Mixing
Does the integral converge? $int_1^infty frac{1}{x(x^2+1)}dx$
$begingroup$
Does the integral converge? $$int_1^infty frac{1}{x(x^2+1)}dx$$
Well, I did show that it converge by finding the indefinite integral first, and getting to $lim =frac{ln 2}{2}$. Which means it converges.
But I thought about another approach, a faster one, I would love to get your opinion.
I thought about using D'Alembert's rule, with the integral $int frac{1}{x}$, and saying that they are positive for every $x in [1,infty]$, and because the limit now will be finite, I can say that because $int frac{1}{x}$ diverges then $int_1^infty frac{1}{x(x^2+1)}dx$ too.
calculus improper-integrals
edited Jan 16 at 16:37
user376343
3,8133829
asked Jan 20 '16 at 15:29
Ilan Aizelman WSIlan Aizelman WS
1,603819
$endgroup$
add a comment |
$begingroup$
Does the integral converge? $$int_1^infty frac{1}{x(x^2+1)}dx$$
Well, I did show that it converge by finding the indefinite integral first, and getting to $lim =frac{ln 2}{2}$. Which means it converges.
But I thought about another approach, a faster one, I would love to get your opinion.
I thought about using D'Alembert's rule, with the integral $int frac{1}{x}$, and saying that they are positive for every $x in [1,infty]$, and because the limit now will be finite, I can say that because $int frac{1}{x}$ diverges then $int_1^infty frac{1}{x(x^2+1)}dx$ too.
calculus improper-integrals
edited Jan 16 at 16:37
user376343
3,8133829
asked Jan 20 '16 at 15:29
Ilan Aizelman WSIlan Aizelman WS
1,603819
$endgroup$
$begingroup$
Relate math.stackexchange.com/questions/960736
$endgroup$
– Nosrati
Dec 11 '18 at 11:02
$begingroup$
Possible duplicate of Convergence of an integral $int_1^{+infty} frac{1}{xsqrt[3]{x^2+1}}mathrm dx$
$endgroup$
– Nosrati
Dec 11 '18 at 11:03
add a comment |
$begingroup$
Does the integral converge? $$int_1^infty frac{1}{x(x^2+1)}dx$$
Well, I did show that it converge by finding the indefinite integral first, and getting to $lim =frac{ln 2}{2}$. Which means it converges.
But I thought about another approach, a faster one, I would love to get your opinion.
I thought about using D'Alembert's rule, with the integral $int frac{1}{x}$, and saying that they are positive for every $x in [1,infty]$, and because the limit now will be finite, I can say that because $int frac{1}{x}$ diverges then $int_1^infty frac{1}{x(x^2+1)}dx$ too.
calculus improper-integrals
edited Jan 16 at 16:37
user376343
3,8133829
asked Jan 20 '16 at 15:29
Ilan Aizelman WSIlan Aizelman WS
1,603819
$endgroup$
Does the integral converge? $$int_1^infty frac{1}{x(x^2+1)}dx$$
Well, I did show that it converge by finding the indefinite integral first, and getting to $lim =frac{ln 2}{2}$. Which means it converges.
But I thought about another approach, a faster one, I would love to get your opinion.
I thought about using D'Alembert's rule, with the integral $int frac{1}{x}$, and saying that they are positive for every $x in [1,infty]$, and because the limit now will be finite, I can say that because $int frac{1}{x}$ diverges then $int_1^infty frac{1}{x(x^2+1)}dx$ too.
calculus improper-integrals
calculus improper-integrals
edited Jan 16 at 16:37
user376343
3,8133829
asked Jan 20 '16 at 15:29
Ilan Aizelman WSIlan Aizelman WS
1,603819
edited Jan 16 at 16:37
user376343
3,8133829
asked Jan 20 '16 at 15:29
Ilan Aizelman WSIlan Aizelman WS
1,603819
edited Jan 16 at 16:37
user376343
3,8133829
edited Jan 16 at 16:37
user376343
3,8133829
edited Jan 16 at 16:37
user376343
3,8133829
3,8133829
asked Jan 20 '16 at 15:29
Ilan Aizelman WSIlan Aizelman WS
1,603819
asked Jan 20 '16 at 15:29
Ilan Aizelman WSIlan Aizelman WS
1,603819
asked Jan 20 '16 at 15:29
Ilan Aizelman WSIlan Aizelman WS
1,603819
1,603819
$begingroup$
Relate math.stackexchange.com/questions/960736
$endgroup$
– Nosrati
Dec 11 '18 at 11:02
$begingroup$
Possible duplicate of Convergence of an integral $int_1^{+infty} frac{1}{xsqrt[3]{x^2+1}}mathrm dx$
$endgroup$
– Nosrati
Dec 11 '18 at 11:03
add a comment |
$begingroup$
Relate math.stackexchange.com/questions/960736
$endgroup$
– Nosrati
Dec 11 '18 at 11:02
$begingroup$
Possible duplicate of Convergence of an integral $int_1^{+infty} frac{1}{xsqrt[3]{x^2+1}}mathrm dx$
$endgroup$
– Nosrati
Dec 11 '18 at 11:03
$begingroup$
Relate math.stackexchange.com/questions/960736
$endgroup$
– Nosrati
Dec 11 '18 at 11:02
$begingroup$
Relate math.stackexchange.com/questions/960736
$endgroup$
– Nosrati
Dec 11 '18 at 11:02
$begingroup$
Possible duplicate of Convergence of an integral $int_1^{+infty} frac{1}{xsqrt[3]{x^2+1}}mathrm dx$
$endgroup$
– Nosrati
Dec 11 '18 at 11:03
$begingroup$
Possible duplicate of Convergence of an integral $int_1^{+infty} frac{1}{xsqrt[3]{x^2+1}}mathrm dx$
$endgroup$
– Nosrati
Dec 11 '18 at 11:03
add a comment |
2 Answers
2
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Since $xgeq 1$ in the domain of integration, we know that $$frac{1}{x(x^2+1)}leqfrac{1}{x^2+1}.$$ It is well known that $$int_1^infty frac{1}{1+x^2},dxleqint_{-infty}^inftyfrac{1}{1+x^2},dx=pi<infty.$$ So the integral converges. You cannot say that since $intfrac{1}{x},dx$ diverges, $int_1^infty frac{1}{x(1+x^2)},dx$ does too since $1/xgeqfrac{1}{x(x^2+1)}$ on the domain of integration.
answered Jan 20 '16 at 15:32
Alex SAlex S
18k12160
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add a comment |
$begingroup$
Hint : $dfrac{1}{x(x^2+1)} < dfrac{1}{x^2+1}$.
answered Jan 20 '16 at 15:34
DeepSeaDeepSea
71.3k54487
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
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$begingroup$
Since $xgeq 1$ in the domain of integration, we know that $$frac{1}{x(x^2+1)}leqfrac{1}{x^2+1}.$$ It is well known that $$int_1^infty frac{1}{1+x^2},dxleqint_{-infty}^inftyfrac{1}{1+x^2},dx=pi<infty.$$ So the integral converges. You cannot say that since $intfrac{1}{x},dx$ diverges, $int_1^infty frac{1}{x(1+x^2)},dx$ does too since $1/xgeqfrac{1}{x(x^2+1)}$ on the domain of integration.
answered Jan 20 '16 at 15:32
Alex SAlex S
18k12160
$endgroup$
add a comment |
$begingroup$
Since $xgeq 1$ in the domain of integration, we know that $$frac{1}{x(x^2+1)}leqfrac{1}{x^2+1}.$$ It is well known that $$int_1^infty frac{1}{1+x^2},dxleqint_{-infty}^inftyfrac{1}{1+x^2},dx=pi<infty.$$ So the integral converges. You cannot say that since $intfrac{1}{x},dx$ diverges, $int_1^infty frac{1}{x(1+x^2)},dx$ does too since $1/xgeqfrac{1}{x(x^2+1)}$ on the domain of integration.
answered Jan 20 '16 at 15:32
Alex SAlex S
18k12160
$endgroup$
add a comment |
$begingroup$
Since $xgeq 1$ in the domain of integration, we know that $$frac{1}{x(x^2+1)}leqfrac{1}{x^2+1}.$$ It is well known that $$int_1^infty frac{1}{1+x^2},dxleqint_{-infty}^inftyfrac{1}{1+x^2},dx=pi<infty.$$ So the integral converges. You cannot say that since $intfrac{1}{x},dx$ diverges, $int_1^infty frac{1}{x(1+x^2)},dx$ does too since $1/xgeqfrac{1}{x(x^2+1)}$ on the domain of integration.
answered Jan 20 '16 at 15:32
Alex SAlex S
18k12160
$endgroup$
Since $xgeq 1$ in the domain of integration, we know that $$frac{1}{x(x^2+1)}leqfrac{1}{x^2+1}.$$ It is well known that $$int_1^infty frac{1}{1+x^2},dxleqint_{-infty}^inftyfrac{1}{1+x^2},dx=pi<infty.$$ So the integral converges. You cannot say that since $intfrac{1}{x},dx$ diverges, $int_1^infty frac{1}{x(1+x^2)},dx$ does too since $1/xgeqfrac{1}{x(x^2+1)}$ on the domain of integration.
answered Jan 20 '16 at 15:32
Alex SAlex S
18k12160
answered Jan 20 '16 at 15:32
Alex SAlex S
18k12160
answered Jan 20 '16 at 15:32
Alex SAlex S
18k12160
answered Jan 20 '16 at 15:32
Alex SAlex S
18k12160
18k12160
add a comment |
add a comment |
$begingroup$
Hint : $dfrac{1}{x(x^2+1)} < dfrac{1}{x^2+1}$.
answered Jan 20 '16 at 15:34
DeepSeaDeepSea
71.3k54487
$endgroup$
add a comment |
$begingroup$
Hint : $dfrac{1}{x(x^2+1)} < dfrac{1}{x^2+1}$.
answered Jan 20 '16 at 15:34
DeepSeaDeepSea
71.3k54487
$endgroup$
add a comment |
$begingroup$
Hint : $dfrac{1}{x(x^2+1)} < dfrac{1}{x^2+1}$.
answered Jan 20 '16 at 15:34
DeepSeaDeepSea
71.3k54487
$endgroup$
Hint : $dfrac{1}{x(x^2+1)} < dfrac{1}{x^2+1}$.
answered Jan 20 '16 at 15:34
DeepSeaDeepSea
71.3k54487
answered Jan 20 '16 at 15:34
DeepSeaDeepSea
71.3k54487
answered Jan 20 '16 at 15:34
DeepSeaDeepSea
71.3k54487
answered Jan 20 '16 at 15:34
DeepSeaDeepSea
71.3k54487
71.3k54487
add a comment |
add a comment |
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$begingroup$
Relate math.stackexchange.com/questions/960736
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– Nosrati
Dec 11 '18 at 11:02
$begingroup$
Possible duplicate of Convergence of an integral $int_1^{+infty} frac{1}{xsqrt[3]{x^2+1}}mathrm dx$
$endgroup$
– Nosrati
Dec 11 '18 at 11:03