Principal value of $int^infty_-inftyfrac1x^2dx$: counter-intuitive?
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In attempt to evaluate
$$text Pint^infty_-inftyfrac1x^2dx$$
we consider $$oint_Cfrac1z^2dz$$
where $C$ is an infinitely large semicircle on the upper half plane centered at the origin, with a small indent at the origin.
Obviously, the indent integral and the large arc integral tends to zero. By Cauchy’s integral theorem, the contour integral is also zero. Thus,
$$text Pint^infty_-inftyfrac1x^2dx=0$$
But the integrand is always positive. How could the integral equal zero? How can this counter-intuitive result be explained?
calculus improper-integrals cauchy-principal-value
edited Aug 19 at 8:55
OppoInfinity
508
asked Aug 19 at 6:25
Szeto
4,2661522
add a comment |Â
up vote
0
down vote
favorite
In attempt to evaluate
$$text Pint^infty_-inftyfrac1x^2dx$$
we consider $$oint_Cfrac1z^2dz$$
where $C$ is an infinitely large semicircle on the upper half plane centered at the origin, with a small indent at the origin.
Obviously, the indent integral and the large arc integral tends to zero. By Cauchy’s integral theorem, the contour integral is also zero. Thus,
$$text Pint^infty_-inftyfrac1x^2dx=0$$
But the integrand is always positive. How could the integral equal zero? How can this counter-intuitive result be explained?
calculus improper-integrals cauchy-principal-value
edited Aug 19 at 8:55
OppoInfinity
508
asked Aug 19 at 6:25
Szeto
4,2661522
1
$1/z^2$ is not continuous on the semi disk.
– xbh
Aug 19 at 6:29
1
The indent integral does not tend to zero.
– pks
Aug 19 at 6:52
I don't even see any relations between the Cauchy integral theorem and the principal value, yet. Where does your conclusion of the principal value come from? Anyway, the principal value of $frac1x^2$ does not exist using its definition since you already determined that the integrand is always positive.
– YoungMath
Aug 19 at 9:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In attempt to evaluate
$$text Pint^infty_-inftyfrac1x^2dx$$
we consider $$oint_Cfrac1z^2dz$$
where $C$ is an infinitely large semicircle on the upper half plane centered at the origin, with a small indent at the origin.
Obviously, the indent integral and the large arc integral tends to zero. By Cauchy’s integral theorem, the contour integral is also zero. Thus,
$$text Pint^infty_-inftyfrac1x^2dx=0$$
But the integrand is always positive. How could the integral equal zero? How can this counter-intuitive result be explained?
calculus improper-integrals cauchy-principal-value
edited Aug 19 at 8:55
OppoInfinity
508
asked Aug 19 at 6:25
Szeto
4,2661522
In attempt to evaluate
$$text Pint^infty_-inftyfrac1x^2dx$$
we consider $$oint_Cfrac1z^2dz$$
where $C$ is an infinitely large semicircle on the upper half plane centered at the origin, with a small indent at the origin.
Obviously, the indent integral and the large arc integral tends to zero. By Cauchy’s integral theorem, the contour integral is also zero. Thus,
$$text Pint^infty_-inftyfrac1x^2dx=0$$
But the integrand is always positive. How could the integral equal zero? How can this counter-intuitive result be explained?
calculus improper-integrals cauchy-principal-value
edited Aug 19 at 8:55
OppoInfinity
508
asked Aug 19 at 6:25
Szeto
4,2661522
edited Aug 19 at 8:55
OppoInfinity
508
edited Aug 19 at 8:55
OppoInfinity
508
edited Aug 19 at 8:55
OppoInfinity
508
508
asked Aug 19 at 6:25
Szeto
4,2661522
asked Aug 19 at 6:25
Szeto
4,2661522
asked Aug 19 at 6:25
Szeto
4,2661522
4,2661522
1
$1/z^2$ is not continuous on the semi disk.
– xbh
Aug 19 at 6:29
1
The indent integral does not tend to zero.
– pks
Aug 19 at 6:52
I don't even see any relations between the Cauchy integral theorem and the principal value, yet. Where does your conclusion of the principal value come from? Anyway, the principal value of $frac1x^2$ does not exist using its definition since you already determined that the integrand is always positive.
– YoungMath
Aug 19 at 9:49
add a comment |Â
1
$1/z^2$ is not continuous on the semi disk.
– xbh
Aug 19 at 6:29
1
The indent integral does not tend to zero.
– pks
Aug 19 at 6:52
I don't even see any relations between the Cauchy integral theorem and the principal value, yet. Where does your conclusion of the principal value come from? Anyway, the principal value of $frac1x^2$ does not exist using its definition since you already determined that the integrand is always positive.
– YoungMath
Aug 19 at 9:49
1
1
$1/z^2$ is not continuous on the semi disk.
– xbh
Aug 19 at 6:29
$1/z^2$ is not continuous on the semi disk.
– xbh
Aug 19 at 6:29
1
1
The indent integral does not tend to zero.
– pks
Aug 19 at 6:52
The indent integral does not tend to zero.
– pks
Aug 19 at 6:52
I don't even see any relations between the Cauchy integral theorem and the principal value, yet. Where does your conclusion of the principal value come from? Anyway, the principal value of $frac1x^2$ does not exist using its definition since you already determined that the integrand is always positive.
– YoungMath
Aug 19 at 9:49
I don't even see any relations between the Cauchy integral theorem and the principal value, yet. Where does your conclusion of the principal value come from? Anyway, the principal value of $frac1x^2$ does not exist using its definition since you already determined that the integrand is always positive.
– YoungMath
Aug 19 at 9:49
add a comment |Â
1 Answer
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The issue is first in your obvious step, which then makes it wrong to take the limit after applying Cauchy's theorem. The (counterclockwise) integral over the small semicircle is
$$ int_partial B(0,epsilon)cap Im z>0 frac dzz^2 = int_0^pi fraci epsilon e^ithetaepsilon^2 e^i2theta dtheta = frac 1epsilon int_0^pi ie^-itheta dtheta = frac2epsilon to infty $$
However you might note that the (counterclockwise) integral over the larger semicircle is similarly $int_partial B(0,R)cap Im z>0 = frac2R$, and on the line segments we have
$$ int_[-R,R]setminus[-epsilon,epsilon] fracdzz^2 = 2int_epsilon^R fracdxx^2 = frac2epsilon- frac2R$$
which verifies Cauchy's theorem, before you take the limits.
answered Aug 19 at 9:20
Calvin Khor
8,28711133
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The issue is first in your obvious step, which then makes it wrong to take the limit after applying Cauchy's theorem. The (counterclockwise) integral over the small semicircle is
$$ int_partial B(0,epsilon)cap Im z>0 frac dzz^2 = int_0^pi fraci epsilon e^ithetaepsilon^2 e^i2theta dtheta = frac 1epsilon int_0^pi ie^-itheta dtheta = frac2epsilon to infty $$
However you might note that the (counterclockwise) integral over the larger semicircle is similarly $int_partial B(0,R)cap Im z>0 = frac2R$, and on the line segments we have
$$ int_[-R,R]setminus[-epsilon,epsilon] fracdzz^2 = 2int_epsilon^R fracdxx^2 = frac2epsilon- frac2R$$
which verifies Cauchy's theorem, before you take the limits.
answered Aug 19 at 9:20
Calvin Khor
8,28711133
add a comment |Â
up vote
1
down vote
The issue is first in your obvious step, which then makes it wrong to take the limit after applying Cauchy's theorem. The (counterclockwise) integral over the small semicircle is
$$ int_partial B(0,epsilon)cap Im z>0 frac dzz^2 = int_0^pi fraci epsilon e^ithetaepsilon^2 e^i2theta dtheta = frac 1epsilon int_0^pi ie^-itheta dtheta = frac2epsilon to infty $$
However you might note that the (counterclockwise) integral over the larger semicircle is similarly $int_partial B(0,R)cap Im z>0 = frac2R$, and on the line segments we have
$$ int_[-R,R]setminus[-epsilon,epsilon] fracdzz^2 = 2int_epsilon^R fracdxx^2 = frac2epsilon- frac2R$$
which verifies Cauchy's theorem, before you take the limits.
answered Aug 19 at 9:20
Calvin Khor
8,28711133
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The issue is first in your obvious step, which then makes it wrong to take the limit after applying Cauchy's theorem. The (counterclockwise) integral over the small semicircle is
$$ int_partial B(0,epsilon)cap Im z>0 frac dzz^2 = int_0^pi fraci epsilon e^ithetaepsilon^2 e^i2theta dtheta = frac 1epsilon int_0^pi ie^-itheta dtheta = frac2epsilon to infty $$
However you might note that the (counterclockwise) integral over the larger semicircle is similarly $int_partial B(0,R)cap Im z>0 = frac2R$, and on the line segments we have
$$ int_[-R,R]setminus[-epsilon,epsilon] fracdzz^2 = 2int_epsilon^R fracdxx^2 = frac2epsilon- frac2R$$
which verifies Cauchy's theorem, before you take the limits.
answered Aug 19 at 9:20
Calvin Khor
8,28711133
The issue is first in your obvious step, which then makes it wrong to take the limit after applying Cauchy's theorem. The (counterclockwise) integral over the small semicircle is
$$ int_partial B(0,epsilon)cap Im z>0 frac dzz^2 = int_0^pi fraci epsilon e^ithetaepsilon^2 e^i2theta dtheta = frac 1epsilon int_0^pi ie^-itheta dtheta = frac2epsilon to infty $$
However you might note that the (counterclockwise) integral over the larger semicircle is similarly $int_partial B(0,R)cap Im z>0 = frac2R$, and on the line segments we have
$$ int_[-R,R]setminus[-epsilon,epsilon] fracdzz^2 = 2int_epsilon^R fracdxx^2 = frac2epsilon- frac2R$$
which verifies Cauchy's theorem, before you take the limits.
answered Aug 19 at 9:20
Calvin Khor
8,28711133
answered Aug 19 at 9:20
Calvin Khor
8,28711133
answered Aug 19 at 9:20
Calvin Khor
8,28711133
answered Aug 19 at 9:20
Calvin Khor
8,28711133
8,28711133
add a comment |Â
add a comment |Â
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1
$1/z^2$ is not continuous on the semi disk.
– xbh
Aug 19 at 6:29
1
The indent integral does not tend to zero.
– pks
Aug 19 at 6:52
I don't even see any relations between the Cauchy integral theorem and the principal value, yet. Where does your conclusion of the principal value come from? Anyway, the principal value of $frac1x^2$ does not exist using its definition since you already determined that the integrand is always positive.
– YoungMath
Aug 19 at 9:49