Does the following integral depending on $alpha$ converge : $int_{R^2} frac{1}{(1+x^4+y^4)^alpha}$?
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Does the following integral depending on $alpha$ converge : $$int_{R^2} frac{1}{(1+x^4+y^4)^alpha}$$ ?
I know that for $alpha le 0$ it diverges. It is pretty trivial since you can compare the function to $1$.
For $alpha ge 0$ I tried to change to polar coordinates and I got to
$$int_{0}^{2pi} dtheta int_{0}^{infty} frac{rdr}{(1+frac{r^4}{2}(2-sin^2(2theta))^alpha}$$
I'm not pretty sure how to continue from here. I think I can bound it from both sides since $0 le sin^2(2theta) le 1$ . I don't know exactly for which values of $alpha$ I should use each bound.
Help would be appreciated.
integration multivariable-calculus improper-integrals
asked Jan 8 at 22:13
Gabi GGabi G
408110
$endgroup$
add a comment |
$begingroup$
Does the following integral depending on $alpha$ converge : $$int_{R^2} frac{1}{(1+x^4+y^4)^alpha}$$ ?
I know that for $alpha le 0$ it diverges. It is pretty trivial since you can compare the function to $1$.
For $alpha ge 0$ I tried to change to polar coordinates and I got to
$$int_{0}^{2pi} dtheta int_{0}^{infty} frac{rdr}{(1+frac{r^4}{2}(2-sin^2(2theta))^alpha}$$
I'm not pretty sure how to continue from here. I think I can bound it from both sides since $0 le sin^2(2theta) le 1$ . I don't know exactly for which values of $alpha$ I should use each bound.
Help would be appreciated.
integration multivariable-calculus improper-integrals
asked Jan 8 at 22:13
Gabi GGabi G
408110
$endgroup$
6
$begingroup$
Take a wider view of the situation: clearly the problem of convergence, if any, occurs when $rtoinfty$, and then, $$1+x^4+y^4approx r^4$$ hence the integral converges iff $$int_1^inftyfrac1{(r^4)^alpha}rdr$$ converges, that is, iff $$alpha>frac12$$
$endgroup$
– Did
Jan 8 at 22:19
$begingroup$
Thanks! It really helped me
$endgroup$
– Gabi G
Jan 8 at 23:03
add a comment |
$begingroup$
Does the following integral depending on $alpha$ converge : $$int_{R^2} frac{1}{(1+x^4+y^4)^alpha}$$ ?
I know that for $alpha le 0$ it diverges. It is pretty trivial since you can compare the function to $1$.
For $alpha ge 0$ I tried to change to polar coordinates and I got to
$$int_{0}^{2pi} dtheta int_{0}^{infty} frac{rdr}{(1+frac{r^4}{2}(2-sin^2(2theta))^alpha}$$
I'm not pretty sure how to continue from here. I think I can bound it from both sides since $0 le sin^2(2theta) le 1$ . I don't know exactly for which values of $alpha$ I should use each bound.
Help would be appreciated.
integration multivariable-calculus improper-integrals
asked Jan 8 at 22:13
Gabi GGabi G
408110
$endgroup$
Does the following integral depending on $alpha$ converge : $$int_{R^2} frac{1}{(1+x^4+y^4)^alpha}$$ ?
I know that for $alpha le 0$ it diverges. It is pretty trivial since you can compare the function to $1$.
For $alpha ge 0$ I tried to change to polar coordinates and I got to
$$int_{0}^{2pi} dtheta int_{0}^{infty} frac{rdr}{(1+frac{r^4}{2}(2-sin^2(2theta))^alpha}$$
I'm not pretty sure how to continue from here. I think I can bound it from both sides since $0 le sin^2(2theta) le 1$ . I don't know exactly for which values of $alpha$ I should use each bound.
Help would be appreciated.
integration multivariable-calculus improper-integrals
integration multivariable-calculus improper-integrals
asked Jan 8 at 22:13
Gabi GGabi G
408110
asked Jan 8 at 22:13
Gabi GGabi G
408110
asked Jan 8 at 22:13
Gabi GGabi G
408110
asked Jan 8 at 22:13
Gabi GGabi G
408110
asked Jan 8 at 22:13
Gabi GGabi G
408110
408110
6
$begingroup$
Take a wider view of the situation: clearly the problem of convergence, if any, occurs when $rtoinfty$, and then, $$1+x^4+y^4approx r^4$$ hence the integral converges iff $$int_1^inftyfrac1{(r^4)^alpha}rdr$$ converges, that is, iff $$alpha>frac12$$
$endgroup$
– Did
Jan 8 at 22:19
$begingroup$
Thanks! It really helped me
$endgroup$
– Gabi G
Jan 8 at 23:03
add a comment |
6
$begingroup$
Take a wider view of the situation: clearly the problem of convergence, if any, occurs when $rtoinfty$, and then, $$1+x^4+y^4approx r^4$$ hence the integral converges iff $$int_1^inftyfrac1{(r^4)^alpha}rdr$$ converges, that is, iff $$alpha>frac12$$
$endgroup$
– Did
Jan 8 at 22:19
$begingroup$
Thanks! It really helped me
$endgroup$
– Gabi G
Jan 8 at 23:03
6
6
$begingroup$
Take a wider view of the situation: clearly the problem of convergence, if any, occurs when $rtoinfty$, and then, $$1+x^4+y^4approx r^4$$ hence the integral converges iff $$int_1^inftyfrac1{(r^4)^alpha}rdr$$ converges, that is, iff $$alpha>frac12$$
$endgroup$
– Did
Jan 8 at 22:19
$begingroup$
Take a wider view of the situation: clearly the problem of convergence, if any, occurs when $rtoinfty$, and then, $$1+x^4+y^4approx r^4$$ hence the integral converges iff $$int_1^inftyfrac1{(r^4)^alpha}rdr$$ converges, that is, iff $$alpha>frac12$$
$endgroup$
– Did
Jan 8 at 22:19
$begingroup$
Thanks! It really helped me
$endgroup$
– Gabi G
Jan 8 at 23:03
$begingroup$
Thanks! It really helped me
$endgroup$
– Gabi G
Jan 8 at 23:03
add a comment |
0
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6
$begingroup$
Take a wider view of the situation: clearly the problem of convergence, if any, occurs when $rtoinfty$, and then, $$1+x^4+y^4approx r^4$$ hence the integral converges iff $$int_1^inftyfrac1{(r^4)^alpha}rdr$$ converges, that is, iff $$alpha>frac12$$
$endgroup$
– Did
Jan 8 at 22:19
$begingroup$
Thanks! It really helped me
$endgroup$
– Gabi G
Jan 8 at 23:03