Question

Can anyone help me with this?

For which \[p \in R \] is the function

\[f _{p} (x)= \frac{ 1 }{ (x-1)\log ^{p}_{}(x) } \]
improperly Riemann integrable on the interval \[\left[ e,\infty \right] \] ?

Answer 1

What would you guess be?
What definitions apply here? Any relevant theorems?

Answer 2

for p=0, it is not.

Answer 3

Could you explain me how?

Answer 4

Notice that
\[
\lim_{x->\infty} \frac {x \ln^p(x) }{(x-1) \ln^p(x)}=1

\]

Answer 5

Now for
\[
\int_e^b \frac{dx}{ x \ln^p(x)}\\
u=\ln(x)\\
du =\frac {dx} x\\
\int_e^b \frac{dx}{ x \ln^p(x)}=\int_1^{\ln(b)}\frac {du}{u^p}
\]
The last integral converges if \( p>1\) and diverges if \( p\le1\)
Hence the original integral do the same by the limit convergence theorem.

Answer 6

http://math.duke.edu/~cbray/Stanford/2003-2004/Math%2042/limitcomp.pdf

Answer 7

Did you understand the proof?

Answer 8

yeah..sort of :) thank you so much... :)

Answer 9

YW

Answer 10

can you help for this as well ?

For which p∈R is the infinite series
\[\sum_{k=2}^{\infty} =\frac{ k(-1)^{k} }{ (k-1)^{^{2}}\log ^{^{p}}(k) }\]

convergent? conditionally convergent? absolutely convergent?

Answer 11

The rules specify that you should post it as a new question.

Answer 12

ohh..right.. sorry about that..