Let $\alpha>0$ and define $f(x)=\ln(x)/(x-1)^{\alpha}$ for all $x>1$.

Before I got this problem, I was asked to determine the values of $\alpha$ such that for each of $I_{1}:=\int_{2}^{\infty}f(x)\,\mathrm{d}x$ and $I_{2}:=\int_{1}^{2}f(x)\,\mathrm{d}x$ converge.

Let's start with $I_{1}$. Since $\ln(x)\leq (x/2)+\ln(2)-1$ for all $x\in [2,\infty)$, so we have $$I_{1}=\int_{2}^{\infty}f(x)\, \mathrm{d}x\leq \int_{1}^{\infty}\frac{m+1}{2m^{\alpha}}\, \mathrm{d}m+\int_{1}^{\infty}\frac{\ln(2)-1}{m^{\alpha}}\, \mathrm{d}m. $$ We see that the second term converges if $\alpha >1$. Using the integration by parts in order to analyze the first term, we see that $$\int_{1}^{\infty}\frac{m+1}{2m^{\alpha}}\, \mathrm{d}m=\lim_{n\to\infty}\left [ \frac{1}{2m^{\alpha}} \right ]_{0}^{n}+\frac{1}{\alpha-1}\int_{1}^{\infty}\frac{1}{m^{\alpha-1}}\,\mathrm{d}m $$ which converges if $\alpha>2$. Altogether, the integral $I_{1}$ converges if $\alpha>2$.

The similar method can be applied with another integral $I_{2}$, where $\ln(x)< x-1$ for all $x\in (1,2]$, $$ I_{2}=\int_{1}^{2}f(x)\, \mathrm{d}x< \int_{1}^{2}\frac{x-1}{(x-1)^{\alpha}}\, \mathrm{d}x=\int_{0}^{1}\frac{1}{m^{\alpha-1}}\, \mathrm{d}m $$ which converges if $\alpha<2$. Now I have no idea what to do for the last integral, $I:=I_{1}+I_{2}$. Does this problem cheat me?

  • $\begingroup$ wouldn't that just mean the last integral never converges? $\endgroup$ Apr 28, 2015 at 22:39
  • $\begingroup$ I do not understand why the problem is saying about determining the values of $\alpha$ that make the integral $I$ converge (that is, it is assumed to converge with the unknown values of $\alpha$, which is our goal to find). I even have thought that it's testing me. $\endgroup$
    – UnknownW
    Apr 29, 2015 at 0:57

1 Answer 1


Your analysis of $I_2$ is fine, because the point is that $\log{x}$ has a simple zero at $x=1$, near which it looks like $x-1$.

For $I_1$, you can do a lot better. You need to know that $\log{x} = o(x^{\varepsilon})$ for every $\varepsilon>0$; this you can show as a consequence of $y^{n} e^{-k y} \to 0 $ as $y \to \infty$ for any $k>0$. Then sticking this in the integral, you have $$ \int_2^{\infty} \frac{\log{x}}{(x-1)^{\alpha}} \, dx < \int_2^{\infty} \frac{Kx^{\varepsilon}}{(x-1)^{\alpha}} \, dx. $$ It is again easy to check that this integral converges for $\varepsilon-\alpha < -1$, i.e. $\alpha > 1+\varepsilon$. Since it suffices that this inequality holds for some $\varepsilon>0$, the condition is in fact $\alpha>1$.

Finally, $I_1$ converges exactly when $\alpha>1$ and $I_2$ converges exactly when $\alpha<2$ hence $I$ converges exactly when $1<\alpha<2$.

  • $\begingroup$ +1 to this nice solution. I took it upon myself to add a concluding sentence (just delete it if you do not want it). $\endgroup$
    – Did
    Apr 29, 2015 at 12:01

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