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Consider if $int_1^infty[ln(1+frac1x)-sin(frac1x)]text dx $ converges.
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Consider if $int_1^infty[ln(1+frac1x)-sin(frac1x)]text dx $ converges.
I took a substitution $t=frac1x $. Turn it into considering if $int_0^1fracln(1+t)-sin(t)t^2text dt $ converges.
And now since $lim_tto0 fracln(1+t)-sin(t)t^2=-frac12 $, then it has no singular point hence it's an ordinary integrate. It follows that it converges.
Is it right ? Or any test would be more convenient?
calculus
asked Jul 21 at 3:01
Jaqen Chou
1476
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Consider if $int_1^infty[ln(1+frac1x)-sin(frac1x)]text dx $ converges.
I took a substitution $t=frac1x $. Turn it into considering if $int_0^1fracln(1+t)-sin(t)t^2text dt $ converges.
And now since $lim_tto0 fracln(1+t)-sin(t)t^2=-frac12 $, then it has no singular point hence it's an ordinary integrate. It follows that it converges.
Is it right ? Or any test would be more convenient?
calculus
asked Jul 21 at 3:01
Jaqen Chou
1476
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider if $int_1^infty[ln(1+frac1x)-sin(frac1x)]text dx $ converges.
I took a substitution $t=frac1x $. Turn it into considering if $int_0^1fracln(1+t)-sin(t)t^2text dt $ converges.
And now since $lim_tto0 fracln(1+t)-sin(t)t^2=-frac12 $, then it has no singular point hence it's an ordinary integrate. It follows that it converges.
Is it right ? Or any test would be more convenient?
calculus
asked Jul 21 at 3:01
Jaqen Chou
1476
Consider if $int_1^infty[ln(1+frac1x)-sin(frac1x)]text dx $ converges.
I took a substitution $t=frac1x $. Turn it into considering if $int_0^1fracln(1+t)-sin(t)t^2text dt $ converges.
And now since $lim_tto0 fracln(1+t)-sin(t)t^2=-frac12 $, then it has no singular point hence it's an ordinary integrate. It follows that it converges.
Is it right ? Or any test would be more convenient?
calculus
asked Jul 21 at 3:01
Jaqen Chou
1476
asked Jul 21 at 3:01
Jaqen Chou
1476
asked Jul 21 at 3:01
Jaqen Chou
1476
asked Jul 21 at 3:01
Jaqen Chou
1476
1476
add a comment |Â
add a comment |Â
1 Answer
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That works, although I think perhaps the best way to think about this is in terms of Taylor series. $ln(1 + 1/x) = 1/x + O(1/x^2)$ and $sin(1/x) = 1/x + O(1/x^3)$, so $ln(1 + 1/x) - sin(1/x) = O(1/x^2)$; since $int_1^infty 1/x^2 ,dx < infty$, the integral converges.
answered Jul 21 at 3:15
Marcus M
8,1731847
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
That works, although I think perhaps the best way to think about this is in terms of Taylor series. $ln(1 + 1/x) = 1/x + O(1/x^2)$ and $sin(1/x) = 1/x + O(1/x^3)$, so $ln(1 + 1/x) - sin(1/x) = O(1/x^2)$; since $int_1^infty 1/x^2 ,dx < infty$, the integral converges.
answered Jul 21 at 3:15
Marcus M
8,1731847
add a comment |Â
up vote
3
down vote
accepted
That works, although I think perhaps the best way to think about this is in terms of Taylor series. $ln(1 + 1/x) = 1/x + O(1/x^2)$ and $sin(1/x) = 1/x + O(1/x^3)$, so $ln(1 + 1/x) - sin(1/x) = O(1/x^2)$; since $int_1^infty 1/x^2 ,dx < infty$, the integral converges.
answered Jul 21 at 3:15
Marcus M
8,1731847
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
That works, although I think perhaps the best way to think about this is in terms of Taylor series. $ln(1 + 1/x) = 1/x + O(1/x^2)$ and $sin(1/x) = 1/x + O(1/x^3)$, so $ln(1 + 1/x) - sin(1/x) = O(1/x^2)$; since $int_1^infty 1/x^2 ,dx < infty$, the integral converges.
answered Jul 21 at 3:15
Marcus M
8,1731847
That works, although I think perhaps the best way to think about this is in terms of Taylor series. $ln(1 + 1/x) = 1/x + O(1/x^2)$ and $sin(1/x) = 1/x + O(1/x^3)$, so $ln(1 + 1/x) - sin(1/x) = O(1/x^2)$; since $int_1^infty 1/x^2 ,dx < infty$, the integral converges.
answered Jul 21 at 3:15
Marcus M
8,1731847
answered Jul 21 at 3:15
Marcus M
8,1731847
answered Jul 21 at 3:15
Marcus M
8,1731847
answered Jul 21 at 3:15
Marcus M
8,1731847
8,1731847
add a comment |Â
add a comment |Â
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