0

Let $\alpha>0$ and

$$ f(x)=\frac{\ln x}{(x-1)^{\alpha}} $$

for $x>1$

i found that for $\int_2^{\infty}f(x) dx$ the integral is convergence for $\alpha > 2$

but for which $\alpha$ is $f(x)$ convergence for $\int_1^2f(x) dx$ and $\int_1^{\infty}f(x) dx$

Jamgreen
  • 809

2 Answers2

0

Note that $\ln(x)$ is eventually smaller than any positive power.

We have that $$\int^\infty_1 \frac{\ln n}{n^\alpha} < \sum^\infty_{n=1} \frac{\ln n}{n^\alpha}$$

We know that the RHS is convergent for any $\alpha > 1$, because $\sum^\infty_{n=1} \frac{1}{n^\alpha}$ is convergent and $\ln(x)$ is eventually smaller than any positive power.

wythagoras
  • 25,026
  • ... and so does $(x-1)^{\alpha}$? I mean, I see this coming into play, but I do not see it coming crucially into play. –  May 04 '15 at 15:50
  • i have found that $\frac{\ln x}{(x-1)^{\alpha}} \leq \frac{x-1}{(x-1)^{\alpha}}=\frac{1}{(x-1)^{\alpha-1}}$ but i only know that integral $\int_0^1$ convergence for $\alpha < 2$ not for $\int_1^2$ – Jamgreen May 04 '15 at 15:54
  • @G.Sassatelli I hope that my edit helps – wythagoras May 04 '15 at 15:58
  • @wythagoras I am very sorry: for some reason I read "number" instead of "power". This indoubtedly helps, and it helped back then too (there's a type-o on the LHS). It does not solve the issue around $1$, though. –  May 04 '15 at 16:06
  • @G.Sassatelli It does. Suppose we have $\alpha = 1.0001$. For sufficiently large $n$ we have $\ln n < n^{0.00009}$, now we can use that. For the $n$ for which it doesn't hold the sum will be finite anyways. – wythagoras May 04 '15 at 16:09
0

$$\int_1^2 \frac{\log x}{(x-1)^{\alpha}}\,dx=\lim_{\epsilon \to 0+}\int_{1+\epsilon}^2 \frac{\log x}{(x-1)^{\alpha}}\,dx$$

For $\alpha =1$, the integral clearly diverges logarithmically.

Now, for $\alpha \ne 1$, integration by parts reveals that

$$\begin{align} \int_{1+\epsilon}^2 \frac{\log x}{(x-1)^{\alpha}}\,dx&=\left(\frac{1}{1-\alpha}(x-1)^{1-\alpha}\log x\right)|_{1-\epsilon}^2-\frac{1}{1-\alpha}\int_{1+\epsilon}^2 \frac{(x-1)^{1-\alpha}}{x} dx\\\\ &=\frac{1}{1-\alpha}(\log 2-\epsilon^{\alpha}\log(1-\epsilon))-\frac{1}{1-\alpha}\int_{1+\epsilon}^2 \frac{(x-1)^{1-\alpha}}{x} dx \end{align}$$

The first term on the right-hand side clearly converges as $\epsilon \to 0^+$ for all $0<\alpha$, while the integral clearly converges as $\epsilon \to 0^+$ for all $\alpha<1$. Thus, the integral $\int_1^2 \frac{\log x}{(x-1)^{\alpha}}\,dx$ converges only when $0<\alpha<1$ and diverges otherwise.

NOTE:

Since $\int_2^{\infty} \frac{\log x}{(x-1)^{\alpha}}\,dx$ converges only when $alpha >1$, the integral $\int_1^{\infty} \frac{\log x}{(x-1)^{\alpha}}\,dx$ diverges for all $\alpha$.

Mark Viola
  • 179,405