Why does the integral of 1/x diverge?
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Why does the integration below diverge?
beginequation
int_-infty^inftyfrac1xdx
endequation
I know this integral diverge from $-infty$ to $0$ (or $0$ to $infty$). But I don't understand why these two integrals are not the same. $frac1x$ is an odd function, so I think the integral $-infty$ to $infty$ is to be $0$.
improper-integrals
asked Aug 19 at 5:07
Orient
605
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up vote
0
down vote
favorite
Why does the integration below diverge?
beginequation
int_-infty^inftyfrac1xdx
endequation
I know this integral diverge from $-infty$ to $0$ (or $0$ to $infty$). But I don't understand why these two integrals are not the same. $frac1x$ is an odd function, so I think the integral $-infty$ to $infty$ is to be $0$.
improper-integrals
asked Aug 19 at 5:07
Orient
605
6
Lookup Cauchy principal value. Your improper integral diverges, but its Cauchy principal value is indeed $0$.
– dxiv
Aug 19 at 5:10
1
In order the integral $ beginequation int_-infty^inftyfrac1xdx endequation$ to converge, the other two $ beginequation int_-infty^0frac1xdx endequation$ and $ beginequation int_0^inftyfrac1xdx endequation$ have to coverge.
– dmtri
Aug 19 at 5:51
Similar (with $x$ instead of $1/x$): math.stackexchange.com/questions/2493443/…
– Hans Lundmark
Aug 19 at 7:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Why does the integration below diverge?
beginequation
int_-infty^inftyfrac1xdx
endequation
I know this integral diverge from $-infty$ to $0$ (or $0$ to $infty$). But I don't understand why these two integrals are not the same. $frac1x$ is an odd function, so I think the integral $-infty$ to $infty$ is to be $0$.
improper-integrals
asked Aug 19 at 5:07
Orient
605
Why does the integration below diverge?
beginequation
int_-infty^inftyfrac1xdx
endequation
I know this integral diverge from $-infty$ to $0$ (or $0$ to $infty$). But I don't understand why these two integrals are not the same. $frac1x$ is an odd function, so I think the integral $-infty$ to $infty$ is to be $0$.
improper-integrals
asked Aug 19 at 5:07
Orient
605
asked Aug 19 at 5:07
Orient
605
asked Aug 19 at 5:07
Orient
605
asked Aug 19 at 5:07
Orient
605
605
6
Lookup Cauchy principal value. Your improper integral diverges, but its Cauchy principal value is indeed $0$.
– dxiv
Aug 19 at 5:10
1
In order the integral $ beginequation int_-infty^inftyfrac1xdx endequation$ to converge, the other two $ beginequation int_-infty^0frac1xdx endequation$ and $ beginequation int_0^inftyfrac1xdx endequation$ have to coverge.
– dmtri
Aug 19 at 5:51
Similar (with $x$ instead of $1/x$): math.stackexchange.com/questions/2493443/…
– Hans Lundmark
Aug 19 at 7:49
add a comment |Â
6
Lookup Cauchy principal value. Your improper integral diverges, but its Cauchy principal value is indeed $0$.
– dxiv
Aug 19 at 5:10
1
In order the integral $ beginequation int_-infty^inftyfrac1xdx endequation$ to converge, the other two $ beginequation int_-infty^0frac1xdx endequation$ and $ beginequation int_0^inftyfrac1xdx endequation$ have to coverge.
– dmtri
Aug 19 at 5:51
Similar (with $x$ instead of $1/x$): math.stackexchange.com/questions/2493443/…
– Hans Lundmark
Aug 19 at 7:49
6
6
Lookup Cauchy principal value. Your improper integral diverges, but its Cauchy principal value is indeed $0$.
– dxiv
Aug 19 at 5:10
Lookup Cauchy principal value. Your improper integral diverges, but its Cauchy principal value is indeed $0$.
– dxiv
Aug 19 at 5:10
1
1
In order the integral $ beginequation int_-infty^inftyfrac1xdx endequation$ to converge, the other two $ beginequation int_-infty^0frac1xdx endequation$ and $ beginequation int_0^inftyfrac1xdx endequation$ have to coverge.
– dmtri
Aug 19 at 5:51
In order the integral $ beginequation int_-infty^inftyfrac1xdx endequation$ to converge, the other two $ beginequation int_-infty^0frac1xdx endequation$ and $ beginequation int_0^inftyfrac1xdx endequation$ have to coverge.
– dmtri
Aug 19 at 5:51
Similar (with $x$ instead of $1/x$): math.stackexchange.com/questions/2493443/…
– Hans Lundmark
Aug 19 at 7:49
Similar (with $x$ instead of $1/x$): math.stackexchange.com/questions/2493443/…
– Hans Lundmark
Aug 19 at 7:49
add a comment |Â
1 Answer
1
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votes
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2
down vote
accepted
Let's simplify the problem first by integrating from $-1$ to $1$. There is a discontinuity at $0$, so you write the integral as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_epsilon^1frac1xdxright)$$
If you perform this calculation, you obtain zero. But that is not the only possible way to write the integral. You can equally well write it as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_2epsilon^1frac1xdxright)$$
Note that I've change the lower limit of the second integral. Now obviously you will get a nonzero answer. Or I can change the upper limit of the first integral to be $-5epsilon$.
In addition, you get the same issue when you go to $pminfty$. If your integration ranges are symmetric, you get $0$. This symmetric integration is the so called Cauchy principal value that @dxiv mentioned in the comment.
edited Aug 19 at 6:01
zahbaz
7,62021636
answered Aug 19 at 5:57
Andrei
7,6982822
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let's simplify the problem first by integrating from $-1$ to $1$. There is a discontinuity at $0$, so you write the integral as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_epsilon^1frac1xdxright)$$
If you perform this calculation, you obtain zero. But that is not the only possible way to write the integral. You can equally well write it as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_2epsilon^1frac1xdxright)$$
Note that I've change the lower limit of the second integral. Now obviously you will get a nonzero answer. Or I can change the upper limit of the first integral to be $-5epsilon$.
In addition, you get the same issue when you go to $pminfty$. If your integration ranges are symmetric, you get $0$. This symmetric integration is the so called Cauchy principal value that @dxiv mentioned in the comment.
edited Aug 19 at 6:01
zahbaz
7,62021636
answered Aug 19 at 5:57
Andrei
7,6982822
add a comment |Â
up vote
2
down vote
accepted
Let's simplify the problem first by integrating from $-1$ to $1$. There is a discontinuity at $0$, so you write the integral as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_epsilon^1frac1xdxright)$$
If you perform this calculation, you obtain zero. But that is not the only possible way to write the integral. You can equally well write it as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_2epsilon^1frac1xdxright)$$
Note that I've change the lower limit of the second integral. Now obviously you will get a nonzero answer. Or I can change the upper limit of the first integral to be $-5epsilon$.
In addition, you get the same issue when you go to $pminfty$. If your integration ranges are symmetric, you get $0$. This symmetric integration is the so called Cauchy principal value that @dxiv mentioned in the comment.
edited Aug 19 at 6:01
zahbaz
7,62021636
answered Aug 19 at 5:57
Andrei
7,6982822
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let's simplify the problem first by integrating from $-1$ to $1$. There is a discontinuity at $0$, so you write the integral as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_epsilon^1frac1xdxright)$$
If you perform this calculation, you obtain zero. But that is not the only possible way to write the integral. You can equally well write it as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_2epsilon^1frac1xdxright)$$
Note that I've change the lower limit of the second integral. Now obviously you will get a nonzero answer. Or I can change the upper limit of the first integral to be $-5epsilon$.
In addition, you get the same issue when you go to $pminfty$. If your integration ranges are symmetric, you get $0$. This symmetric integration is the so called Cauchy principal value that @dxiv mentioned in the comment.
edited Aug 19 at 6:01
zahbaz
7,62021636
answered Aug 19 at 5:57
Andrei
7,6982822
Let's simplify the problem first by integrating from $-1$ to $1$. There is a discontinuity at $0$, so you write the integral as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_epsilon^1frac1xdxright)$$
If you perform this calculation, you obtain zero. But that is not the only possible way to write the integral. You can equally well write it as
$$int_-1^1frac1xdx=lim_epsilonto0+left(int_-1^-epsilonfrac1xdx+int_2epsilon^1frac1xdxright)$$
Note that I've change the lower limit of the second integral. Now obviously you will get a nonzero answer. Or I can change the upper limit of the first integral to be $-5epsilon$.
In addition, you get the same issue when you go to $pminfty$. If your integration ranges are symmetric, you get $0$. This symmetric integration is the so called Cauchy principal value that @dxiv mentioned in the comment.
edited Aug 19 at 6:01
zahbaz
7,62021636
answered Aug 19 at 5:57
Andrei
7,6982822
edited Aug 19 at 6:01
zahbaz
7,62021636
edited Aug 19 at 6:01
zahbaz
7,62021636
edited Aug 19 at 6:01
zahbaz
7,62021636
7,62021636
answered Aug 19 at 5:57
Andrei
7,6982822
answered Aug 19 at 5:57
Andrei
7,6982822
answered Aug 19 at 5:57
Andrei
7,6982822
7,6982822
add a comment |Â
add a comment |Â
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6
Lookup Cauchy principal value. Your improper integral diverges, but its Cauchy principal value is indeed $0$.
– dxiv
Aug 19 at 5:10
1
In order the integral $ beginequation int_-infty^inftyfrac1xdx endequation$ to converge, the other two $ beginequation int_-infty^0frac1xdx endequation$ and $ beginequation int_0^inftyfrac1xdx endequation$ have to coverge.
– dmtri
Aug 19 at 5:51
Similar (with $x$ instead of $1/x$): math.stackexchange.com/questions/2493443/…
– Hans Lundmark
Aug 19 at 7:49