I have a couple of questions in relation to a document I'm reading regarding Bessel Functions/the Gamma Function and will post them in a second, all listed below,

Question

mendicant_bias · Answer

1.) Why is Gamma written solely as a function of x when it appears to be a function of two variables?

2.) How is I_1 improper if x>1?  If an integrand of one or more than one variable is continuous either before or after integrating, but is discontinuous in only one of those states, not both, how does that affect things? (ex. shortly)

mendicant_bias · Answer

attachment

mendicant_bias · Answer

@SithsAndGiggles

anonymous · Answer

Regarding the first question, $t$ is just an integration variable, or rather a dummy variable. In the same way
$$f(x)=\int_0^x f(t)\,dt$$
is a function of $x$, so is
$$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,dt$$
(Well, not exactly the same way, but the point is that $x$ is a parameter for the integral, while $t$ is the integrating variable.)

anonymous · Answer

For the second question, $I_1$ is said to be improper when $x\color{red}<1$. When this is the case,
$$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,dt=\int_0^\infty \frac{dt}{t^{1-x}e^t}$$
has a singularity at $t=0$.

anonymous · Answer

*upper limit should be 1, not $\infty$, sorry

mendicant_bias · Answer

The singularity is due to division by zero, right?

anonymous · Answer

Correct

mendicant_bias · Answer

(Thinking about how to ask these questions, might take some time)

mendicant_bias · Answer

Is the dummy variable situation like convolution? I'm feeling it is, but am having trouble putting two and two together. 

As in, pretend this expression has already been integrated and is still in the above form, upon evaluating all original x's would remain in place, and all t's would be replaced with x, yes?

anonymous · Answer

Here's a simple example to demonstrate what the result would be. Suppose you have
$$g(x)=\int_0^1 t^x\,dt$$
In computing this integral, we fix $x$, so by the power rule we get
$$g(x)=\left.\frac{t^{x+1}}{x+1}\right|_0^1=\frac{1}{x+1}-0$$
So $\Gamma(x)$ is a function of $x$ that involves an integral with respect to $t$. Integrating over an interval will give you an expression in terms of $x$ only. ($t$ is taken care of exactly like how you would have no variables remaining in the computation of any definite integral with respect to a single variable, i.e. $\int_a^b f(t)\,dt$ is some expression in terms of the limits, not $t$.)