Question

improper integral ...

Answer 1

\[\int\limits_{0}^{1} \frac{ \ln(x)dx }{ x ^{p} } \]
Determine which values of p the following integrals converge

Answer 2

I think integration by parts may help, if you're allowed to use it.

Answer 3

try rewriting as x^-p log(x) then integrating. Parts would be helpful, pick a u.

Answer 4

u=ln(x) ?

Answer 5

du/dx = 1/x
du = dx/x

Answer 6

yes

Answer 7

i got,
f= ln(x)
df = 1/x dx

dg = x^-p
g = x^(1-p) / (1-p)

Answer 8

x^(1-p) log x / (1-p) - (1/1-p) * integral x^-p dx

this should set it up, ill try to draw it if you cannot understand it

Answer 9

i dont really get how you found that

Answer 10

okay, ill work through it, sorry for late reply had early lecture.

find soln to
\[\int\limits_{0}^{1}\frac{ lnx }{ x^p }\]

rewrite as

\[\int\limits(\log(x)x^{-p}\]

integrate by parts
\[\int\limits uv' = uv - \int\limits u'v\]

\[u = \log(x) \]
\[du= \frac{ 1 }{ x }dx\]
\[v' = x^{-p} dx\]
\[v = \frac{ x ^{-1}p }{ 1-p }\]

v is found by integrating v'
\[v' = x^{-p} -> \int\limits v' = \int\limits x^{-p}\]
\[\int\limits \frac{ x^{a+1} }{ a+1 } = \int\limits x^{-p}dx = \frac{ x^{-p-1} }{ -p+1 } = \frac{ x^{-1}p }{ 1-p }\]

plug into parts
\[\int\limits\limits uv' = uv - \int\limits\limits u'v\]

manipulate

\[\log(x) (\frac{ x^{-1} p}{1-p }) = \frac{ x^{1-p} logx}{ 1-p }\]

\[\frac{ x^{1-p} \log(x)}{ 1-p }-\frac{ 1 }{ 1-p } \int\limits x^{-p} = \frac{ x^{1-p} \log(x)}{ 1-p }-\frac{ 1 }{ 1-p }(\frac{ x^{-1} p}{1-p })= \frac{ x^{1-p} \log(x)}{ 1-p }-\frac{ x^{1-p} }{ (1-p)^2}\]


(forgot my +C)
then plug in your bounds and solve