I want to check if the following integral converge or diverge.

$\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt, \alpha ,\beta\in \mathbb{R}}$


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?

  • 2
    $\begingroup$ For $\beta \geq 0$, the behavior depends mainly on $\alpha$ (divergence for $\alpha < 1$, convergence for $\alpha > 1$). For $\beta < 0$ the integral can diverge both in 1 and $\infty$, so I would try to get an equivalent of the integral at 1 and $\infty$. $\endgroup$
    – jvc
    Jan 18 at 10:07
  • $\begingroup$ For $\beta\geq 0$ and $\alpha >1$ do we show the convergence similar as I did it above? @jvc $\endgroup$
    – Mary Star
    Jan 18 at 12:24

$$\int_1^\infty\frac{\log^\beta(t)}{t^\alpha}dt; \alpha,\beta\in\mathbb{R}$$

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.

  • $\begingroup$ In the case of $\alpha>1$ could we do that also in an other way without Laplace Transform ? $\endgroup$
    – Mary Star
    Jan 18 at 18:10
  • $\begingroup$ Having now the integral $\int_0^\infty x^\beta e^{(1-\alpha)x}dx$ couldn't we find bounds (upper or lower) to show convergence or divergence? (I mean instead of using the Laplace Transform.) $\endgroup$
    – Mary Star
    Jan 21 at 20:46
  • 1
    $\begingroup$ I just edited the post with another approach. I'm not sure if it will fully address your question, but that’s my best shot as someone with no formal mathematical background. I hope it helps you. In any case, if you still have questions, I will try my best to answer it. $\endgroup$ Jan 21 at 23:07
  • $\begingroup$ Why does the first integral converge only when $\beta>-1$? The fraction can also be negative but a real number, can it not? $\endgroup$
    – Mary Star
    Jan 21 at 23:58
  • $\begingroup$ No, it can't. Because when $\beta<-1$, $z^{\beta+1}\rightarrow -\infty$ at $z\rightarrow 0$ $\endgroup$ Jan 22 at 0:02

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