Mixing
Convergence of $int_{pi}^{infty}frac{dx}{x^2 (sin^{2}x)^{1/3}}$ [closed]
$begingroup$
The title tells the question.
I have to show that the improper integral
$$int_{pi}^{infty}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$ is finite i.e it is convergent.
Any suggestions on how to proceed ? I am having hard time with this.
Thank you.
improper-integrals
asked Jan 3 at 10:17
hiren_garaihiren_garai
497417
$endgroup$
closed as off-topic by Paul Frost, Cesareo, Davide Giraudo, Johanna, amWhy Jan 4 at 16:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Paul Frost, Cesareo, Davide Giraudo, Johanna, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
The title tells the question.
I have to show that the improper integral
$$int_{pi}^{infty}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$ is finite i.e it is convergent.
Any suggestions on how to proceed ? I am having hard time with this.
Thank you.
improper-integrals
asked Jan 3 at 10:17
hiren_garaihiren_garai
497417
$endgroup$
closed as off-topic by Paul Frost, Cesareo, Davide Giraudo, Johanna, amWhy Jan 4 at 16:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Paul Frost, Cesareo, Davide Giraudo, Johanna, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
The title tells the question.
I have to show that the improper integral
$$int_{pi}^{infty}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$ is finite i.e it is convergent.
Any suggestions on how to proceed ? I am having hard time with this.
Thank you.
improper-integrals
asked Jan 3 at 10:17
hiren_garaihiren_garai
497417
$endgroup$
The title tells the question.
I have to show that the improper integral
$$int_{pi}^{infty}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$ is finite i.e it is convergent.
Any suggestions on how to proceed ? I am having hard time with this.
Thank you.
improper-integrals
improper-integrals
asked Jan 3 at 10:17
hiren_garaihiren_garai
497417
asked Jan 3 at 10:17
hiren_garaihiren_garai
497417
asked Jan 3 at 10:17
hiren_garaihiren_garai
497417
asked Jan 3 at 10:17
hiren_garaihiren_garai
497417
asked Jan 3 at 10:17
hiren_garaihiren_garai
497417
497417
closed as off-topic by Paul Frost, Cesareo, Davide Giraudo, Johanna, amWhy Jan 4 at 16:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Paul Frost, Cesareo, Davide Giraudo, Johanna, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Paul Frost, Cesareo, Davide Giraudo, Johanna, amWhy Jan 4 at 16:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Paul Frost, Cesareo, Davide Giraudo, Johanna, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
Take $n in mathbb N$ and consider
$$I_n = int_{pi}^{n pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$
You have
$$
begin{aligned}0 le I_n &= sum_{k=1}^{n-1}int_{kpi}^{(k+1)pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}\
&le frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{kpi}^{(k+1)pi}frac{dx}{(sin^{2}x)^{1/3}}\
&= frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}
end{aligned}$$
As $sum 1/k^2$ converges, we're left to prove that $int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}=2int_{0}^{pi/2}frac{dx}{(sin^{2}x)^{1/3}}$ converges. This can be done as $sin x simeq x$ around $0$ and $int_{0}^{pi/2}frac{dx}{x^{2/3}}$ converges.
answered Jan 3 at 10:39
mathcounterexamples.netmathcounterexamples.net
25.7k21954
$endgroup$
$begingroup$
Wow ! That's a very good way to solve. Thank you for your time.
$endgroup$
– hiren_garai
Jan 3 at 10:56
add a comment |
$begingroup$
We have
$$ int_{Npi}^{(N+1)pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=int_{0}^{pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=frac{3,Gammaleft(tfrac{1}{3}right)^3}{2^{4/3}pi}=frac{2picdot 3^{3/4}}{text{AGM}(2,sqrt{2+sqrt{3}})} $$
by Euler's Beta function, the reflection formula for the $Gamma$ function and the relation between special values of the $Gamma$ function and the complete elliptic integral of the first kind / the AGM mean. In particular the LHS is less than $frac{58}{25}pi$. Our integral equals
$$ int_{0}^{pi}frac{1}{left(sin^2 thetaright)^{1/3}}sum_{ngeq 1}frac{1}{(theta+npi)^2},dtheta=frac{1}{pi^2}int_{0}^{pi}frac{psi'left(1+tfrac{theta}{pi}right)}{left(sin^2 thetaright)^{1/3}},dtheta $$
or, by the reflection formula for the $psi'$ function,
$$ int_{0}^{pi/2}left[frac{1}{sin^2theta}-frac{1}{theta^2}-frac{1}{(pi-theta)^2}right]frac{dtheta}{left(sin^2thetaright)^{1/3}}$$
where the term between square brackets is approximately constant on $left(0,frac{pi}{2}right)$, bounded between $1-frac{8}{pi^2}$ and $frac{1}{3}-frac{1}{pi^2}$. This proves that the integral is finite and also allows an approximate evaluation with a relative error $<11%$. A greater accuracy is achieved by computing the first coefficients of the Fourier series of the term between square brackets, or by performing a Lagrange interpolation.
The reduced form of the original integral exhibits a strong resemblance with the integrals appearing in the computation of some series due to Ramanujan, related to the values of the Riemann $zeta$ function at rational points.
answered Jan 3 at 18:59
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Thank you for your answer @JackD'Aurizio. But I'm quite new to some of the terms you used in your answer (e.g reflection formula), so basically it will take time to make out. Anyways thank you again for your response.
$endgroup$
– hiren_garai
Jan 4 at 2:47
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Take $n in mathbb N$ and consider
$$I_n = int_{pi}^{n pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$
You have
$$
begin{aligned}0 le I_n &= sum_{k=1}^{n-1}int_{kpi}^{(k+1)pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}\
&le frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{kpi}^{(k+1)pi}frac{dx}{(sin^{2}x)^{1/3}}\
&= frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}
end{aligned}$$
As $sum 1/k^2$ converges, we're left to prove that $int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}=2int_{0}^{pi/2}frac{dx}{(sin^{2}x)^{1/3}}$ converges. This can be done as $sin x simeq x$ around $0$ and $int_{0}^{pi/2}frac{dx}{x^{2/3}}$ converges.
answered Jan 3 at 10:39
mathcounterexamples.netmathcounterexamples.net
25.7k21954
$endgroup$
$begingroup$
Wow ! That's a very good way to solve. Thank you for your time.
$endgroup$
– hiren_garai
Jan 3 at 10:56
add a comment |
$begingroup$
Hint
Take $n in mathbb N$ and consider
$$I_n = int_{pi}^{n pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$
You have
$$
begin{aligned}0 le I_n &= sum_{k=1}^{n-1}int_{kpi}^{(k+1)pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}\
&le frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{kpi}^{(k+1)pi}frac{dx}{(sin^{2}x)^{1/3}}\
&= frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}
end{aligned}$$
As $sum 1/k^2$ converges, we're left to prove that $int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}=2int_{0}^{pi/2}frac{dx}{(sin^{2}x)^{1/3}}$ converges. This can be done as $sin x simeq x$ around $0$ and $int_{0}^{pi/2}frac{dx}{x^{2/3}}$ converges.
answered Jan 3 at 10:39
mathcounterexamples.netmathcounterexamples.net
25.7k21954
$endgroup$
$begingroup$
Wow ! That's a very good way to solve. Thank you for your time.
$endgroup$
– hiren_garai
Jan 3 at 10:56
add a comment |
$begingroup$
Hint
Take $n in mathbb N$ and consider
$$I_n = int_{pi}^{n pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$
You have
$$
begin{aligned}0 le I_n &= sum_{k=1}^{n-1}int_{kpi}^{(k+1)pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}\
&le frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{kpi}^{(k+1)pi}frac{dx}{(sin^{2}x)^{1/3}}\
&= frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}
end{aligned}$$
As $sum 1/k^2$ converges, we're left to prove that $int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}=2int_{0}^{pi/2}frac{dx}{(sin^{2}x)^{1/3}}$ converges. This can be done as $sin x simeq x$ around $0$ and $int_{0}^{pi/2}frac{dx}{x^{2/3}}$ converges.
answered Jan 3 at 10:39
mathcounterexamples.netmathcounterexamples.net
25.7k21954
$endgroup$
Hint
Take $n in mathbb N$ and consider
$$I_n = int_{pi}^{n pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}$$
You have
$$
begin{aligned}0 le I_n &= sum_{k=1}^{n-1}int_{kpi}^{(k+1)pi}frac{dx}{x^2 (sin^{2}x)^{1/3}}\
&le frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{kpi}^{(k+1)pi}frac{dx}{(sin^{2}x)^{1/3}}\
&= frac{1}{pi^2}sum_{k=1}^{n-1} frac{1}{k^2}int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}
end{aligned}$$
As $sum 1/k^2$ converges, we're left to prove that $int_{0}^{pi}frac{dx}{(sin^{2}x)^{1/3}}=2int_{0}^{pi/2}frac{dx}{(sin^{2}x)^{1/3}}$ converges. This can be done as $sin x simeq x$ around $0$ and $int_{0}^{pi/2}frac{dx}{x^{2/3}}$ converges.
answered Jan 3 at 10:39
mathcounterexamples.netmathcounterexamples.net
25.7k21954
answered Jan 3 at 10:39
mathcounterexamples.netmathcounterexamples.net
25.7k21954
answered Jan 3 at 10:39
mathcounterexamples.netmathcounterexamples.net
25.7k21954
answered Jan 3 at 10:39
mathcounterexamples.netmathcounterexamples.net
25.7k21954
25.7k21954
$begingroup$
Wow ! That's a very good way to solve. Thank you for your time.
$endgroup$
– hiren_garai
Jan 3 at 10:56
add a comment |
$begingroup$
Wow ! That's a very good way to solve. Thank you for your time.
$endgroup$
– hiren_garai
Jan 3 at 10:56
$begingroup$
Wow ! That's a very good way to solve. Thank you for your time.
$endgroup$
– hiren_garai
Jan 3 at 10:56
$begingroup$
Wow ! That's a very good way to solve. Thank you for your time.
$endgroup$
– hiren_garai
Jan 3 at 10:56
add a comment |
$begingroup$
We have
$$ int_{Npi}^{(N+1)pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=int_{0}^{pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=frac{3,Gammaleft(tfrac{1}{3}right)^3}{2^{4/3}pi}=frac{2picdot 3^{3/4}}{text{AGM}(2,sqrt{2+sqrt{3}})} $$
by Euler's Beta function, the reflection formula for the $Gamma$ function and the relation between special values of the $Gamma$ function and the complete elliptic integral of the first kind / the AGM mean. In particular the LHS is less than $frac{58}{25}pi$. Our integral equals
$$ int_{0}^{pi}frac{1}{left(sin^2 thetaright)^{1/3}}sum_{ngeq 1}frac{1}{(theta+npi)^2},dtheta=frac{1}{pi^2}int_{0}^{pi}frac{psi'left(1+tfrac{theta}{pi}right)}{left(sin^2 thetaright)^{1/3}},dtheta $$
or, by the reflection formula for the $psi'$ function,
$$ int_{0}^{pi/2}left[frac{1}{sin^2theta}-frac{1}{theta^2}-frac{1}{(pi-theta)^2}right]frac{dtheta}{left(sin^2thetaright)^{1/3}}$$
where the term between square brackets is approximately constant on $left(0,frac{pi}{2}right)$, bounded between $1-frac{8}{pi^2}$ and $frac{1}{3}-frac{1}{pi^2}$. This proves that the integral is finite and also allows an approximate evaluation with a relative error $<11%$. A greater accuracy is achieved by computing the first coefficients of the Fourier series of the term between square brackets, or by performing a Lagrange interpolation.
The reduced form of the original integral exhibits a strong resemblance with the integrals appearing in the computation of some series due to Ramanujan, related to the values of the Riemann $zeta$ function at rational points.
answered Jan 3 at 18:59
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Thank you for your answer @JackD'Aurizio. But I'm quite new to some of the terms you used in your answer (e.g reflection formula), so basically it will take time to make out. Anyways thank you again for your response.
$endgroup$
– hiren_garai
Jan 4 at 2:47
add a comment |
$begingroup$
We have
$$ int_{Npi}^{(N+1)pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=int_{0}^{pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=frac{3,Gammaleft(tfrac{1}{3}right)^3}{2^{4/3}pi}=frac{2picdot 3^{3/4}}{text{AGM}(2,sqrt{2+sqrt{3}})} $$
by Euler's Beta function, the reflection formula for the $Gamma$ function and the relation between special values of the $Gamma$ function and the complete elliptic integral of the first kind / the AGM mean. In particular the LHS is less than $frac{58}{25}pi$. Our integral equals
$$ int_{0}^{pi}frac{1}{left(sin^2 thetaright)^{1/3}}sum_{ngeq 1}frac{1}{(theta+npi)^2},dtheta=frac{1}{pi^2}int_{0}^{pi}frac{psi'left(1+tfrac{theta}{pi}right)}{left(sin^2 thetaright)^{1/3}},dtheta $$
or, by the reflection formula for the $psi'$ function,
$$ int_{0}^{pi/2}left[frac{1}{sin^2theta}-frac{1}{theta^2}-frac{1}{(pi-theta)^2}right]frac{dtheta}{left(sin^2thetaright)^{1/3}}$$
where the term between square brackets is approximately constant on $left(0,frac{pi}{2}right)$, bounded between $1-frac{8}{pi^2}$ and $frac{1}{3}-frac{1}{pi^2}$. This proves that the integral is finite and also allows an approximate evaluation with a relative error $<11%$. A greater accuracy is achieved by computing the first coefficients of the Fourier series of the term between square brackets, or by performing a Lagrange interpolation.
The reduced form of the original integral exhibits a strong resemblance with the integrals appearing in the computation of some series due to Ramanujan, related to the values of the Riemann $zeta$ function at rational points.
answered Jan 3 at 18:59
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Thank you for your answer @JackD'Aurizio. But I'm quite new to some of the terms you used in your answer (e.g reflection formula), so basically it will take time to make out. Anyways thank you again for your response.
$endgroup$
– hiren_garai
Jan 4 at 2:47
add a comment |
$begingroup$
We have
$$ int_{Npi}^{(N+1)pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=int_{0}^{pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=frac{3,Gammaleft(tfrac{1}{3}right)^3}{2^{4/3}pi}=frac{2picdot 3^{3/4}}{text{AGM}(2,sqrt{2+sqrt{3}})} $$
by Euler's Beta function, the reflection formula for the $Gamma$ function and the relation between special values of the $Gamma$ function and the complete elliptic integral of the first kind / the AGM mean. In particular the LHS is less than $frac{58}{25}pi$. Our integral equals
$$ int_{0}^{pi}frac{1}{left(sin^2 thetaright)^{1/3}}sum_{ngeq 1}frac{1}{(theta+npi)^2},dtheta=frac{1}{pi^2}int_{0}^{pi}frac{psi'left(1+tfrac{theta}{pi}right)}{left(sin^2 thetaright)^{1/3}},dtheta $$
or, by the reflection formula for the $psi'$ function,
$$ int_{0}^{pi/2}left[frac{1}{sin^2theta}-frac{1}{theta^2}-frac{1}{(pi-theta)^2}right]frac{dtheta}{left(sin^2thetaright)^{1/3}}$$
where the term between square brackets is approximately constant on $left(0,frac{pi}{2}right)$, bounded between $1-frac{8}{pi^2}$ and $frac{1}{3}-frac{1}{pi^2}$. This proves that the integral is finite and also allows an approximate evaluation with a relative error $<11%$. A greater accuracy is achieved by computing the first coefficients of the Fourier series of the term between square brackets, or by performing a Lagrange interpolation.
The reduced form of the original integral exhibits a strong resemblance with the integrals appearing in the computation of some series due to Ramanujan, related to the values of the Riemann $zeta$ function at rational points.
answered Jan 3 at 18:59
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
We have
$$ int_{Npi}^{(N+1)pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=int_{0}^{pi}frac{dtheta}{left(sin^2thetaright)^{1/3}}=frac{3,Gammaleft(tfrac{1}{3}right)^3}{2^{4/3}pi}=frac{2picdot 3^{3/4}}{text{AGM}(2,sqrt{2+sqrt{3}})} $$
by Euler's Beta function, the reflection formula for the $Gamma$ function and the relation between special values of the $Gamma$ function and the complete elliptic integral of the first kind / the AGM mean. In particular the LHS is less than $frac{58}{25}pi$. Our integral equals
$$ int_{0}^{pi}frac{1}{left(sin^2 thetaright)^{1/3}}sum_{ngeq 1}frac{1}{(theta+npi)^2},dtheta=frac{1}{pi^2}int_{0}^{pi}frac{psi'left(1+tfrac{theta}{pi}right)}{left(sin^2 thetaright)^{1/3}},dtheta $$
or, by the reflection formula for the $psi'$ function,
$$ int_{0}^{pi/2}left[frac{1}{sin^2theta}-frac{1}{theta^2}-frac{1}{(pi-theta)^2}right]frac{dtheta}{left(sin^2thetaright)^{1/3}}$$
where the term between square brackets is approximately constant on $left(0,frac{pi}{2}right)$, bounded between $1-frac{8}{pi^2}$ and $frac{1}{3}-frac{1}{pi^2}$. This proves that the integral is finite and also allows an approximate evaluation with a relative error $<11%$. A greater accuracy is achieved by computing the first coefficients of the Fourier series of the term between square brackets, or by performing a Lagrange interpolation.
The reduced form of the original integral exhibits a strong resemblance with the integrals appearing in the computation of some series due to Ramanujan, related to the values of the Riemann $zeta$ function at rational points.
answered Jan 3 at 18:59
Jack D'AurizioJack D'Aurizio
288k33280659
edited Jan 3 at 19:05
answered Jan 3 at 18:59
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 3 at 18:59
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 3 at 18:59
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
$begingroup$
Thank you for your answer @JackD'Aurizio. But I'm quite new to some of the terms you used in your answer (e.g reflection formula), so basically it will take time to make out. Anyways thank you again for your response.
$endgroup$
– hiren_garai
Jan 4 at 2:47
add a comment |
$begingroup$
Thank you for your answer @JackD'Aurizio. But I'm quite new to some of the terms you used in your answer (e.g reflection formula), so basically it will take time to make out. Anyways thank you again for your response.
$endgroup$
– hiren_garai
Jan 4 at 2:47
$begingroup$
Thank you for your answer @JackD'Aurizio. But I'm quite new to some of the terms you used in your answer (e.g reflection formula), so basically it will take time to make out. Anyways thank you again for your response.
$endgroup$
– hiren_garai
Jan 4 at 2:47
$begingroup$
Thank you for your answer @JackD'Aurizio. But I'm quite new to some of the terms you used in your answer (e.g reflection formula), so basically it will take time to make out. Anyways thank you again for your response.
$endgroup$
– hiren_garai
Jan 4 at 2:47
add a comment |
