Convergence or divergence of improper integral I want to check if the following integral converge or diverge.
$\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt, \alpha ,\beta\in \mathbb{R}}$
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Do we have to take cases for $\alpha$ and $\beta$ ?
If $0<\beta\leq \alpha-2$ we get $$\ln t\leq t \Rightarrow \left (\ln t \right )^{\beta}\leq t^{\beta} \leq t^{\alpha-2} \Rightarrow \frac{\left (\ln t\right )^{\beta}}{t^{\alpha}}\leq \frac{1}{t^2}$$ and since $\int_1^{\infty}\frac{1}{t^2}\, dt=\left [-\frac{1}{t}\right ]_1^{+\infty}=0+1=1<\infty$ the integral $\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt}$ converges also, right?
Could you give me a hint for the other cases?
 A: $$\int_1^\infty\frac{\log^\beta(t)}{t^\alpha}dt; \alpha,\beta\in\mathbb{R}$$
Let $t\rightarrow e^x:$
$$\int_0^\infty x^\beta e^{(1-\alpha)x}dx$$
For $\alpha \leq 1$, the integral will be divergent regardless of the value of $\beta$:
$$\int_0^\infty x^\beta e^{(1-\alpha)x}dx\geq \int_0^\infty x^\beta dx\rightarrow +\infty
 $$
For $\alpha > 1$,the integral can be seem like a Laplace Transform, in wich $s=\alpha-1$:
$$\displaystyle\lim_{s\rightarrow\alpha-1}\int_0^\infty x^\beta e^{-sx}dx=\displaystyle\lim_{s\rightarrow\alpha-1}\mathcal{L}[x^\beta]=\displaystyle\lim_{s\rightarrow\alpha-1}\frac{\Gamma(\beta+1)}{s^{\beta+1}}=\frac{\Gamma(\beta+1)}{\left(\alpha-1\right)^{\beta+1}}; \beta>-1$$
Therefore, the integral will converge for $\alpha>1$ and $\beta>-1$. Otherwise, it will diverge.
Edit: to prove this without resorting to the Laplace Transform, one may show that:
$$\underbrace{\int_0^\infty x^\beta e^{-sx}dx}_{sx\rightarrow z}=\frac{1}{s^{\beta+1}}\int_0^1 z^\beta e^{-z}dz+\frac{1}{s^{\beta+1}}\int_1^\infty z^\beta e^{-z}dz$$
For the first integral:
$$\int_0^1 z^\beta e^{-z}dz\leq\int_0^1 z^\beta dz=\left[\frac{z^{\beta+1}}{{\beta+1}}\right]^1_0=\frac{1}{\beta+1}$$
Hence, $\beta>-1$ for the integral converge.
For the second integral, let $\beta<N; N\in\mathbb{N}$:
$$\int_1^\infty z^\beta e^{-z}dz\leq\int_1^\infty z^N e^{-z}dz\overbrace{=}^{IBP}1+N\int_1^\infty z^{N-1} e^{-z}dz$$
Which can be further proven by Induction that will also converge.
