Mixing
Convergence of $int_0^2frac{1}{x^alpha |ln^beta(x)|}$
$begingroup$
I know that $int_2^{+infty}frac{1}{x^alphaln^beta(x)}$ is
$i)$ Convergent if $alphagt1$, $forallbeta$
$ii)$Convergent if $betagt1$ for $alpha=1$
$iii)$Divegent if $alphalt1$,$forallbeta$
Now i am trying to understand what happens if i have same situation with interval other interval,say $(0,2)$.Can something similar be formulated for $int_0^2frac{1}{x^alpha |ln^beta(x)|}$?
calculus convergence improper-integrals
asked Jan 29 at 23:09
Turan NasibliTuran Nasibli
846
$endgroup$
add a comment |
$begingroup$
I know that $int_2^{+infty}frac{1}{x^alphaln^beta(x)}$ is
$i)$ Convergent if $alphagt1$, $forallbeta$
$ii)$Convergent if $betagt1$ for $alpha=1$
$iii)$Divegent if $alphalt1$,$forallbeta$
Now i am trying to understand what happens if i have same situation with interval other interval,say $(0,2)$.Can something similar be formulated for $int_0^2frac{1}{x^alpha |ln^beta(x)|}$?
calculus convergence improper-integrals
asked Jan 29 at 23:09
Turan NasibliTuran Nasibli
846
$endgroup$
1
$begingroup$
$(ln , x)^{beta}$ is not defined for $x<1$. If you replace $ln, x$ by $ln, |x|$ then the answer is: convergent for $alpha <1$, divergent for $alpha >1$, convergent for $alpha =1,beta >1$. divergent for $alpha=1,beta leq 1$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 23:18
$begingroup$
@KaviRamaMurthy yes,exactly.but can this formulation be applied to $int_0^2 frac{|ln^{alpha-{2over3}}(x)|}{x^alpha}$?.For question: math.stackexchange.com/questions/3091996/…
$endgroup$
– Turan Nasibli
Jan 29 at 23:24
$begingroup$
Your title and the question are at odds.
$endgroup$
– zhw.
Jan 30 at 0:22
add a comment |
$begingroup$
I know that $int_2^{+infty}frac{1}{x^alphaln^beta(x)}$ is
$i)$ Convergent if $alphagt1$, $forallbeta$
$ii)$Convergent if $betagt1$ for $alpha=1$
$iii)$Divegent if $alphalt1$,$forallbeta$
Now i am trying to understand what happens if i have same situation with interval other interval,say $(0,2)$.Can something similar be formulated for $int_0^2frac{1}{x^alpha |ln^beta(x)|}$?
calculus convergence improper-integrals
asked Jan 29 at 23:09
Turan NasibliTuran Nasibli
846
$endgroup$
I know that $int_2^{+infty}frac{1}{x^alphaln^beta(x)}$ is
$i)$ Convergent if $alphagt1$, $forallbeta$
$ii)$Convergent if $betagt1$ for $alpha=1$
$iii)$Divegent if $alphalt1$,$forallbeta$
Now i am trying to understand what happens if i have same situation with interval other interval,say $(0,2)$.Can something similar be formulated for $int_0^2frac{1}{x^alpha |ln^beta(x)|}$?
calculus convergence improper-integrals
calculus convergence improper-integrals
asked Jan 29 at 23:09
Turan NasibliTuran Nasibli
846
asked Jan 29 at 23:09
Turan NasibliTuran Nasibli
846
edited Jan 29 at 23:20
Turan Nasibli
asked Jan 29 at 23:09
Turan NasibliTuran Nasibli
846
asked Jan 29 at 23:09
Turan NasibliTuran Nasibli
846
asked Jan 29 at 23:09
Turan NasibliTuran Nasibli
846
846
1
$begingroup$
$(ln , x)^{beta}$ is not defined for $x<1$. If you replace $ln, x$ by $ln, |x|$ then the answer is: convergent for $alpha <1$, divergent for $alpha >1$, convergent for $alpha =1,beta >1$. divergent for $alpha=1,beta leq 1$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 23:18
$begingroup$
@KaviRamaMurthy yes,exactly.but can this formulation be applied to $int_0^2 frac{|ln^{alpha-{2over3}}(x)|}{x^alpha}$?.For question: math.stackexchange.com/questions/3091996/…
$endgroup$
– Turan Nasibli
Jan 29 at 23:24
$begingroup$
Your title and the question are at odds.
$endgroup$
– zhw.
Jan 30 at 0:22
add a comment |
1
$begingroup$
$(ln , x)^{beta}$ is not defined for $x<1$. If you replace $ln, x$ by $ln, |x|$ then the answer is: convergent for $alpha <1$, divergent for $alpha >1$, convergent for $alpha =1,beta >1$. divergent for $alpha=1,beta leq 1$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 23:18
$begingroup$
@KaviRamaMurthy yes,exactly.but can this formulation be applied to $int_0^2 frac{|ln^{alpha-{2over3}}(x)|}{x^alpha}$?.For question: math.stackexchange.com/questions/3091996/…
$endgroup$
– Turan Nasibli
Jan 29 at 23:24
$begingroup$
Your title and the question are at odds.
$endgroup$
– zhw.
Jan 30 at 0:22
1
1
$begingroup$
$(ln , x)^{beta}$ is not defined for $x<1$. If you replace $ln, x$ by $ln, |x|$ then the answer is: convergent for $alpha <1$, divergent for $alpha >1$, convergent for $alpha =1,beta >1$. divergent for $alpha=1,beta leq 1$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 23:18
$begingroup$
$(ln , x)^{beta}$ is not defined for $x<1$. If you replace $ln, x$ by $ln, |x|$ then the answer is: convergent for $alpha <1$, divergent for $alpha >1$, convergent for $alpha =1,beta >1$. divergent for $alpha=1,beta leq 1$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 23:18
$begingroup$
@KaviRamaMurthy yes,exactly.but can this formulation be applied to $int_0^2 frac{|ln^{alpha-{2over3}}(x)|}{x^alpha}$?.For question: math.stackexchange.com/questions/3091996/…
$endgroup$
– Turan Nasibli
Jan 29 at 23:24
$begingroup$
@KaviRamaMurthy yes,exactly.but can this formulation be applied to $int_0^2 frac{|ln^{alpha-{2over3}}(x)|}{x^alpha}$?.For question: math.stackexchange.com/questions/3091996/…
$endgroup$
– Turan Nasibli
Jan 29 at 23:24
$begingroup$
Your title and the question are at odds.
$endgroup$
– zhw.
Jan 30 at 0:22
$begingroup$
Your title and the question are at odds.
$endgroup$
– zhw.
Jan 30 at 0:22
add a comment |
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1
$begingroup$
$(ln , x)^{beta}$ is not defined for $x<1$. If you replace $ln, x$ by $ln, |x|$ then the answer is: convergent for $alpha <1$, divergent for $alpha >1$, convergent for $alpha =1,beta >1$. divergent for $alpha=1,beta leq 1$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 23:18
$begingroup$
@KaviRamaMurthy yes,exactly.but can this formulation be applied to $int_0^2 frac{|ln^{alpha-{2over3}}(x)|}{x^alpha}$?.For question: math.stackexchange.com/questions/3091996/…
$endgroup$
– Turan Nasibli
Jan 29 at 23:24
$begingroup$
Your title and the question are at odds.
$endgroup$
– zhw.
Jan 30 at 0:22