Mixing
For which values of $alpha$ the integral $int_{R^2} frac{sin(x^2+y^2)}{(x^2+y^2+1)^{alpha}}$ converges?
$begingroup$
For which values of $alpha$ the integral $int_{R^2} frac{sin(x^2+y^2)}{(x^2+y^2+1)^{alpha}}$ converges?
I changed to polar coordinates and got to $int_{R^2} frac{rsin(r^2)}{(r^2+1)^{alpha}}$
I managed to show that for $alpha le0$ the integral doesn't converge and for $alpha ge 1$ it converges.
However, for $0 < alpha < 1$ , I am not sure how to bound it.
Help would be appreciated.
calculus multivariable-calculus improper-integrals
asked Jan 19 at 23:29
Gabi GGabi G
408110
$endgroup$
add a comment |
$begingroup$
For which values of $alpha$ the integral $int_{R^2} frac{sin(x^2+y^2)}{(x^2+y^2+1)^{alpha}}$ converges?
I changed to polar coordinates and got to $int_{R^2} frac{rsin(r^2)}{(r^2+1)^{alpha}}$
I managed to show that for $alpha le0$ the integral doesn't converge and for $alpha ge 1$ it converges.
However, for $0 < alpha < 1$ , I am not sure how to bound it.
Help would be appreciated.
calculus multivariable-calculus improper-integrals
asked Jan 19 at 23:29
Gabi GGabi G
408110
$endgroup$
$begingroup$
Related: math.stackexchange.com/questions/2621173/…
$endgroup$
– Dave
Jan 19 at 23:36
add a comment |
$begingroup$
For which values of $alpha$ the integral $int_{R^2} frac{sin(x^2+y^2)}{(x^2+y^2+1)^{alpha}}$ converges?
I changed to polar coordinates and got to $int_{R^2} frac{rsin(r^2)}{(r^2+1)^{alpha}}$
I managed to show that for $alpha le0$ the integral doesn't converge and for $alpha ge 1$ it converges.
However, for $0 < alpha < 1$ , I am not sure how to bound it.
Help would be appreciated.
calculus multivariable-calculus improper-integrals
asked Jan 19 at 23:29
Gabi GGabi G
408110
$endgroup$
For which values of $alpha$ the integral $int_{R^2} frac{sin(x^2+y^2)}{(x^2+y^2+1)^{alpha}}$ converges?
I changed to polar coordinates and got to $int_{R^2} frac{rsin(r^2)}{(r^2+1)^{alpha}}$
I managed to show that for $alpha le0$ the integral doesn't converge and for $alpha ge 1$ it converges.
However, for $0 < alpha < 1$ , I am not sure how to bound it.
Help would be appreciated.
calculus multivariable-calculus improper-integrals
calculus multivariable-calculus improper-integrals
asked Jan 19 at 23:29
Gabi GGabi G
408110
asked Jan 19 at 23:29
Gabi GGabi G
408110
edited Jan 19 at 23:57
Gabi G
asked Jan 19 at 23:29
Gabi GGabi G
408110
asked Jan 19 at 23:29
Gabi GGabi G
408110
asked Jan 19 at 23:29
Gabi GGabi G
408110
408110
$begingroup$
Related: math.stackexchange.com/questions/2621173/…
$endgroup$
– Dave
Jan 19 at 23:36
add a comment |
$begingroup$
Related: math.stackexchange.com/questions/2621173/…
$endgroup$
– Dave
Jan 19 at 23:36
$begingroup$
Related: math.stackexchange.com/questions/2621173/…
$endgroup$
– Dave
Jan 19 at 23:36
$begingroup$
Related: math.stackexchange.com/questions/2621173/…
$endgroup$
– Dave
Jan 19 at 23:36
add a comment |
1 Answer
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$begingroup$
After switching to polar coordinates the integral becomes
$$ 2piint_{0}^{+infty}frac{rhosin(rho^2)}{(rho^2+1)^{alpha}}drho=piint_{0}^{+infty}frac{sin(x)}{(x+1)^{alpha}},dx $$
which converges for any $alpha>0$ due to Dirichlet's criterion: $sin(x)$ has a bounded antiderivative and $frac{1}{(x+1)^{alpha}}$ decreases to zero.
answered Jan 19 at 23:37
Jack D'AurizioJack D'Aurizio
290k33283664
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
After switching to polar coordinates the integral becomes
$$ 2piint_{0}^{+infty}frac{rhosin(rho^2)}{(rho^2+1)^{alpha}}drho=piint_{0}^{+infty}frac{sin(x)}{(x+1)^{alpha}},dx $$
which converges for any $alpha>0$ due to Dirichlet's criterion: $sin(x)$ has a bounded antiderivative and $frac{1}{(x+1)^{alpha}}$ decreases to zero.
answered Jan 19 at 23:37
Jack D'AurizioJack D'Aurizio
290k33283664
$endgroup$
add a comment |
$begingroup$
After switching to polar coordinates the integral becomes
$$ 2piint_{0}^{+infty}frac{rhosin(rho^2)}{(rho^2+1)^{alpha}}drho=piint_{0}^{+infty}frac{sin(x)}{(x+1)^{alpha}},dx $$
which converges for any $alpha>0$ due to Dirichlet's criterion: $sin(x)$ has a bounded antiderivative and $frac{1}{(x+1)^{alpha}}$ decreases to zero.
answered Jan 19 at 23:37
Jack D'AurizioJack D'Aurizio
290k33283664
$endgroup$
add a comment |
$begingroup$
After switching to polar coordinates the integral becomes
$$ 2piint_{0}^{+infty}frac{rhosin(rho^2)}{(rho^2+1)^{alpha}}drho=piint_{0}^{+infty}frac{sin(x)}{(x+1)^{alpha}},dx $$
which converges for any $alpha>0$ due to Dirichlet's criterion: $sin(x)$ has a bounded antiderivative and $frac{1}{(x+1)^{alpha}}$ decreases to zero.
answered Jan 19 at 23:37
Jack D'AurizioJack D'Aurizio
290k33283664
$endgroup$
After switching to polar coordinates the integral becomes
$$ 2piint_{0}^{+infty}frac{rhosin(rho^2)}{(rho^2+1)^{alpha}}drho=piint_{0}^{+infty}frac{sin(x)}{(x+1)^{alpha}},dx $$
which converges for any $alpha>0$ due to Dirichlet's criterion: $sin(x)$ has a bounded antiderivative and $frac{1}{(x+1)^{alpha}}$ decreases to zero.
answered Jan 19 at 23:37
Jack D'AurizioJack D'Aurizio
290k33283664
answered Jan 19 at 23:37
Jack D'AurizioJack D'Aurizio
290k33283664
answered Jan 19 at 23:37
Jack D'AurizioJack D'Aurizio
290k33283664
answered Jan 19 at 23:37
Jack D'AurizioJack D'Aurizio
290k33283664
290k33283664
add a comment |
add a comment |
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Related: math.stackexchange.com/questions/2621173/…
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– Dave
Jan 19 at 23:36