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Prove that $int_1^inftyfracsin^2(x)xdx$ diverges. [on hold]

Convergence tests for improper multiple integralsShow that the improper integral $int_1^infty f(x) dx$ exists iff $sum_1^infty a_n$ converges.Does the $n$th term test work for integrals?Is $int_1^inftyfracx cos(x)^21+x^3$ convergent or divergent?Show that $int_1^infty fracln xleft(1+x^2right)^lambdamathrm dx$ is convergent only for $lambda > frac12$Prove that the function $f(x)=fraccos^2xsqrtx^4+1$ is improperly integrable on $(0,infty)$.How to prove convergence of $int_0^inftyfracsin(x)xdx$ without evaluating itf is $C^1$ and decreasing then $int_1^inftyf(x)sin(x)dx$ converges.Finding the derivative of $f(x)=int_1^inftyfrace^-xyy^2dy,:::xin(0,infty)$improper integral problem check

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Def: $int_1^infty f(x)dx:= lim_brightarrowinftyint_1^b f(x)dx$.

Motivation: Just a routine homework problem to check the student's understanding of comparison tests for convergence of improper integrals.

I gave my own answer below.

integration analysis

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edited yesterday

Z Kane

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put on hold as off-topic by user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  According to Wolfram Alpha, it diverges.
  $endgroup$
  – Viktor Glombik
  2 days ago
  

  
  
  
  

add a comment | 

0

0

$begingroup$

Def: $int_1^infty f(x)dx:= lim_brightarrowinftyint_1^b f(x)dx$.

Motivation: Just a routine homework problem to check the student's understanding of comparison tests for convergence of improper integrals.

I gave my own answer below.

integration analysis

share|cite|improve this question

edited yesterday

Z Kane

asked 2 days ago

Z KaneZ Kane

285

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Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman yesterday

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." – user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  According to Wolfram Alpha, it diverges.
  $endgroup$
  – Viktor Glombik
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Def: $int_1^infty f(x)dx:= lim_brightarrowinftyint_1^b f(x)dx$.

Motivation: Just a routine homework problem to check the student's understanding of comparison tests for convergence of improper integrals.

I gave my own answer below.

integration analysis

share|cite|improve this question

edited yesterday

Z Kane

asked 2 days ago

Z KaneZ Kane

285

New contributor

Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

Z Kane

asked 2 days ago

Z KaneZ Kane

285

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Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday

Z Kane

asked 2 days ago

Z KaneZ Kane

285

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asked 2 days ago

Z KaneZ Kane

285

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asked 2 days ago

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3 Answers
3

active

oldest

votes

3

$begingroup$

hint

$$sin^2(x)=frac1-cos(2x)2$$

with

$$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and

$$int_1^+inftyfracdx2x$$
divergent.
the sum is divergent near $+infty$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

hamam_Abdallahhamam_Abdallah

38.2k21634

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
  $endgroup$
  – Tomislav Ostojich
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
  $endgroup$
  – Z Kane
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ZKane Sorry, no. But stackexchange or google could help.
  $endgroup$
  – hamam_Abdallah
  2 days ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.

Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.

Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$

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answered 2 days ago

rabotarabota

14.2k32782

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
  $endgroup$
  – peter a g
  2 days ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).

$$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$

And the first term on the right is finite since

$$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$

And the second term on the right converges for a similar reason

$$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Z KaneZ Kane

285

New contributor

Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

3

$begingroup$

hint

$$sin^2(x)=frac1-cos(2x)2$$

with

$$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and

$$int_1^+inftyfracdx2x$$
divergent.
the sum is divergent near $+infty$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

hamam_Abdallahhamam_Abdallah

38.2k21634

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
  $endgroup$
  – Tomislav Ostojich
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
  $endgroup$
  – Z Kane
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ZKane Sorry, no. But stackexchange or google could help.
  $endgroup$
  – hamam_Abdallah
  2 days ago
  

  
  
  
  

add a comment | 

3

$begingroup$

hint

$$sin^2(x)=frac1-cos(2x)2$$

with

$$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and

$$int_1^+inftyfracdx2x$$
divergent.
the sum is divergent near $+infty$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

hamam_Abdallahhamam_Abdallah

38.2k21634

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
  $endgroup$
  – Tomislav Ostojich
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
  $endgroup$
  – Z Kane
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ZKane Sorry, no. But stackexchange or google could help.
  $endgroup$
  – hamam_Abdallah
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

hint

$$sin^2(x)=frac1-cos(2x)2$$

with

$$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and

$$int_1^+inftyfracdx2x$$
divergent.
the sum is divergent near $+infty$.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

hamam_Abdallahhamam_Abdallah

38.2k21634

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

hamam_Abdallahhamam_Abdallah

38.2k21634

answered 2 days ago

hamam_Abdallahhamam_Abdallah

38.2k21634

answered 2 days ago

hamam_Abdallahhamam_Abdallah

38.2k21634

3

$begingroup$

Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.

Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.

Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$

share|cite|improve this answer

answered 2 days ago

rabotarabota

14.2k32782

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
  $endgroup$
  – peter a g
  2 days ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.

Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.

Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$

share|cite|improve this answer

answered 2 days ago

rabotarabota

14.2k32782

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
  $endgroup$
  – peter a g
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.

Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.

Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$

share|cite|improve this answer

answered 2 days ago

rabotarabota

14.2k32782

$endgroup$

share|cite|improve this answer

answered 2 days ago

rabotarabota

14.2k32782

answered 2 days ago

rabotarabota

14.2k32782

answered 2 days ago

rabotarabota

14.2k32782

0

$begingroup$

I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).

$$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$

And the first term on the right is finite since

$$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$

And the second term on the right converges for a similar reason

$$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Z KaneZ Kane

285

New contributor

Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).

$$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$

And the first term on the right is finite since

$$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$

And the second term on the right converges for a similar reason

$$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Z KaneZ Kane

285

New contributor

Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).

$$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$

And the first term on the right is finite since

$$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$

And the second term on the right converges for a similar reason

$$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Z KaneZ Kane

285

New contributor

Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Z KaneZ Kane

285

New contributor

Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

Z KaneZ Kane

285

New contributor

Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

Z KaneZ Kane

285