Mixing
Convergence of an improper integral,depending on parameters
$begingroup$
I have to show how the convergence of $$int_2^infty frac {1}{x^alpha (ln x)^beta} mathrm{ d}x $$ depends on parameters $$alpha,betagt0$$
And considering the case $alphagt1$,my textbook says $xgeq2$ implies $ln xgeq ln 2$ and hence $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ $$forall xgeq2$$
And it is easy to see that $frac {1}{x^alpha (ln2)^beta}$ converges,so does $frac {1}{x^alpha (ln x)^beta}$,by Comparison Theorem.
But $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$
Implies $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ $forall xgeq2$ and $alphagt1$.
But I struggle to see why $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ and therefore $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ doesn't hold for $0ltalphaleq1$
improper-integrals
asked Jan 8 at 7:39
Turan NəsibliTuran Nəsibli
726
$endgroup$
add a comment |
$begingroup$
I have to show how the convergence of $$int_2^infty frac {1}{x^alpha (ln x)^beta} mathrm{ d}x $$ depends on parameters $$alpha,betagt0$$
And considering the case $alphagt1$,my textbook says $xgeq2$ implies $ln xgeq ln 2$ and hence $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ $$forall xgeq2$$
And it is easy to see that $frac {1}{x^alpha (ln2)^beta}$ converges,so does $frac {1}{x^alpha (ln x)^beta}$,by Comparison Theorem.
But $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$
Implies $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ $forall xgeq2$ and $alphagt1$.
But I struggle to see why $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ and therefore $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ doesn't hold for $0ltalphaleq1$
improper-integrals
asked Jan 8 at 7:39
Turan NəsibliTuran Nəsibli
726
$endgroup$
add a comment |
$begingroup$
I have to show how the convergence of $$int_2^infty frac {1}{x^alpha (ln x)^beta} mathrm{ d}x $$ depends on parameters $$alpha,betagt0$$
And considering the case $alphagt1$,my textbook says $xgeq2$ implies $ln xgeq ln 2$ and hence $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ $$forall xgeq2$$
And it is easy to see that $frac {1}{x^alpha (ln2)^beta}$ converges,so does $frac {1}{x^alpha (ln x)^beta}$,by Comparison Theorem.
But $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$
Implies $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ $forall xgeq2$ and $alphagt1$.
But I struggle to see why $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ and therefore $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ doesn't hold for $0ltalphaleq1$
improper-integrals
asked Jan 8 at 7:39
Turan NəsibliTuran Nəsibli
726
$endgroup$
I have to show how the convergence of $$int_2^infty frac {1}{x^alpha (ln x)^beta} mathrm{ d}x $$ depends on parameters $$alpha,betagt0$$
And considering the case $alphagt1$,my textbook says $xgeq2$ implies $ln xgeq ln 2$ and hence $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ $$forall xgeq2$$
And it is easy to see that $frac {1}{x^alpha (ln2)^beta}$ converges,so does $frac {1}{x^alpha (ln x)^beta}$,by Comparison Theorem.
But $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$
Implies $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ $forall xgeq2$ and $alphagt1$.
But I struggle to see why $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ and therefore $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ doesn't hold for $0ltalphaleq1$
improper-integrals
improper-integrals
asked Jan 8 at 7:39
Turan NəsibliTuran Nəsibli
726
asked Jan 8 at 7:39
Turan NəsibliTuran Nəsibli
726
asked Jan 8 at 7:39
Turan NəsibliTuran Nəsibli
726
asked Jan 8 at 7:39
Turan NəsibliTuran Nəsibli
726
asked Jan 8 at 7:39
Turan NəsibliTuran Nəsibli
726
726
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.
answered Jan 8 at 7:41
MindlackMindlack
3,44217
$endgroup$
$begingroup$
Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
$endgroup$
– Turan Nəsibli
Jan 8 at 8:36
$begingroup$
Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
$endgroup$
– Mindlack
Jan 8 at 9:38
$begingroup$
So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
$endgroup$
– Turan Nəsibli
Jan 8 at 10:47
$begingroup$
First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
$endgroup$
– Mindlack
Jan 8 at 11:00
$begingroup$
I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
$endgroup$
– Turan Nəsibli
Jan 8 at 11:06
add a comment |
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1 Answer
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$begingroup$
Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.
answered Jan 8 at 7:41
MindlackMindlack
3,44217
$endgroup$
$begingroup$
Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
$endgroup$
– Turan Nəsibli
Jan 8 at 8:36
$begingroup$
Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
$endgroup$
– Mindlack
Jan 8 at 9:38
$begingroup$
So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
$endgroup$
– Turan Nəsibli
Jan 8 at 10:47
$begingroup$
First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
$endgroup$
– Mindlack
Jan 8 at 11:00
$begingroup$
I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
$endgroup$
– Turan Nəsibli
Jan 8 at 11:06
add a comment |
$begingroup$
Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.
answered Jan 8 at 7:41
MindlackMindlack
3,44217
$endgroup$
$begingroup$
Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
$endgroup$
– Turan Nəsibli
Jan 8 at 8:36
$begingroup$
Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
$endgroup$
– Mindlack
Jan 8 at 9:38
$begingroup$
So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
$endgroup$
– Turan Nəsibli
Jan 8 at 10:47
$begingroup$
First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
$endgroup$
– Mindlack
Jan 8 at 11:00
$begingroup$
I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
$endgroup$
– Turan Nəsibli
Jan 8 at 11:06
add a comment |
$begingroup$
Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.
answered Jan 8 at 7:41
MindlackMindlack
3,44217
$endgroup$
Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.
answered Jan 8 at 7:41
MindlackMindlack
3,44217
edited Jan 8 at 9:36
answered Jan 8 at 7:41
MindlackMindlack
3,44217
answered Jan 8 at 7:41
MindlackMindlack
3,44217
answered Jan 8 at 7:41
MindlackMindlack
3,44217
3,44217
$begingroup$
Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
$endgroup$
– Turan Nəsibli
Jan 8 at 8:36
$begingroup$
Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
$endgroup$
– Mindlack
Jan 8 at 9:38
$begingroup$
So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
$endgroup$
– Turan Nəsibli
Jan 8 at 10:47
$begingroup$
First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
$endgroup$
– Mindlack
Jan 8 at 11:00
$begingroup$
I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
$endgroup$
– Turan Nəsibli
Jan 8 at 11:06
add a comment |
$begingroup$
Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
$endgroup$
– Turan Nəsibli
Jan 8 at 8:36
$begingroup$
Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
$endgroup$
– Mindlack
Jan 8 at 9:38
$begingroup$
So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
$endgroup$
– Turan Nəsibli
Jan 8 at 10:47
$begingroup$
First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
$endgroup$
– Mindlack
Jan 8 at 11:00
$begingroup$
I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
$endgroup$
– Turan Nəsibli
Jan 8 at 11:06
$begingroup$
Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
$endgroup$
– Turan Nəsibli
Jan 8 at 8:36
$begingroup$
Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
$endgroup$
– Turan Nəsibli
Jan 8 at 8:36
$begingroup$
Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
$endgroup$
– Mindlack
Jan 8 at 9:38
$begingroup$
Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
$endgroup$
– Mindlack
Jan 8 at 9:38
$begingroup$
So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
$endgroup$
– Turan Nəsibli
Jan 8 at 10:47
$begingroup$
So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
$endgroup$
– Turan Nəsibli
Jan 8 at 10:47
$begingroup$
First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
$endgroup$
– Mindlack
Jan 8 at 11:00
$begingroup$
First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
$endgroup$
– Mindlack
Jan 8 at 11:00
$begingroup$
I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
$endgroup$
– Turan Nəsibli
Jan 8 at 11:06
$begingroup$
I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
$endgroup$
– Turan Nəsibli
Jan 8 at 11:06
add a comment |
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