integrationorthogonal-polynomialsspecial functions
After a fair bit of effort, I managed to prove that
$$\int_0^\infty t^\alpha \exp(-t) L_n^{\alpha+1}(t)\mathrm dt=\Gamma(\alpha+1)$$
where $L_n^\alpha (t)$ is a generalized Laguerre polynomial, with the usual restriction of $\Re \alpha > -1$. My proof was rather complicated (involving lots of integration by parts); I'm wondering if there may be a simpler way of proving this proposition.
Very roughly speaking: Laguerre polynomials look like the family of trigonometric functions ($\sin nx$ or $\cos nx$) in the region where the weight is concentrated. That is, they have moderate size and slowly increasing oscillation with $n$. Outside of this region, they look pretty much like any random collection of polynomials.
When $\alpha=0$, the weight $e^{-x}$ is concentrated near $0$. On the graph below the weight is shown in black, and the first five Laguerre polynomials $L_n$ are in various colors. You can see that they look like $1-\sin nx$ near $0$, and then wildly diverge when the weight becomes small.
For comparison, I did the same with $\alpha=2$. The polynomials have civilized appearance where the weight is not too small, roughly in the interval $[1,4]$. They wildly diverge both to the left and to the right.
do these polynomials still converge using the Least Square approximant
When $\alpha=0$, the Laguerre polynomial form on orthonormal basis of $L^2([0,\infty),e^{-x})$ (reference here). Therefore, for any function $f\in L^2([0,\infty),e^{-x})$ we have $\sum c_n L_n\to f$ in the norm, where $c_n=\langle f,L_n\rangle$.
When $\alpha>0$, the Laguerre polynomial form on orthogonal (but not normalized) basis* of $L^2([0,\infty),x^\alpha e^{-x})$. Therefore, for any function $f\in L^2([0,\infty),x^\alpha e^{-x})$ we have $\sum c_n L_n^{(\alpha)}\to f$ in the norm, where $c_n=\langle f,L_n^{(\alpha)}\rangle / \|L_n^{(\alpha)}\|$. (The value of $\|L_n^{(\alpha)}\|$ is given on Wikipedia).
(*) Disclaimer: I don't have a reference for the the completeness of $\{L_n^{(\alpha)}\}$, but I just put a bounty on unanswered question On the completeness of the generalized Laguerre polynomials in the hope someone does. Or you can consult the books G. Polya Orthogonal polynomials and G. Sansone Orthogonal functions, in case they treat $L_n^{(\alpha)}$.
There is no trouble at the end of the day, thanks to the fact that there are two, not just one, branch points at $p=0, \tfrac{1}{\alpha+1}$. One can take $[0,\tfrac{1}{\alpha+1}]$ to be the branch cut of the integrand in Eq. (3) and thereby make it analytic except along this cut. At the same time, the contour $\gamma$ should be chosen such that it encloses the whole branch cut.
Notice that in Eq. (2), the integral in $t$ is undefined for ${\rm Re}(\tfrac{1}{p})\ge\alpha+1$, and the branch cut falls within this region. It is consistent with our previous observation that the contour $\gamma$ should avoid the cut.
Now $\gamma$ can be deformed into an arbitrarily large circle. It certainly cannot smoothly pass through the simple pole at $p=1$ but has to leave behind a small clockwise contour around it. The contribution of this contour can be evaluated using the residue at the pole, and the result is $2\pi i \alpha^{-\nu-1}\Gamma(\nu+1)$. (I actually suspect that your definition of the Marcum Q-function is missing a factor of $1/2\pi i$.)
One still has to evaluate the integral over the large circle, which doesn't vanish. This can be done by making a change of variable $z = 1/p$. Then the integral will be turned into one over a contour encircling an essential singularity at $z=0$. Evaluating the residue will give the remaining part of the result.
Update: On the above, I chose the branch cut of a function of the form $(p-p_{0})^{\alpha}(p-p_{1})^{\beta}$ such that it exists only between the two algebraic branch points. However, we in general ought to have two infinitely long cuts emanating from the branch points. To have a single cut only between the two branch points, the two original branch cuts should be aligned so that they coincide and must somehow cancel each other in the overlapping region. It is possible only when $\alpha+\beta$ is an integer.
To illustrate this point, suppose that $p_{0}, p_{1} \in \mathbb{R}$ and that $p_{0}<p_{1}$. We take the branch cuts to be $[p_{0},\infty)$ and $[p_{1},\infty)$. Then, consider polar representations \begin{equation} p-p_{0} = r_{0} e^{i\theta_{0}}, \quad p-p_{1} = r_{1} e^{i\theta_{1}}. \end{equation} The angles $\theta_{0}$ and $\theta_{1}$ have discontinuous jumps $\Delta\theta_{0}=\Delta\theta_{1}=2\pi$ at the respective branch cuts, i.e., $[p_{0},\infty)$ and $[p_{1},\infty)$. Now let's consider \begin{equation} (p-p_{0})^{\alpha}(p-p_{1})^{\beta} = r_{0}^{\alpha} r_{1}^{\beta} e^{i(\alpha\theta_{0}+\beta\theta_{1})}. \end{equation} At the overlapping region, i.e., $[p_{1},\infty)$, the argument of this quantity jumps by \begin{equation} \alpha\Delta\theta_{0} + \beta\Delta\theta_{1} = 2\pi (\alpha+\beta). \end{equation} Therefore, when $\alpha+\beta$ is an integer, there is no discontinuity in $e^{i(\alpha\theta_{0}+\beta\theta_{1})}$ and hence in $(p-p_{0})^{\alpha}(p-p_{1})^{\beta}$ across $[p_{1},\infty)$. Two overlapping cuts in this region cancel each other and the function is made analytic there.
This kind of cancellation can happen whenever branch cuts from any number of algebraic branch points coincide and the powers add up to an integer.
Best Answer
Here, it is useful to know the generating function ($|z|<1$) $$ \frac{\exp\left(-\frac{xz}{1-z}\right)}{(1-z)^{\alpha+1}} = \sum_{n=0}^\infty L^\alpha_n (x) z^n .$$ Thereby, we can evaluate the integral for all $n$ simultaneously $$g=\sum_n z^n \int_0^\infty dt\, t^\alpha e^{-t} L_n^{\alpha+1}(t) = \int_0^\infty dt\, t^\alpha e^{-t} \frac{\exp\left(-\frac{tz}{1-z}\right)}{(1-z)^{\alpha+2}}. $$ The integral can be brought onto the integral for the $\Gamma$ function by a change of variables $$(1-z)^{\alpha+2} g= \int_0^\infty dt\, t^\alpha e^{ - \frac{t}{1-z} } = (1-z)^{\alpha+1} \int_0^\infty dt \, t^\alpha e^{ -t} =(1-z)^{\alpha+1} \Gamma(\alpha+1). $$ Thus, we have $g = \Gamma(\alpha+1)/(1-z)$ and $$\int_0^\infty dt\, t^\alpha e^{-t} L_n^{\alpha+1}(t) = \Gamma(\alpha+1) \qquad \forall n.$$