The question is: exists a natural number $n \geq 2$such that

$$ \displaystyle\int_{0}^{+ \infty} \displaystyle\frac{\ln r}{(1 + r^2)^{n}} r dr< \infty ?$$

I am trying to do this : i know that exists $A>0$ such that $ln r < r , \forall r > A$

note that $\displaystyle\int_{0}^{+ \infty} \displaystyle\frac{\ln r}{(1 + r^2)^{n}} r dr = \displaystyle\int_{0}^{A} \displaystyle\frac{\ln r}{(1 + r^2)^{n}} r dr + \displaystyle\int_{A}^{+ \infty} \displaystyle\frac{\ln r}{(1 + r^2)^{n}} r dr$

The second integral converges for all $n \geq 2$ because for $r > A $ we have $\displaystyle\frac{\ln r}{(1 + r^2)^{n}} r < \displaystyle\frac{r^2}{(1+r^2)^n} < \displaystyle\frac{r^2}{r^4} = \displaystyle\frac{1}{r^2}$ .

and the integral $\displaystyle\int_{A}^{+ \infty} \displaystyle\frac{1}{r^2} dr$ converges. The problem is the first integral..

Someone can give me a hint? Any help is welcome .

Thanks in advance

  • 3
    $\begingroup$ For the first: the denominator is harmless (tends to $1$), the numerator is $r\ln r$, how does that behave? $\endgroup$ – Daniel Fischer Nov 6 '13 at 20:15

As Daniel fischer pointed out, we have to investigate the convergence of $$I:=\int_0^1 r\ln r\mathrm dr.$$ Defining $f(r):=r\ln r$ for $r\in (0,1]$, we notice that $\lim_{r\to 0}f(r)= \lim_{t\to -\infty}t\cdot e^{-t}=0$, hence the function $f$ is continuous and bounded on the interval $(0,1]$. This proves the convergence of $I$.


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