Mixing
Definite integral over singularity
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It can be shown both heuristically and mathematically that the definite integral of an odd function over a symmetric interval is 0.
$$int_-a^aOdd(x)dx=0$$
What I want to know is what is stopping me from evaluating
$$int_-a^a+Deltafrac1xdx$$
$$=int_-a^afrac1xdx+int_a^a+Deltafrac1xdx$$
$$=int_a^a+Deltafrac1xdx$$
Which for $a>0$ would return a finite number with no issue.
calculus integration functions definite-integrals
asked Jul 28 at 22:56
Colin Hicks
36028
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up vote
1
down vote
favorite
It can be shown both heuristically and mathematically that the definite integral of an odd function over a symmetric interval is 0.
$$int_-a^aOdd(x)dx=0$$
What I want to know is what is stopping me from evaluating
$$int_-a^a+Deltafrac1xdx$$
$$=int_-a^afrac1xdx+int_a^a+Deltafrac1xdx$$
$$=int_a^a+Deltafrac1xdx$$
Which for $a>0$ would return a finite number with no issue.
calculus integration functions definite-integrals
asked Jul 28 at 22:56
Colin Hicks
36028
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
It can be shown both heuristically and mathematically that the definite integral of an odd function over a symmetric interval is 0.
$$int_-a^aOdd(x)dx=0$$
What I want to know is what is stopping me from evaluating
$$int_-a^a+Deltafrac1xdx$$
$$=int_-a^afrac1xdx+int_a^a+Deltafrac1xdx$$
$$=int_a^a+Deltafrac1xdx$$
Which for $a>0$ would return a finite number with no issue.
calculus integration functions definite-integrals
asked Jul 28 at 22:56
Colin Hicks
36028
It can be shown both heuristically and mathematically that the definite integral of an odd function over a symmetric interval is 0.
$$int_-a^aOdd(x)dx=0$$
What I want to know is what is stopping me from evaluating
$$int_-a^a+Deltafrac1xdx$$
$$=int_-a^afrac1xdx+int_a^a+Deltafrac1xdx$$
$$=int_a^a+Deltafrac1xdx$$
Which for $a>0$ would return a finite number with no issue.
calculus integration functions definite-integrals
asked Jul 28 at 22:56
Colin Hicks
36028
edited Jul 28 at 23:13
asked Jul 28 at 22:56
Colin Hicks
36028
asked Jul 28 at 22:56
Colin Hicks
36028
asked Jul 28 at 22:56
Colin Hicks
36028
36028
add a comment |Â
add a comment |Â
1 Answer
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up vote
4
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accepted
The problem is that
$$int limits_-a^a frac1x , mathrmd x = int limits_-a^0 frac1x , mathrmd x + int limits_0^a frac1x , mathrmd x = - infty + infty $$
is undefined, which reflects the fact the integral does not exist in the sense of the usual definitions of integrals (Riemann or Lebesgue).
However, intuitively we would like say that these infinities are equally large and should therefore cancel (just like the finite numbers would in the case of an odd function with a well-defined integral).
This intuition is captured by the definition of the Cauchy principal value
of such an integral. It is indeed zero in our case, so we can write
$$ mathrmP.mathrmV. int limits_-a^a+Delta frac1x , mathrmd x = int limits_a^a+Delta frac1x , mathrmd x , . $$
answered Jul 28 at 23:06
ComplexYetTrivial
2,777624
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The problem is that
$$int limits_-a^a frac1x , mathrmd x = int limits_-a^0 frac1x , mathrmd x + int limits_0^a frac1x , mathrmd x = - infty + infty $$
is undefined, which reflects the fact the integral does not exist in the sense of the usual definitions of integrals (Riemann or Lebesgue).
However, intuitively we would like say that these infinities are equally large and should therefore cancel (just like the finite numbers would in the case of an odd function with a well-defined integral).
This intuition is captured by the definition of the Cauchy principal value
of such an integral. It is indeed zero in our case, so we can write
$$ mathrmP.mathrmV. int limits_-a^a+Delta frac1x , mathrmd x = int limits_a^a+Delta frac1x , mathrmd x , . $$
answered Jul 28 at 23:06
ComplexYetTrivial
2,777624
add a comment |Â
up vote
4
down vote
accepted
The problem is that
$$int limits_-a^a frac1x , mathrmd x = int limits_-a^0 frac1x , mathrmd x + int limits_0^a frac1x , mathrmd x = - infty + infty $$
is undefined, which reflects the fact the integral does not exist in the sense of the usual definitions of integrals (Riemann or Lebesgue).
However, intuitively we would like say that these infinities are equally large and should therefore cancel (just like the finite numbers would in the case of an odd function with a well-defined integral).
This intuition is captured by the definition of the Cauchy principal value
of such an integral. It is indeed zero in our case, so we can write
$$ mathrmP.mathrmV. int limits_-a^a+Delta frac1x , mathrmd x = int limits_a^a+Delta frac1x , mathrmd x , . $$
answered Jul 28 at 23:06
ComplexYetTrivial
2,777624
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The problem is that
$$int limits_-a^a frac1x , mathrmd x = int limits_-a^0 frac1x , mathrmd x + int limits_0^a frac1x , mathrmd x = - infty + infty $$
is undefined, which reflects the fact the integral does not exist in the sense of the usual definitions of integrals (Riemann or Lebesgue).
However, intuitively we would like say that these infinities are equally large and should therefore cancel (just like the finite numbers would in the case of an odd function with a well-defined integral).
This intuition is captured by the definition of the Cauchy principal value
of such an integral. It is indeed zero in our case, so we can write
$$ mathrmP.mathrmV. int limits_-a^a+Delta frac1x , mathrmd x = int limits_a^a+Delta frac1x , mathrmd x , . $$
answered Jul 28 at 23:06
ComplexYetTrivial
2,777624
The problem is that
$$int limits_-a^a frac1x , mathrmd x = int limits_-a^0 frac1x , mathrmd x + int limits_0^a frac1x , mathrmd x = - infty + infty $$
is undefined, which reflects the fact the integral does not exist in the sense of the usual definitions of integrals (Riemann or Lebesgue).
However, intuitively we would like say that these infinities are equally large and should therefore cancel (just like the finite numbers would in the case of an odd function with a well-defined integral).
This intuition is captured by the definition of the Cauchy principal value
of such an integral. It is indeed zero in our case, so we can write
$$ mathrmP.mathrmV. int limits_-a^a+Delta frac1x , mathrmd x = int limits_a^a+Delta frac1x , mathrmd x , . $$
answered Jul 28 at 23:06
ComplexYetTrivial
2,777624
answered Jul 28 at 23:06
ComplexYetTrivial
2,777624
answered Jul 28 at 23:06
ComplexYetTrivial
2,777624
answered Jul 28 at 23:06
ComplexYetTrivial
2,777624
2,777624
add a comment |Â
add a comment |Â
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