求使积分$I(\alpha,a) = \int_0^\infty \frac{x^{\alpha-1}dx}{e^{a x}-1}$存在的a与α值

Hey there! Let's break down this convergence problem step by step—we can split the integral into two parts to check behavior near 0 and near infinity separately, since those are the two spots where the integrand might blow up or decay too slowly.

We can split the integral into two segments to analyze endpoint behavior independently:
$$I(\alpha,a) = \int_0^\varepsilon \frac{x^{{\alpha-1}dx}{e}{a x}-1} + \int_\varepsilon^\infty \frac{x^{{\alpha-1}dx}{e}{a x}-1}$$
where $\varepsilon$ is any small positive real number.

1. Convergence as $x \to 0^+$ #

When $x$ approaches 0, use the Taylor expansion of the exponential function: $e^{ax} = 1 + ax + \frac{(ax)^2}{2} + \dots$. The dominant term gives $e^{ax}-1 \approx ax$, so the integrand simplifies to:
$$\frac{x^{{\alpha-1}}{e}{ax}-1} \approx \frac{x^{\alpha-1}}{ax} = \frac{1}{a}x^{\alpha-2}$$

For the integral $\int_0^\varepsilon x^p dx$ (where $p = \alpha-2$) to converge, we need $\text{Re}(p) > -1$. Substituting $p$ gives:
$$\text{Re}(\alpha-2) > -1 \implies \text{Re}(\alpha) > 1$$

If $\text{Re}(\alpha) \leq 1$, the integral near 0 will diverge, making the entire integral $I(\alpha,a)$ undefined.

2. Convergence as $x \to \infty$ #

As $x$ tends to infinity, $e^{ax}$ grows far faster than any polynomial, so $e^{ax}-1 \approx e^{ax}$. The integrand becomes:
$$\frac{x^{{\alpha-1}}{e}{ax}-1} \approx x^{\alpha-1}e{-ax}$$

Since $a>0$ (given in the problem), $ax$ is a positive real number going to infinity. The exponential decay of $e^{-ax}$ will overpower any polynomial growth (even from the complex $\alpha$ term). Specifically, $|x^{\alpha-1}e{-ax}| = x^{{\text{Re}(\alpha)-1}e}{-ax}$, and the integral $\int_\varepsilon^\infty x^{{\text{Re}(\alpha)-1}e}{-ax}dx$ always converges—substitute $t=ax$ and it becomes $\frac{1}{a^{\text{Re}(\alpha)}} \int_{a\varepsilon}^\infty t^{{\text{Re}(\alpha)-1}e}{-t}dt$, a segment of the Gamma function that converges at infinity for any $\text{Re}(\alpha)$.

Final Conclusion #

The integral $I(\alpha,a)$ exists and is finite if and only if:

• $a > 0$ (your given condition, a necessary base requirement)
• $\alpha \in \mathbb{C}$ with $\text{Re}(\alpha) > 1$

As a bonus: When these conditions are met, the integral can be expressed using the Riemann Zeta function and Gamma function: $I(\alpha,a) = \frac{\Gamma(\alpha)\zeta(\alpha)}{a^\alpha}$.

内容的提问来源于stack exchange，提问作者LorenzoCasalena93