{\Gamma(n-1)\Gamma(n+1)} (\frac{x}{2})^{2n-2}=\sum^{\infty}_{n=2}\frac{(-1)^n}{\Gamma(n-1)\Gamma(n+1)} (\frac{x}{2})^{2n-2}$$
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.

 

[LagrangeEuler]

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.

 

[mathhabibi]

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.

 

[mathhabibi]

Also, the index of the bessel function can't be a negative integer, see here

 

[pasmith]

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 $+\infty$ and $-\infty$ as being distinct because the real line is ordered. The complex plane is not.

 

[pasmith]

You can't put $\alpha = -|m|$ for $m \in \mathbb{Z}$ directly into the definition $$J_\alpha(x) = \sum_{n=0}^\infty \frac{(-1)^n}{n!\Gamma(n + \alpha + 1)}\left(\frac{x}{2}\right)^{2n+\alpha}$$ but that doesn't mean that $J_{-m}$ is not defined.

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

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

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