$sin^2x/x^4$: investigate the convergence of the improper integral












-1












$begingroup$


Investigate the convergence of the improper integral
$$
int_0^{infty}frac{sin^2x}{x^4},dx
$$
Was ill, hard to understand






integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What bounds for the integral?
    $endgroup$
    – egreg
    Dec 5 '18 at 22:21










  • $begingroup$
    0 +00 they are/
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:23










  • $begingroup$
    next time you make a question, try to show what you had tried, otherwise it is likely you dont get answers and the question could be closed fast
    $endgroup$
    – Masacroso
    Dec 5 '18 at 23:04
















-1












$begingroup$


Investigate the convergence of the improper integral
$$
int_0^{infty}frac{sin^2x}{x^4},dx
$$
Was ill, hard to understand






integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What bounds for the integral?
    $endgroup$
    – egreg
    Dec 5 '18 at 22:21










  • $begingroup$
    0 +00 they are/
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:23










  • $begingroup$
    next time you make a question, try to show what you had tried, otherwise it is likely you dont get answers and the question could be closed fast
    $endgroup$
    – Masacroso
    Dec 5 '18 at 23:04














-1












-1








-1





$begingroup$


Investigate the convergence of the improper integral
$$
int_0^{infty}frac{sin^2x}{x^4},dx
$$
Was ill, hard to understand






integration







share|cite|improve this question











$endgroup$




Investigate the convergence of the improper integral
$$
int_0^{infty}frac{sin^2x}{x^4},dx
$$
Was ill, hard to understand






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 22:25









egreg

180k1485202




180k1485202










asked Dec 5 '18 at 22:16









Siberian PlayerSiberian Player

62




62








  • 2




    $begingroup$
    What bounds for the integral?
    $endgroup$
    – egreg
    Dec 5 '18 at 22:21










  • $begingroup$
    0 +00 they are/
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:23










  • $begingroup$
    next time you make a question, try to show what you had tried, otherwise it is likely you dont get answers and the question could be closed fast
    $endgroup$
    – Masacroso
    Dec 5 '18 at 23:04














  • 2




    $begingroup$
    What bounds for the integral?
    $endgroup$
    – egreg
    Dec 5 '18 at 22:21










  • $begingroup$
    0 +00 they are/
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:23










  • $begingroup$
    next time you make a question, try to show what you had tried, otherwise it is likely you dont get answers and the question could be closed fast
    $endgroup$
    – Masacroso
    Dec 5 '18 at 23:04








2




2




$begingroup$
What bounds for the integral?
$endgroup$
– egreg
Dec 5 '18 at 22:21




$begingroup$
What bounds for the integral?
$endgroup$
– egreg
Dec 5 '18 at 22:21












$begingroup$
0 +00 they are/
$endgroup$
– Siberian Player
Dec 5 '18 at 22:23




$begingroup$
0 +00 they are/
$endgroup$
– Siberian Player
Dec 5 '18 at 22:23












$begingroup$
next time you make a question, try to show what you had tried, otherwise it is likely you dont get answers and the question could be closed fast
$endgroup$
– Masacroso
Dec 5 '18 at 23:04




$begingroup$
next time you make a question, try to show what you had tried, otherwise it is likely you dont get answers and the question could be closed fast
$endgroup$
– Masacroso
Dec 5 '18 at 23:04










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint: consider separately
$$
int_0^{1}frac{sin^2x}{x^4},dx
qquadtext{and}qquad
int_1^{infty}frac{sin^2x}{x^4},dx
$$
The latter converges because, for $xge1$,
$$
frac{sin^2x}{x^4}lefrac{1}{x^4}
$$
The former can be written as
$$
int_0^{1}frac{sin^2x}{x^2}frac{1}{x^2},dx
$$
Can you go on?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Probably no.....
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:31










  • $begingroup$
    @SiberianPlayer: $int_{0}^{1}frac{dx}{x^2}$ is divergent, and so it is your integral.
    $endgroup$
    – Jack D'Aurizio
    Dec 5 '18 at 22:50











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Hint: consider separately
$$
int_0^{1}frac{sin^2x}{x^4},dx
qquadtext{and}qquad
int_1^{infty}frac{sin^2x}{x^4},dx
$$
The latter converges because, for $xge1$,
$$
frac{sin^2x}{x^4}lefrac{1}{x^4}
$$
The former can be written as
$$
int_0^{1}frac{sin^2x}{x^2}frac{1}{x^2},dx
$$
Can you go on?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Probably no.....
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:31










  • $begingroup$
    @SiberianPlayer: $int_{0}^{1}frac{dx}{x^2}$ is divergent, and so it is your integral.
    $endgroup$
    – Jack D'Aurizio
    Dec 5 '18 at 22:50
















2












$begingroup$

Hint: consider separately
$$
int_0^{1}frac{sin^2x}{x^4},dx
qquadtext{and}qquad
int_1^{infty}frac{sin^2x}{x^4},dx
$$
The latter converges because, for $xge1$,
$$
frac{sin^2x}{x^4}lefrac{1}{x^4}
$$
The former can be written as
$$
int_0^{1}frac{sin^2x}{x^2}frac{1}{x^2},dx
$$
Can you go on?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Probably no.....
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:31










  • $begingroup$
    @SiberianPlayer: $int_{0}^{1}frac{dx}{x^2}$ is divergent, and so it is your integral.
    $endgroup$
    – Jack D'Aurizio
    Dec 5 '18 at 22:50














2












2








2





$begingroup$

Hint: consider separately
$$
int_0^{1}frac{sin^2x}{x^4},dx
qquadtext{and}qquad
int_1^{infty}frac{sin^2x}{x^4},dx
$$
The latter converges because, for $xge1$,
$$
frac{sin^2x}{x^4}lefrac{1}{x^4}
$$
The former can be written as
$$
int_0^{1}frac{sin^2x}{x^2}frac{1}{x^2},dx
$$
Can you go on?






share|cite|improve this answer









$endgroup$



Hint: consider separately
$$
int_0^{1}frac{sin^2x}{x^4},dx
qquadtext{and}qquad
int_1^{infty}frac{sin^2x}{x^4},dx
$$
The latter converges because, for $xge1$,
$$
frac{sin^2x}{x^4}lefrac{1}{x^4}
$$
The former can be written as
$$
int_0^{1}frac{sin^2x}{x^2}frac{1}{x^2},dx
$$
Can you go on?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 22:27









egregegreg

180k1485202




180k1485202












  • $begingroup$
    Probably no.....
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:31










  • $begingroup$
    @SiberianPlayer: $int_{0}^{1}frac{dx}{x^2}$ is divergent, and so it is your integral.
    $endgroup$
    – Jack D'Aurizio
    Dec 5 '18 at 22:50


















  • $begingroup$
    Probably no.....
    $endgroup$
    – Siberian Player
    Dec 5 '18 at 22:31










  • $begingroup$
    @SiberianPlayer: $int_{0}^{1}frac{dx}{x^2}$ is divergent, and so it is your integral.
    $endgroup$
    – Jack D'Aurizio
    Dec 5 '18 at 22:50
















$begingroup$
Probably no.....
$endgroup$
– Siberian Player
Dec 5 '18 at 22:31




$begingroup$
Probably no.....
$endgroup$
– Siberian Player
Dec 5 '18 at 22:31












$begingroup$
@SiberianPlayer: $int_{0}^{1}frac{dx}{x^2}$ is divergent, and so it is your integral.
$endgroup$
– Jack D'Aurizio
Dec 5 '18 at 22:50




$begingroup$
@SiberianPlayer: $int_{0}^{1}frac{dx}{x^2}$ is divergent, and so it is your integral.
$endgroup$
– Jack D'Aurizio
Dec 5 '18 at 22:50


















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