Improper integral $frac{1}{x^2(1+x)}$
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$int_{0}^{infty}frac{1}{x^2(x+1) }dx$ converges.
What if I write the same integral as
$int_{0}^{infty}frac{-1}{x}dx +int_{0}^{infty}frac{1}{x+1}dx +int_{0}^{infty}frac{1}{x^2}dx$.
And we know $int_{0}^{infty}frac{1}{x}dx$ diverges hence above integral diverges.
Where am I wrong?
real-analysis
edited Dec 27 '18 at 16:47
mathpadawan
2,019422
asked Dec 27 '18 at 16:31
Khan RumaisaKhan Rumaisa
12
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add a comment |
$begingroup$
$int_{0}^{infty}frac{1}{x^2(x+1) }dx$ converges.
What if I write the same integral as
$int_{0}^{infty}frac{-1}{x}dx +int_{0}^{infty}frac{1}{x+1}dx +int_{0}^{infty}frac{1}{x^2}dx$.
And we know $int_{0}^{infty}frac{1}{x}dx$ diverges hence above integral diverges.
Where am I wrong?
real-analysis
edited Dec 27 '18 at 16:47
mathpadawan
2,019422
asked Dec 27 '18 at 16:31
Khan RumaisaKhan Rumaisa
12
$endgroup$
$begingroup$
Please use mathjax in your question
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– Prakhar Nagpal
Dec 27 '18 at 16:33
1
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Note: the first (main) integral doesn't converge in $0$
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– Damien
Dec 27 '18 at 17:45
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Unless, I am missing something $int_{0}^{infty}frac{1}{x^2(x+1) }dx=-infty$, and therefore is not convergent. Also, though you did not state so explicitly, the integral is being broken into parts by partial decomposition. Wolfram Seems to Agree: wolframalpha.com/input/…
$endgroup$
– The Matt
Dec 27 '18 at 17:53
add a comment |
$begingroup$
$int_{0}^{infty}frac{1}{x^2(x+1) }dx$ converges.
What if I write the same integral as
$int_{0}^{infty}frac{-1}{x}dx +int_{0}^{infty}frac{1}{x+1}dx +int_{0}^{infty}frac{1}{x^2}dx$.
And we know $int_{0}^{infty}frac{1}{x}dx$ diverges hence above integral diverges.
Where am I wrong?
real-analysis
edited Dec 27 '18 at 16:47
mathpadawan
2,019422
asked Dec 27 '18 at 16:31
Khan RumaisaKhan Rumaisa
12
$endgroup$
$int_{0}^{infty}frac{1}{x^2(x+1) }dx$ converges.
What if I write the same integral as
$int_{0}^{infty}frac{-1}{x}dx +int_{0}^{infty}frac{1}{x+1}dx +int_{0}^{infty}frac{1}{x^2}dx$.
And we know $int_{0}^{infty}frac{1}{x}dx$ diverges hence above integral diverges.
Where am I wrong?
real-analysis
real-analysis
edited Dec 27 '18 at 16:47
mathpadawan
2,019422
asked Dec 27 '18 at 16:31
Khan RumaisaKhan Rumaisa
12
edited Dec 27 '18 at 16:47
mathpadawan
2,019422
asked Dec 27 '18 at 16:31
Khan RumaisaKhan Rumaisa
12
edited Dec 27 '18 at 16:47
mathpadawan
2,019422
edited Dec 27 '18 at 16:47
mathpadawan
2,019422
edited Dec 27 '18 at 16:47
mathpadawan
2,019422
2,019422
asked Dec 27 '18 at 16:31
Khan RumaisaKhan Rumaisa
12
asked Dec 27 '18 at 16:31
Khan RumaisaKhan Rumaisa
12
asked Dec 27 '18 at 16:31
Khan RumaisaKhan Rumaisa
12
12
$begingroup$
Please use mathjax in your question
$endgroup$
– Prakhar Nagpal
Dec 27 '18 at 16:33
1
$begingroup$
Note: the first (main) integral doesn't converge in $0$
$endgroup$
– Damien
Dec 27 '18 at 17:45
$begingroup$
Unless, I am missing something $int_{0}^{infty}frac{1}{x^2(x+1) }dx=-infty$, and therefore is not convergent. Also, though you did not state so explicitly, the integral is being broken into parts by partial decomposition. Wolfram Seems to Agree: wolframalpha.com/input/…
$endgroup$
– The Matt
Dec 27 '18 at 17:53
add a comment |
$begingroup$
Please use mathjax in your question
$endgroup$
– Prakhar Nagpal
Dec 27 '18 at 16:33
1
$begingroup$
Note: the first (main) integral doesn't converge in $0$
$endgroup$
– Damien
Dec 27 '18 at 17:45
$begingroup$
Unless, I am missing something $int_{0}^{infty}frac{1}{x^2(x+1) }dx=-infty$, and therefore is not convergent. Also, though you did not state so explicitly, the integral is being broken into parts by partial decomposition. Wolfram Seems to Agree: wolframalpha.com/input/…
$endgroup$
– The Matt
Dec 27 '18 at 17:53
$begingroup$
Please use mathjax in your question
$endgroup$
– Prakhar Nagpal
Dec 27 '18 at 16:33
$begingroup$
Please use mathjax in your question
$endgroup$
– Prakhar Nagpal
Dec 27 '18 at 16:33
1
1
$begingroup$
Note: the first (main) integral doesn't converge in $0$
$endgroup$
– Damien
Dec 27 '18 at 17:45
$begingroup$
Note: the first (main) integral doesn't converge in $0$
$endgroup$
– Damien
Dec 27 '18 at 17:45
$begingroup$
Unless, I am missing something $int_{0}^{infty}frac{1}{x^2(x+1) }dx=-infty$, and therefore is not convergent. Also, though you did not state so explicitly, the integral is being broken into parts by partial decomposition. Wolfram Seems to Agree: wolframalpha.com/input/…
$endgroup$
– The Matt
Dec 27 '18 at 17:53
$begingroup$
Unless, I am missing something $int_{0}^{infty}frac{1}{x^2(x+1) }dx=-infty$, and therefore is not convergent. Also, though you did not state so explicitly, the integral is being broken into parts by partial decomposition. Wolfram Seems to Agree: wolframalpha.com/input/…
$endgroup$
– The Matt
Dec 27 '18 at 17:53
add a comment |
1 Answer
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It is not, in general, the case that $int (f+g) = int f + int g$. The usual proof of this result is valid only when two of the integrals involved (hence all of them) converge. A simple corollary of this gives that $int (f+g+h) = int f + int g + int h$ when three (hence all four) of those integrals converge. In your case, however, as you note, not all of your integrals converge, hence this result is not applicable.
answered Dec 27 '18 at 16:46
user3482749user3482749
4,3291119
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1 Answer
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1 Answer
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$begingroup$
It is not, in general, the case that $int (f+g) = int f + int g$. The usual proof of this result is valid only when two of the integrals involved (hence all of them) converge. A simple corollary of this gives that $int (f+g+h) = int f + int g + int h$ when three (hence all four) of those integrals converge. In your case, however, as you note, not all of your integrals converge, hence this result is not applicable.
answered Dec 27 '18 at 16:46
user3482749user3482749
4,3291119
$endgroup$
add a comment |
$begingroup$
It is not, in general, the case that $int (f+g) = int f + int g$. The usual proof of this result is valid only when two of the integrals involved (hence all of them) converge. A simple corollary of this gives that $int (f+g+h) = int f + int g + int h$ when three (hence all four) of those integrals converge. In your case, however, as you note, not all of your integrals converge, hence this result is not applicable.
answered Dec 27 '18 at 16:46
user3482749user3482749
4,3291119
$endgroup$
add a comment |
$begingroup$
It is not, in general, the case that $int (f+g) = int f + int g$. The usual proof of this result is valid only when two of the integrals involved (hence all of them) converge. A simple corollary of this gives that $int (f+g+h) = int f + int g + int h$ when three (hence all four) of those integrals converge. In your case, however, as you note, not all of your integrals converge, hence this result is not applicable.
answered Dec 27 '18 at 16:46
user3482749user3482749
4,3291119
$endgroup$
It is not, in general, the case that $int (f+g) = int f + int g$. The usual proof of this result is valid only when two of the integrals involved (hence all of them) converge. A simple corollary of this gives that $int (f+g+h) = int f + int g + int h$ when three (hence all four) of those integrals converge. In your case, however, as you note, not all of your integrals converge, hence this result is not applicable.
answered Dec 27 '18 at 16:46
user3482749user3482749
4,3291119
answered Dec 27 '18 at 16:46
user3482749user3482749
4,3291119
answered Dec 27 '18 at 16:46
user3482749user3482749
4,3291119
answered Dec 27 '18 at 16:46
user3482749user3482749
4,3291119
4,3291119
add a comment |
add a comment |
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$begingroup$
Please use mathjax in your question
$endgroup$
– Prakhar Nagpal
Dec 27 '18 at 16:33
1
$begingroup$
Note: the first (main) integral doesn't converge in $0$
$endgroup$
– Damien
Dec 27 '18 at 17:45
$begingroup$
Unless, I am missing something $int_{0}^{infty}frac{1}{x^2(x+1) }dx=-infty$, and therefore is not convergent. Also, though you did not state so explicitly, the integral is being broken into parts by partial decomposition. Wolfram Seems to Agree: wolframalpha.com/input/…
$endgroup$
– The Matt
Dec 27 '18 at 17:53