Mixing
Does $int_e^infty{frac{ln(x)^alpha}{x}dx}$ exists?
$begingroup$
My task is to determine wetherthe integral $int_e^infty{frac{ln(x)^alpha}{x},dx}$ does exist or not in depencence of $alphain mathbb{R}$.
For that I wrote $b$ instead of $infty$ and then calculated the integral using substitution, which should be $frac{1}{alpha+1}((ln(b))^{alpha+1}-1)$. Now for the improper integral I get $limlimits_{b->infty}(int_e^b(frac{ln(x)^alpha}{x}dx))=frac{1}{alpha+1}cdot(limlimits_{b->infty}ln(b)^{alpha+1}-1)$
And because $limlimits_{c->infty}ln(c)=infty$ and $limlimits_{c->infty} c^{alpha+1} = left{
begin{array}{ll}
infty & alphageq -1 \
1 & alpha=-1 \
0 & alpha<-1
end{array}
right.
$
So the limit only exists for $alpha<-1$, is this right?
improper-integrals
edited Jan 12 at 13:11
Bernard
121k740116
asked Jan 12 at 13:04
YefexemYefexem
1
$endgroup$
add a comment |
$begingroup$
My task is to determine wetherthe integral $int_e^infty{frac{ln(x)^alpha}{x},dx}$ does exist or not in depencence of $alphain mathbb{R}$.
For that I wrote $b$ instead of $infty$ and then calculated the integral using substitution, which should be $frac{1}{alpha+1}((ln(b))^{alpha+1}-1)$. Now for the improper integral I get $limlimits_{b->infty}(int_e^b(frac{ln(x)^alpha}{x}dx))=frac{1}{alpha+1}cdot(limlimits_{b->infty}ln(b)^{alpha+1}-1)$
And because $limlimits_{c->infty}ln(c)=infty$ and $limlimits_{c->infty} c^{alpha+1} = left{
begin{array}{ll}
infty & alphageq -1 \
1 & alpha=-1 \
0 & alpha<-1
end{array}
right.
$
So the limit only exists for $alpha<-1$, is this right?
improper-integrals
edited Jan 12 at 13:11
Bernard
121k740116
asked Jan 12 at 13:04
YefexemYefexem
1
$endgroup$
$begingroup$
This is the right result
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 13:20
$begingroup$
Why do you consider in the end the lim of a power of $c $ instead that of $ln c $?
$endgroup$
– user
Jan 12 at 13:25
$begingroup$
because ln(c) is infinity anyway
$endgroup$
– Yefexem
Jan 12 at 13:28
$begingroup$
Does: $$int{frac{(ln x)^{alpha}}{x}dx}=frac{(ln x)^{alpha +1}}{alpha+1}+C$$ help? (FYI: I used IBP)
$endgroup$
– Rhys Hughes
Jan 12 at 13:38
add a comment |
$begingroup$
My task is to determine wetherthe integral $int_e^infty{frac{ln(x)^alpha}{x},dx}$ does exist or not in depencence of $alphain mathbb{R}$.
For that I wrote $b$ instead of $infty$ and then calculated the integral using substitution, which should be $frac{1}{alpha+1}((ln(b))^{alpha+1}-1)$. Now for the improper integral I get $limlimits_{b->infty}(int_e^b(frac{ln(x)^alpha}{x}dx))=frac{1}{alpha+1}cdot(limlimits_{b->infty}ln(b)^{alpha+1}-1)$
And because $limlimits_{c->infty}ln(c)=infty$ and $limlimits_{c->infty} c^{alpha+1} = left{
begin{array}{ll}
infty & alphageq -1 \
1 & alpha=-1 \
0 & alpha<-1
end{array}
right.
$
So the limit only exists for $alpha<-1$, is this right?
improper-integrals
edited Jan 12 at 13:11
Bernard
121k740116
asked Jan 12 at 13:04
YefexemYefexem
1
$endgroup$
My task is to determine wetherthe integral $int_e^infty{frac{ln(x)^alpha}{x},dx}$ does exist or not in depencence of $alphain mathbb{R}$.
For that I wrote $b$ instead of $infty$ and then calculated the integral using substitution, which should be $frac{1}{alpha+1}((ln(b))^{alpha+1}-1)$. Now for the improper integral I get $limlimits_{b->infty}(int_e^b(frac{ln(x)^alpha}{x}dx))=frac{1}{alpha+1}cdot(limlimits_{b->infty}ln(b)^{alpha+1}-1)$
And because $limlimits_{c->infty}ln(c)=infty$ and $limlimits_{c->infty} c^{alpha+1} = left{
begin{array}{ll}
infty & alphageq -1 \
1 & alpha=-1 \
0 & alpha<-1
end{array}
right.
$
So the limit only exists for $alpha<-1$, is this right?
improper-integrals
improper-integrals
edited Jan 12 at 13:11
Bernard
121k740116
asked Jan 12 at 13:04
YefexemYefexem
1
edited Jan 12 at 13:11
Bernard
121k740116
asked Jan 12 at 13:04
YefexemYefexem
1
edited Jan 12 at 13:11
Bernard
121k740116
edited Jan 12 at 13:11
Bernard
121k740116
edited Jan 12 at 13:11
Bernard
121k740116
121k740116
asked Jan 12 at 13:04
YefexemYefexem
1
asked Jan 12 at 13:04
YefexemYefexem
1
asked Jan 12 at 13:04
YefexemYefexem
1
1
$begingroup$
This is the right result
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 13:20
$begingroup$
Why do you consider in the end the lim of a power of $c $ instead that of $ln c $?
$endgroup$
– user
Jan 12 at 13:25
$begingroup$
because ln(c) is infinity anyway
$endgroup$
– Yefexem
Jan 12 at 13:28
$begingroup$
Does: $$int{frac{(ln x)^{alpha}}{x}dx}=frac{(ln x)^{alpha +1}}{alpha+1}+C$$ help? (FYI: I used IBP)
$endgroup$
– Rhys Hughes
Jan 12 at 13:38
add a comment |
$begingroup$
This is the right result
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 13:20
$begingroup$
Why do you consider in the end the lim of a power of $c $ instead that of $ln c $?
$endgroup$
– user
Jan 12 at 13:25
$begingroup$
because ln(c) is infinity anyway
$endgroup$
– Yefexem
Jan 12 at 13:28
$begingroup$
Does: $$int{frac{(ln x)^{alpha}}{x}dx}=frac{(ln x)^{alpha +1}}{alpha+1}+C$$ help? (FYI: I used IBP)
$endgroup$
– Rhys Hughes
Jan 12 at 13:38
$begingroup$
This is the right result
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 13:20
$begingroup$
This is the right result
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 13:20
$begingroup$
Why do you consider in the end the lim of a power of $c $ instead that of $ln c $?
$endgroup$
– user
Jan 12 at 13:25
$begingroup$
Why do you consider in the end the lim of a power of $c $ instead that of $ln c $?
$endgroup$
– user
Jan 12 at 13:25
$begingroup$
because ln(c) is infinity anyway
$endgroup$
– Yefexem
Jan 12 at 13:28
$begingroup$
because ln(c) is infinity anyway
$endgroup$
– Yefexem
Jan 12 at 13:28
$begingroup$
Does: $$int{frac{(ln x)^{alpha}}{x}dx}=frac{(ln x)^{alpha +1}}{alpha+1}+C$$ help? (FYI: I used IBP)
$endgroup$
– Rhys Hughes
Jan 12 at 13:38
$begingroup$
Does: $$int{frac{(ln x)^{alpha}}{x}dx}=frac{(ln x)^{alpha +1}}{alpha+1}+C$$ help? (FYI: I used IBP)
$endgroup$
– Rhys Hughes
Jan 12 at 13:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$I=int_e^inftyfrac{ln^a(x)}{x}dx$$
$$u=ln(x),,dx=xdu$$
$$I=int_1^infty u^adu=left[frac{u^{a+1}}{a+1}right]_1^infty$$
and this is clearly divergent
answered Jan 12 at 13:46
Henry LeeHenry Lee
1,966219
$endgroup$
$begingroup$
Except if $alpha < -1$, I think.
$endgroup$
– Claude Leibovici
Jan 12 at 15:39
$begingroup$
@ClaudeLeibovici yes that sounds right
$endgroup$
– Henry Lee
Jan 12 at 15:41
add a comment |
Your Answer
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$begingroup$
$$I=int_e^inftyfrac{ln^a(x)}{x}dx$$
$$u=ln(x),,dx=xdu$$
$$I=int_1^infty u^adu=left[frac{u^{a+1}}{a+1}right]_1^infty$$
and this is clearly divergent
answered Jan 12 at 13:46
Henry LeeHenry Lee
1,966219
$endgroup$
$begingroup$
Except if $alpha < -1$, I think.
$endgroup$
– Claude Leibovici
Jan 12 at 15:39
$begingroup$
@ClaudeLeibovici yes that sounds right
$endgroup$
– Henry Lee
Jan 12 at 15:41
add a comment |
$begingroup$
$$I=int_e^inftyfrac{ln^a(x)}{x}dx$$
$$u=ln(x),,dx=xdu$$
$$I=int_1^infty u^adu=left[frac{u^{a+1}}{a+1}right]_1^infty$$
and this is clearly divergent
answered Jan 12 at 13:46
Henry LeeHenry Lee
1,966219
$endgroup$
$begingroup$
Except if $alpha < -1$, I think.
$endgroup$
– Claude Leibovici
Jan 12 at 15:39
$begingroup$
@ClaudeLeibovici yes that sounds right
$endgroup$
– Henry Lee
Jan 12 at 15:41
add a comment |
$begingroup$
$$I=int_e^inftyfrac{ln^a(x)}{x}dx$$
$$u=ln(x),,dx=xdu$$
$$I=int_1^infty u^adu=left[frac{u^{a+1}}{a+1}right]_1^infty$$
and this is clearly divergent
answered Jan 12 at 13:46
Henry LeeHenry Lee
1,966219
$endgroup$
$$I=int_e^inftyfrac{ln^a(x)}{x}dx$$
$$u=ln(x),,dx=xdu$$
$$I=int_1^infty u^adu=left[frac{u^{a+1}}{a+1}right]_1^infty$$
and this is clearly divergent
answered Jan 12 at 13:46
Henry LeeHenry Lee
1,966219
answered Jan 12 at 13:46
Henry LeeHenry Lee
1,966219
answered Jan 12 at 13:46
Henry LeeHenry Lee
1,966219
answered Jan 12 at 13:46
Henry LeeHenry Lee
1,966219
1,966219
$begingroup$
Except if $alpha < -1$, I think.
$endgroup$
– Claude Leibovici
Jan 12 at 15:39
$begingroup$
@ClaudeLeibovici yes that sounds right
$endgroup$
– Henry Lee
Jan 12 at 15:41
add a comment |
$begingroup$
Except if $alpha < -1$, I think.
$endgroup$
– Claude Leibovici
Jan 12 at 15:39
$begingroup$
@ClaudeLeibovici yes that sounds right
$endgroup$
– Henry Lee
Jan 12 at 15:41
$begingroup$
Except if $alpha < -1$, I think.
$endgroup$
– Claude Leibovici
Jan 12 at 15:39
$begingroup$
Except if $alpha < -1$, I think.
$endgroup$
– Claude Leibovici
Jan 12 at 15:39
$begingroup$
@ClaudeLeibovici yes that sounds right
$endgroup$
– Henry Lee
Jan 12 at 15:41
$begingroup$
@ClaudeLeibovici yes that sounds right
$endgroup$
– Henry Lee
Jan 12 at 15:41
add a comment |
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$begingroup$
This is the right result
$endgroup$
– Dr. Sonnhard Graubner
Jan 12 at 13:20
$begingroup$
Why do you consider in the end the lim of a power of $c $ instead that of $ln c $?
$endgroup$
– user
Jan 12 at 13:25
$begingroup$
because ln(c) is infinity anyway
$endgroup$
– Yefexem
Jan 12 at 13:28
$begingroup$
Does: $$int{frac{(ln x)^{alpha}}{x}dx}=frac{(ln x)^{alpha +1}}{alpha+1}+C$$ help? (FYI: I used IBP)
$endgroup$
– Rhys Hughes
Jan 12 at 13:38