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Determine convergence of an improper integral
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I'm trying to see for which values of a does the integral
$$
int _0^infty :fracsinleft(xright)left(1+xright)^a
$$
converge.
I know (from this answer: For what values of $alpha$, does this integral converge?) that for $$1le a<2$$ the integral
$$
int _0^infty :fracsinleft(xright)left(xright)^a$$
converges, but didn't find it very useful...
real-analysis improper-integrals
edited Aug 1 at 8:39
Bernard
110k635102
asked Aug 1 at 7:05
Gil Bar Koltun
1
add a comment |Â
up vote
-3
down vote
favorite
I'm trying to see for which values of a does the integral
$$
int _0^infty :fracsinleft(xright)left(1+xright)^a
$$
converge.
I know (from this answer: For what values of $alpha$, does this integral converge?) that for $$1le a<2$$ the integral
$$
int _0^infty :fracsinleft(xright)left(xright)^a$$
converges, but didn't find it very useful...
real-analysis improper-integrals
edited Aug 1 at 8:39
Bernard
110k635102
asked Aug 1 at 7:05
Gil Bar Koltun
1
add a comment |Â
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
I'm trying to see for which values of a does the integral
$$
int _0^infty :fracsinleft(xright)left(1+xright)^a
$$
converge.
I know (from this answer: For what values of $alpha$, does this integral converge?) that for $$1le a<2$$ the integral
$$
int _0^infty :fracsinleft(xright)left(xright)^a$$
converges, but didn't find it very useful...
real-analysis improper-integrals
edited Aug 1 at 8:39
Bernard
110k635102
asked Aug 1 at 7:05
Gil Bar Koltun
1
I'm trying to see for which values of a does the integral
$$
int _0^infty :fracsinleft(xright)left(1+xright)^a
$$
converge.
I know (from this answer: For what values of $alpha$, does this integral converge?) that for $$1le a<2$$ the integral
$$
int _0^infty :fracsinleft(xright)left(xright)^a$$
converges, but didn't find it very useful...
real-analysis improper-integrals
edited Aug 1 at 8:39
Bernard
110k635102
asked Aug 1 at 7:05
Gil Bar Koltun
1
edited Aug 1 at 8:39
Bernard
110k635102
edited Aug 1 at 8:39
Bernard
110k635102
edited Aug 1 at 8:39
Bernard
110k635102
110k635102
asked Aug 1 at 7:05
Gil Bar Koltun
1
asked Aug 1 at 7:05
Gil Bar Koltun
1
asked Aug 1 at 7:05
Gil Bar Koltun
1
1
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
The function $$fracsin x(1+x)^alpha$$ is well defined at $0$ and its values is $0$. Hence you must check the behaviour at infinity.
Since
$$fracsin x(1+x)^alpha simfracsin xx^alpha$$
when $x$ runs to infinity, your case is very similar to the one here: For what values of $alpha$, does this integral converge?
Note that $$int_0^infty fracsin x(1+x)^alphadx le int_0^infty frac1(1+x)^alphadx$$
Take $y=x+1$, then $dx=dy$ and your integral becomes $$int_1^infty fracdyy^alpha$$ which converges if and only if $alpha >1$.
If $alpha<1$ then your integral becomes
$$int_0^infty (1+x)^-alphasin x dx$$ and it clearly diverges. It remains only to deal with the case $alpha=1$. In such case the integral converges to an improper Riemann integral.
answered Aug 1 at 7:50
Legoman
4,64421033
Why is it clear that for $alpha<1$ the integral $$int_0^infty(1+x)^-alphasinx,dx$$ diverges? Further, in the third display the absolute value sign is missing.
– user539887
Aug 1 at 8:44
$sin x le 1 $ for any $x$.
– Legoman
Aug 1 at 9:53
From the fact that $f(x)le g(x)$ for all $xin[0,infty)$ and the improper interval $int_0^infty g(x),dx$ is convergent it does not follows that $int_0^infty f(x),dx$ is convergent (even if we assume $g$ to be positive).
– user539887
Aug 1 at 10:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The function $$fracsin x(1+x)^alpha$$ is well defined at $0$ and its values is $0$. Hence you must check the behaviour at infinity.
Since
$$fracsin x(1+x)^alpha simfracsin xx^alpha$$
when $x$ runs to infinity, your case is very similar to the one here: For what values of $alpha$, does this integral converge?
Note that $$int_0^infty fracsin x(1+x)^alphadx le int_0^infty frac1(1+x)^alphadx$$
Take $y=x+1$, then $dx=dy$ and your integral becomes $$int_1^infty fracdyy^alpha$$ which converges if and only if $alpha >1$.
If $alpha<1$ then your integral becomes
$$int_0^infty (1+x)^-alphasin x dx$$ and it clearly diverges. It remains only to deal with the case $alpha=1$. In such case the integral converges to an improper Riemann integral.
answered Aug 1 at 7:50
Legoman
4,64421033
Why is it clear that for $alpha<1$ the integral $$int_0^infty(1+x)^-alphasinx,dx$$ diverges? Further, in the third display the absolute value sign is missing.
– user539887
Aug 1 at 8:44
$sin x le 1 $ for any $x$.
– Legoman
Aug 1 at 9:53
From the fact that $f(x)le g(x)$ for all $xin[0,infty)$ and the improper interval $int_0^infty g(x),dx$ is convergent it does not follows that $int_0^infty f(x),dx$ is convergent (even if we assume $g$ to be positive).
– user539887
Aug 1 at 10:27
add a comment |Â
up vote
0
down vote
The function $$fracsin x(1+x)^alpha$$ is well defined at $0$ and its values is $0$. Hence you must check the behaviour at infinity.
Since
$$fracsin x(1+x)^alpha simfracsin xx^alpha$$
when $x$ runs to infinity, your case is very similar to the one here: For what values of $alpha$, does this integral converge?
Note that $$int_0^infty fracsin x(1+x)^alphadx le int_0^infty frac1(1+x)^alphadx$$
Take $y=x+1$, then $dx=dy$ and your integral becomes $$int_1^infty fracdyy^alpha$$ which converges if and only if $alpha >1$.
If $alpha<1$ then your integral becomes
$$int_0^infty (1+x)^-alphasin x dx$$ and it clearly diverges. It remains only to deal with the case $alpha=1$. In such case the integral converges to an improper Riemann integral.
answered Aug 1 at 7:50
Legoman
4,64421033
Why is it clear that for $alpha<1$ the integral $$int_0^infty(1+x)^-alphasinx,dx$$ diverges? Further, in the third display the absolute value sign is missing.
– user539887
Aug 1 at 8:44
$sin x le 1 $ for any $x$.
– Legoman
Aug 1 at 9:53
From the fact that $f(x)le g(x)$ for all $xin[0,infty)$ and the improper interval $int_0^infty g(x),dx$ is convergent it does not follows that $int_0^infty f(x),dx$ is convergent (even if we assume $g$ to be positive).
– user539887
Aug 1 at 10:27
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The function $$fracsin x(1+x)^alpha$$ is well defined at $0$ and its values is $0$. Hence you must check the behaviour at infinity.
Since
$$fracsin x(1+x)^alpha simfracsin xx^alpha$$
when $x$ runs to infinity, your case is very similar to the one here: For what values of $alpha$, does this integral converge?
Note that $$int_0^infty fracsin x(1+x)^alphadx le int_0^infty frac1(1+x)^alphadx$$
Take $y=x+1$, then $dx=dy$ and your integral becomes $$int_1^infty fracdyy^alpha$$ which converges if and only if $alpha >1$.
If $alpha<1$ then your integral becomes
$$int_0^infty (1+x)^-alphasin x dx$$ and it clearly diverges. It remains only to deal with the case $alpha=1$. In such case the integral converges to an improper Riemann integral.
answered Aug 1 at 7:50
Legoman
4,64421033
The function $$fracsin x(1+x)^alpha$$ is well defined at $0$ and its values is $0$. Hence you must check the behaviour at infinity.
Since
$$fracsin x(1+x)^alpha simfracsin xx^alpha$$
when $x$ runs to infinity, your case is very similar to the one here: For what values of $alpha$, does this integral converge?
Note that $$int_0^infty fracsin x(1+x)^alphadx le int_0^infty frac1(1+x)^alphadx$$
Take $y=x+1$, then $dx=dy$ and your integral becomes $$int_1^infty fracdyy^alpha$$ which converges if and only if $alpha >1$.
If $alpha<1$ then your integral becomes
$$int_0^infty (1+x)^-alphasin x dx$$ and it clearly diverges. It remains only to deal with the case $alpha=1$. In such case the integral converges to an improper Riemann integral.
answered Aug 1 at 7:50
Legoman
4,64421033
answered Aug 1 at 7:50
Legoman
4,64421033
answered Aug 1 at 7:50
Legoman
4,64421033
answered Aug 1 at 7:50
Legoman
4,64421033
4,64421033
Why is it clear that for $alpha<1$ the integral $$int_0^infty(1+x)^-alphasinx,dx$$ diverges? Further, in the third display the absolute value sign is missing.
– user539887
Aug 1 at 8:44
$sin x le 1 $ for any $x$.
– Legoman
Aug 1 at 9:53
From the fact that $f(x)le g(x)$ for all $xin[0,infty)$ and the improper interval $int_0^infty g(x),dx$ is convergent it does not follows that $int_0^infty f(x),dx$ is convergent (even if we assume $g$ to be positive).
– user539887
Aug 1 at 10:27
add a comment |Â
Why is it clear that for $alpha<1$ the integral $$int_0^infty(1+x)^-alphasinx,dx$$ diverges? Further, in the third display the absolute value sign is missing.
– user539887
Aug 1 at 8:44
$sin x le 1 $ for any $x$.
– Legoman
Aug 1 at 9:53
From the fact that $f(x)le g(x)$ for all $xin[0,infty)$ and the improper interval $int_0^infty g(x),dx$ is convergent it does not follows that $int_0^infty f(x),dx$ is convergent (even if we assume $g$ to be positive).
– user539887
Aug 1 at 10:27
Why is it clear that for $alpha<1$ the integral $$int_0^infty(1+x)^-alphasinx,dx$$ diverges? Further, in the third display the absolute value sign is missing.
– user539887
Aug 1 at 8:44
Why is it clear that for $alpha<1$ the integral $$int_0^infty(1+x)^-alphasinx,dx$$ diverges? Further, in the third display the absolute value sign is missing.
– user539887
Aug 1 at 8:44
$sin x le 1 $ for any $x$.
– Legoman
Aug 1 at 9:53
$sin x le 1 $ for any $x$.
– Legoman
Aug 1 at 9:53
From the fact that $f(x)le g(x)$ for all $xin[0,infty)$ and the improper interval $int_0^infty g(x),dx$ is convergent it does not follows that $int_0^infty f(x),dx$ is convergent (even if we assume $g$ to be positive).
– user539887
Aug 1 at 10:27
From the fact that $f(x)le g(x)$ for all $xin[0,infty)$ and the improper interval $int_0^infty g(x),dx$ is convergent it does not follows that $int_0^infty f(x),dx$ is convergent (even if we assume $g$ to be positive).
– user539887
Aug 1 at 10:27
add a comment |Â
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