Prove this improper integral is finite
up vote
-1
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I tried to expand the term in the integral but it turns out to be of order 1/x and diverges...
Any help is appreciated! Thanks in advance!
$int_{0}^{infty}sqrt{log (1+1/x^2)}dx<infty?$
calculus integration complex-analysis residue-calculus
asked Nov 18 at 8:30
user49229
144
add a comment |
up vote
-1
down vote
favorite
I tried to expand the term in the integral but it turns out to be of order 1/x and diverges...
Any help is appreciated! Thanks in advance!
$int_{0}^{infty}sqrt{log (1+1/x^2)}dx<infty?$
calculus integration complex-analysis residue-calculus
asked Nov 18 at 8:30
user49229
144
What have you tried?
– Makina
Nov 18 at 8:33
I tried to expand it but it turned out to be of order 1/x and diverges..
– user49229
Nov 18 at 8:36
@user49229 Edit that into the question.
– J.G.
Nov 18 at 8:40
You cannot prove it is finite because it is not, exactly for $$sqrt{logleft(1+tfrac{1}{x^2}right)}simfrac{1}{x}.$$
– Jack D'Aurizio
Nov 18 at 16:34
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I tried to expand the term in the integral but it turns out to be of order 1/x and diverges...
Any help is appreciated! Thanks in advance!
$int_{0}^{infty}sqrt{log (1+1/x^2)}dx<infty?$
calculus integration complex-analysis residue-calculus
asked Nov 18 at 8:30
user49229
144
I tried to expand the term in the integral but it turns out to be of order 1/x and diverges...
Any help is appreciated! Thanks in advance!
$int_{0}^{infty}sqrt{log (1+1/x^2)}dx<infty?$
calculus integration complex-analysis residue-calculus
calculus integration complex-analysis residue-calculus
asked Nov 18 at 8:30
user49229
144
asked Nov 18 at 8:30
user49229
144
edited Nov 18 at 8:41
asked Nov 18 at 8:30
user49229
144
asked Nov 18 at 8:30
user49229
144
asked Nov 18 at 8:30
user49229
144
144
What have you tried?
– Makina
Nov 18 at 8:33
I tried to expand it but it turned out to be of order 1/x and diverges..
– user49229
Nov 18 at 8:36
@user49229 Edit that into the question.
– J.G.
Nov 18 at 8:40
You cannot prove it is finite because it is not, exactly for $$sqrt{logleft(1+tfrac{1}{x^2}right)}simfrac{1}{x}.$$
– Jack D'Aurizio
Nov 18 at 16:34
add a comment |
What have you tried?
– Makina
Nov 18 at 8:33
I tried to expand it but it turned out to be of order 1/x and diverges..
– user49229
Nov 18 at 8:36
@user49229 Edit that into the question.
– J.G.
Nov 18 at 8:40
You cannot prove it is finite because it is not, exactly for $$sqrt{logleft(1+tfrac{1}{x^2}right)}simfrac{1}{x}.$$
– Jack D'Aurizio
Nov 18 at 16:34
What have you tried?
– Makina
Nov 18 at 8:33
What have you tried?
– Makina
Nov 18 at 8:33
I tried to expand it but it turned out to be of order 1/x and diverges..
– user49229
Nov 18 at 8:36
I tried to expand it but it turned out to be of order 1/x and diverges..
– user49229
Nov 18 at 8:36
@user49229 Edit that into the question.
– J.G.
Nov 18 at 8:40
@user49229 Edit that into the question.
– J.G.
Nov 18 at 8:40
You cannot prove it is finite because it is not, exactly for $$sqrt{logleft(1+tfrac{1}{x^2}right)}simfrac{1}{x}.$$
– Jack D'Aurizio
Nov 18 at 16:34
You cannot prove it is finite because it is not, exactly for $$sqrt{logleft(1+tfrac{1}{x^2}right)}simfrac{1}{x}.$$
– Jack D'Aurizio
Nov 18 at 16:34
add a comment |
1 Answer
1
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oldest
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up vote
0
down vote
This expression is good to check the integrability at $infty$. For the integrability in 0, try to set $1/x^2=t$.
answered Nov 18 at 8:48
Marco
1909
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This expression is good to check the integrability at $infty$. For the integrability in 0, try to set $1/x^2=t$.
answered Nov 18 at 8:48
Marco
1909
add a comment |
up vote
0
down vote
This expression is good to check the integrability at $infty$. For the integrability in 0, try to set $1/x^2=t$.
answered Nov 18 at 8:48
Marco
1909
add a comment |
up vote
0
down vote
up vote
0
down vote
This expression is good to check the integrability at $infty$. For the integrability in 0, try to set $1/x^2=t$.
answered Nov 18 at 8:48
Marco
1909
This expression is good to check the integrability at $infty$. For the integrability in 0, try to set $1/x^2=t$.
answered Nov 18 at 8:48
Marco
1909
answered Nov 18 at 8:48
Marco
1909
answered Nov 18 at 8:48
Marco
1909
answered Nov 18 at 8:48
Marco
1909
1909
add a comment |
add a comment |
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What have you tried?
– Makina
Nov 18 at 8:33
I tried to expand it but it turned out to be of order 1/x and diverges..
– user49229
Nov 18 at 8:36
@user49229 Edit that into the question.
– J.G.
Nov 18 at 8:40
You cannot prove it is finite because it is not, exactly for $$sqrt{logleft(1+tfrac{1}{x^2}right)}simfrac{1}{x}.$$
– Jack D'Aurizio
Nov 18 at 16:34