Is the integral $\int_0^{+\infty} x^{2a-1} e^{-4x} dx$ bounded?

Question

Let $0<a<\frac12$. I am trying to understand if there exists a constant $C>0$ (possibly depending on $a$) such that integral $$\int_0^{+\infty} x^{2a-1} e^{-4x} dx$$ is so that $$\int_0^{+\infty} x^{2a-1} e^{-4x} dx\le C.$$

I see that there are no problems at infinity but probably there is a problem at $0$ when $a\to 0$.

I have also computed directly the integral, which returns $$-\frac{\Gamma(2a, 4x)}{4a},$$ where $\Gamma(2a, 4x)$ stands for the incomplete Gamma function. However, I can not manage the computations to prove if it is bounded or not.

Anyone could please help me?

Answer 1

On $0 < x < 1$, you have $$x^{2a-1}e^{-4x} > x^{2a-1}e^{-4}.$$

Therefore, $$\int_0^\infty x^{2a-1}e^{-4x}\,dx \geq \int_0^1x^{2a-1}e^{-4x}\,dx \geq \int_0^1x^{2a-1}e^{-4}\,dx = \frac{1}{2a}e^{-4}$$

This lower bound tends to $\infty$ as $a \to 0$, so there is no such $C$ independent of $a$.

Answer 2

The integral converges around $0$ if $x^{2a-1+1}>0$ that is if $2a>0\Rightarrow a>0.$ At infinity it is bounded for arbitrary $k$ as $e^{-x}$ is sufficient enough for it to make it converge. Therefore it converges for $a>0.$

Answer 3

$\newcommand{\M}[1]{\left\{ \mathcal{M} #1 \right\} }$ About your computation: you don't need the incomplete Gamma, and in fact, as indicated by @Jean-Claude Arbaut, your evaluation is incorrect. The fastest way to compute this is to use the table of Mellin transforms and their properties: $$\int_{0}^{\infty}x^{2a-1}e^{-4x}dx=\left\{ \mathcal{M} e^{-4x} \right\}(2a)=4^{-2a}\left\{\mathcal{M} e^{-x} \right\}(2a)=4^{-2a}\Gamma(2a)$$

Here we have used:

1. The definition of the Mellin transform: $$\M{ f(x)}(s)=\int_{0}^{\infty}x^{s-1}f(x)dx$$

2. The property $$\M{ f(ax) }=a^{-s}\M{f(x)}$$

proven by simple change of variables and integration by parts.

3. The fact that, by definition of the Mellin transform and Gamma function, $$\M{e^{-x}}=\Gamma(s)$$