# For which $\alpha$, $\beta$ does $\int\limits_1^{\infty} x^{\alpha} \cdot (\ln x)^\beta dx$ converge? [duplicate]

For which $\alpha$ and $\beta$ does the following integral converge ?: $$\int_{1}^{\infty}x^{\alpha}\,\ln^{\beta}\left(x\right)\,\mathrm{d}x$$

Here is my analysis:

I noticed that the function under the integral is of the following type: $\,\mathrm{f}\left(x\right)\,\mathrm{g}\left(x\right)$. If we had $\,\mathrm{g}\left(x\right) = 1$, so $\displaystyle{\int_{1}^{\infty}x^{\alpha}\,\mathrm{d}x}$ converges when $\alpha < -1$.

When $\beta > 0$ the power $\alpha$ should somehow overcome that value. If $\,\mathrm{g}\left(x\right)$ was simply $\,\mathrm{g}\left(x\right) = x^{\beta}$, then the relationship between $\alpha$ and $\beta$ would have been linear, so $\alpha + \beta < -1$. I cannot come up with the relationship in case when $\,\mathrm{g}\left(x\right) = \ln^{\beta}\left(x\right)$.

• Is it really the same? – Egor Okhterov Jul 16 '16 at 17:19

Recall that $\log(x)\le x-1<x$ for all $x>0$. Then, for any $\gamma>0$, $\log(x)<\frac{x^\gamma}{\gamma}$. Therefore, we have for any $\gamma>0$ and $\beta>0$ we have
$$\log^\beta(x) <\frac{x^{\beta\gamma}}{\gamma^\beta} \tag 1$$
Since $(1)$ is true for any positive $\gamma$, it is true for $0<\gamma\le -\frac{1+\alpha}{\beta}$ for $\beta>0$ and $\alpha <-1$.
Therefore, the integral converges for $\beta\ge 0$ and $\alpha<-1$ and diverges for $\alpha \ge -1$ (since $x^\alpha \log^\beta(x) \ge 1/x$ for sufficiently large $x$ when $\alpha \ge -1$ and $\beta \ge 0$ ).
The case for $\beta<0$ is left as an exercise for the reader.
• Why $\log(x)<\frac{x^\gamma}{\gamma}$? – Egor Okhterov Jul 28 '16 at 19:12
• @Pixar $\log(y)\le y-1<y$. Now, let $y=x^\gamma$. – Mark Viola Jul 28 '16 at 19:21