Mixing
Show that this integral is convergent for n in natural numbers:
$begingroup$
Is $int_{0}^{∞} frac{x^{2n−1}}{(1 + x^2)^{n+3}}dx$ convergent for n in natural numbers? I've already found the indefinite integral which is that but I don't know how to continue.
calculus integration convergence definite-integrals
asked Feb 2 at 12:26
Gerard NavarroGerard Navarro
1
$endgroup$
add a comment |
$begingroup$
Is $int_{0}^{∞} frac{x^{2n−1}}{(1 + x^2)^{n+3}}dx$ convergent for n in natural numbers? I've already found the indefinite integral which is that but I don't know how to continue.
calculus integration convergence definite-integrals
asked Feb 2 at 12:26
Gerard NavarroGerard Navarro
1
$endgroup$
$begingroup$
I'm quite sure your indefinite integral is wrong. To prove convergence, just find an equivalent of the integrand when $xto+infty$.
$endgroup$
– Nicolas FRANCOIS
Feb 2 at 12:34
add a comment |
$begingroup$
Is $int_{0}^{∞} frac{x^{2n−1}}{(1 + x^2)^{n+3}}dx$ convergent for n in natural numbers? I've already found the indefinite integral which is that but I don't know how to continue.
calculus integration convergence definite-integrals
asked Feb 2 at 12:26
Gerard NavarroGerard Navarro
1
$endgroup$
Is $int_{0}^{∞} frac{x^{2n−1}}{(1 + x^2)^{n+3}}dx$ convergent for n in natural numbers? I've already found the indefinite integral which is that but I don't know how to continue.
calculus integration convergence definite-integrals
calculus integration convergence definite-integrals
asked Feb 2 at 12:26
Gerard NavarroGerard Navarro
1
asked Feb 2 at 12:26
Gerard NavarroGerard Navarro
1
edited Feb 2 at 12:37
Gerard Navarro
asked Feb 2 at 12:26
Gerard NavarroGerard Navarro
1
asked Feb 2 at 12:26
Gerard NavarroGerard Navarro
1
asked Feb 2 at 12:26
Gerard NavarroGerard Navarro
1
1
$begingroup$
I'm quite sure your indefinite integral is wrong. To prove convergence, just find an equivalent of the integrand when $xto+infty$.
$endgroup$
– Nicolas FRANCOIS
Feb 2 at 12:34
add a comment |
$begingroup$
I'm quite sure your indefinite integral is wrong. To prove convergence, just find an equivalent of the integrand when $xto+infty$.
$endgroup$
– Nicolas FRANCOIS
Feb 2 at 12:34
$begingroup$
I'm quite sure your indefinite integral is wrong. To prove convergence, just find an equivalent of the integrand when $xto+infty$.
$endgroup$
– Nicolas FRANCOIS
Feb 2 at 12:34
$begingroup$
I'm quite sure your indefinite integral is wrong. To prove convergence, just find an equivalent of the integrand when $xto+infty$.
$endgroup$
– Nicolas FRANCOIS
Feb 2 at 12:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: observe that for $x to +infty$, we have
$$frac{x^{2n-1}}{(1+x^2)^{n+3}} sim frac{1}{x^{2n+6 - 2n +1}}=frac{1}{x^7}$$
answered Feb 2 at 12:50
HarnakHarnak
1,369512
$endgroup$
$begingroup$
But this integral does not converge from $0$ to $∞$, does it?
$endgroup$
– Gerard Navarro
Feb 2 at 13:00
$begingroup$
Actually, it does. As you can see using the asymptotic test. I believe the indefinite integral you provided is not correct.
$endgroup$
– Harnak
Feb 2 at 13:05
add a comment |
$begingroup$
$int_1^{infty} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_1^{infty} frac {x^{2n-1}} {(x^{2})^{n+3}}dx=int_1^{infty} x^{-7}dx <infty$. Can you show that $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx<infty$?
answered Feb 2 at 12:58
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
$endgroup$
$begingroup$
Could it be $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_0^{1} frac {x^{2n-1}} {(1)^{n+3}}dx=number$?
$endgroup$
– Gerard Navarro
Feb 2 at 13:27
$begingroup$
@GerardNavarro Yes. But you can also use the fact any continuous function on $[0,1]$ is integrable.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 23:09
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: observe that for $x to +infty$, we have
$$frac{x^{2n-1}}{(1+x^2)^{n+3}} sim frac{1}{x^{2n+6 - 2n +1}}=frac{1}{x^7}$$
answered Feb 2 at 12:50
HarnakHarnak
1,369512
$endgroup$
$begingroup$
But this integral does not converge from $0$ to $∞$, does it?
$endgroup$
– Gerard Navarro
Feb 2 at 13:00
$begingroup$
Actually, it does. As you can see using the asymptotic test. I believe the indefinite integral you provided is not correct.
$endgroup$
– Harnak
Feb 2 at 13:05
add a comment |
$begingroup$
Hint: observe that for $x to +infty$, we have
$$frac{x^{2n-1}}{(1+x^2)^{n+3}} sim frac{1}{x^{2n+6 - 2n +1}}=frac{1}{x^7}$$
answered Feb 2 at 12:50
HarnakHarnak
1,369512
$endgroup$
$begingroup$
But this integral does not converge from $0$ to $∞$, does it?
$endgroup$
– Gerard Navarro
Feb 2 at 13:00
$begingroup$
Actually, it does. As you can see using the asymptotic test. I believe the indefinite integral you provided is not correct.
$endgroup$
– Harnak
Feb 2 at 13:05
add a comment |
$begingroup$
Hint: observe that for $x to +infty$, we have
$$frac{x^{2n-1}}{(1+x^2)^{n+3}} sim frac{1}{x^{2n+6 - 2n +1}}=frac{1}{x^7}$$
answered Feb 2 at 12:50
HarnakHarnak
1,369512
$endgroup$
Hint: observe that for $x to +infty$, we have
$$frac{x^{2n-1}}{(1+x^2)^{n+3}} sim frac{1}{x^{2n+6 - 2n +1}}=frac{1}{x^7}$$
answered Feb 2 at 12:50
HarnakHarnak
1,369512
answered Feb 2 at 12:50
HarnakHarnak
1,369512
answered Feb 2 at 12:50
HarnakHarnak
1,369512
answered Feb 2 at 12:50
HarnakHarnak
1,369512
1,369512
$begingroup$
But this integral does not converge from $0$ to $∞$, does it?
$endgroup$
– Gerard Navarro
Feb 2 at 13:00
$begingroup$
Actually, it does. As you can see using the asymptotic test. I believe the indefinite integral you provided is not correct.
$endgroup$
– Harnak
Feb 2 at 13:05
add a comment |
$begingroup$
But this integral does not converge from $0$ to $∞$, does it?
$endgroup$
– Gerard Navarro
Feb 2 at 13:00
$begingroup$
Actually, it does. As you can see using the asymptotic test. I believe the indefinite integral you provided is not correct.
$endgroup$
– Harnak
Feb 2 at 13:05
$begingroup$
But this integral does not converge from $0$ to $∞$, does it?
$endgroup$
– Gerard Navarro
Feb 2 at 13:00
$begingroup$
But this integral does not converge from $0$ to $∞$, does it?
$endgroup$
– Gerard Navarro
Feb 2 at 13:00
$begingroup$
Actually, it does. As you can see using the asymptotic test. I believe the indefinite integral you provided is not correct.
$endgroup$
– Harnak
Feb 2 at 13:05
$begingroup$
Actually, it does. As you can see using the asymptotic test. I believe the indefinite integral you provided is not correct.
$endgroup$
– Harnak
Feb 2 at 13:05
add a comment |
$begingroup$
$int_1^{infty} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_1^{infty} frac {x^{2n-1}} {(x^{2})^{n+3}}dx=int_1^{infty} x^{-7}dx <infty$. Can you show that $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx<infty$?
answered Feb 2 at 12:58
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
$endgroup$
$begingroup$
Could it be $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_0^{1} frac {x^{2n-1}} {(1)^{n+3}}dx=number$?
$endgroup$
– Gerard Navarro
Feb 2 at 13:27
$begingroup$
@GerardNavarro Yes. But you can also use the fact any continuous function on $[0,1]$ is integrable.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 23:09
add a comment |
$begingroup$
$int_1^{infty} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_1^{infty} frac {x^{2n-1}} {(x^{2})^{n+3}}dx=int_1^{infty} x^{-7}dx <infty$. Can you show that $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx<infty$?
answered Feb 2 at 12:58
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
$endgroup$
$begingroup$
Could it be $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_0^{1} frac {x^{2n-1}} {(1)^{n+3}}dx=number$?
$endgroup$
– Gerard Navarro
Feb 2 at 13:27
$begingroup$
@GerardNavarro Yes. But you can also use the fact any continuous function on $[0,1]$ is integrable.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 23:09
add a comment |
$begingroup$
$int_1^{infty} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_1^{infty} frac {x^{2n-1}} {(x^{2})^{n+3}}dx=int_1^{infty} x^{-7}dx <infty$. Can you show that $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx<infty$?
answered Feb 2 at 12:58
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
$endgroup$
$int_1^{infty} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_1^{infty} frac {x^{2n-1}} {(x^{2})^{n+3}}dx=int_1^{infty} x^{-7}dx <infty$. Can you show that $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx<infty$?
answered Feb 2 at 12:58
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
answered Feb 2 at 12:58
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
answered Feb 2 at 12:58
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
answered Feb 2 at 12:58
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
74.5k53270
$begingroup$
Could it be $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_0^{1} frac {x^{2n-1}} {(1)^{n+3}}dx=number$?
$endgroup$
– Gerard Navarro
Feb 2 at 13:27
$begingroup$
@GerardNavarro Yes. But you can also use the fact any continuous function on $[0,1]$ is integrable.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 23:09
add a comment |
$begingroup$
Could it be $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_0^{1} frac {x^{2n-1}} {(1)^{n+3}}dx=number$?
$endgroup$
– Gerard Navarro
Feb 2 at 13:27
$begingroup$
@GerardNavarro Yes. But you can also use the fact any continuous function on $[0,1]$ is integrable.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 23:09
$begingroup$
Could it be $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_0^{1} frac {x^{2n-1}} {(1)^{n+3}}dx=number$?
$endgroup$
– Gerard Navarro
Feb 2 at 13:27
$begingroup$
Could it be $int_0^{1} frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx leq int_0^{1} frac {x^{2n-1}} {(1)^{n+3}}dx=number$?
$endgroup$
– Gerard Navarro
Feb 2 at 13:27
$begingroup$
@GerardNavarro Yes. But you can also use the fact any continuous function on $[0,1]$ is integrable.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 23:09
$begingroup$
@GerardNavarro Yes. But you can also use the fact any continuous function on $[0,1]$ is integrable.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 23:09
add a comment |
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$begingroup$
I'm quite sure your indefinite integral is wrong. To prove convergence, just find an equivalent of the integrand when $xto+infty$.
$endgroup$
– Nicolas FRANCOIS
Feb 2 at 12:34