Mixing
Improper integral outside the domain of the function
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Improper Integral
Wolfram Alpha Answer
Obviously improper as the limits run outside the domain of the function. But does it converge? Any machine algorithm I feed this to says it does, and will provide a complex number as the answer. The definition of convergence relies on the limit existing and being finite - which apparently is the case in this instance. But how can I explain that it technically converges when the function isn't even defined on negatives?
Or perhaps is the function defined on negatives when we include the complex domain? It's kind of freaking me out that I can't get any machine algorithm to tell me this diverges...
integration
asked Jan 11 at 4:13
Michael SchusterMichael Schuster
1
$endgroup$
add a comment |
$begingroup$
Improper Integral
Wolfram Alpha Answer
Obviously improper as the limits run outside the domain of the function. But does it converge? Any machine algorithm I feed this to says it does, and will provide a complex number as the answer. The definition of convergence relies on the limit existing and being finite - which apparently is the case in this instance. But how can I explain that it technically converges when the function isn't even defined on negatives?
Or perhaps is the function defined on negatives when we include the complex domain? It's kind of freaking me out that I can't get any machine algorithm to tell me this diverges...
integration
asked Jan 11 at 4:13
Michael SchusterMichael Schuster
1
$endgroup$
1
$begingroup$
If it makes you feel better, treat $intlimits_{-1}^2sqrt{x}~dx$ as $intlimits_{-1}^0sqrt{x}~dx~+intlimits_{0}^2sqrt{x}~dx$ and recognize that assuming you allow for complex numbers that $intlimits_{-1}^0 sqrt{x}~dx = intlimits_0^1icdotsqrt{x}~dx$. If you don't allow for complex numbers, then the result will of course be undefined.
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– JMoravitz
Jan 11 at 4:25
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OK JM, thank you - Will throw an asterisk in there -->Diverges *(unless we allow for complex numbers...)
$endgroup$
– Michael Schuster
Jan 11 at 4:32
add a comment |
$begingroup$
Improper Integral
Wolfram Alpha Answer
Obviously improper as the limits run outside the domain of the function. But does it converge? Any machine algorithm I feed this to says it does, and will provide a complex number as the answer. The definition of convergence relies on the limit existing and being finite - which apparently is the case in this instance. But how can I explain that it technically converges when the function isn't even defined on negatives?
Or perhaps is the function defined on negatives when we include the complex domain? It's kind of freaking me out that I can't get any machine algorithm to tell me this diverges...
integration
asked Jan 11 at 4:13
Michael SchusterMichael Schuster
1
$endgroup$
Improper Integral
Wolfram Alpha Answer
Obviously improper as the limits run outside the domain of the function. But does it converge? Any machine algorithm I feed this to says it does, and will provide a complex number as the answer. The definition of convergence relies on the limit existing and being finite - which apparently is the case in this instance. But how can I explain that it technically converges when the function isn't even defined on negatives?
Or perhaps is the function defined on negatives when we include the complex domain? It's kind of freaking me out that I can't get any machine algorithm to tell me this diverges...
integration
integration
asked Jan 11 at 4:13
Michael SchusterMichael Schuster
1
asked Jan 11 at 4:13
Michael SchusterMichael Schuster
1
asked Jan 11 at 4:13
Michael SchusterMichael Schuster
1
asked Jan 11 at 4:13
Michael SchusterMichael Schuster
1
asked Jan 11 at 4:13
Michael SchusterMichael Schuster
1
1
1
$begingroup$
If it makes you feel better, treat $intlimits_{-1}^2sqrt{x}~dx$ as $intlimits_{-1}^0sqrt{x}~dx~+intlimits_{0}^2sqrt{x}~dx$ and recognize that assuming you allow for complex numbers that $intlimits_{-1}^0 sqrt{x}~dx = intlimits_0^1icdotsqrt{x}~dx$. If you don't allow for complex numbers, then the result will of course be undefined.
$endgroup$
– JMoravitz
Jan 11 at 4:25
$begingroup$
OK JM, thank you - Will throw an asterisk in there -->Diverges *(unless we allow for complex numbers...)
$endgroup$
– Michael Schuster
Jan 11 at 4:32
add a comment |
1
$begingroup$
If it makes you feel better, treat $intlimits_{-1}^2sqrt{x}~dx$ as $intlimits_{-1}^0sqrt{x}~dx~+intlimits_{0}^2sqrt{x}~dx$ and recognize that assuming you allow for complex numbers that $intlimits_{-1}^0 sqrt{x}~dx = intlimits_0^1icdotsqrt{x}~dx$. If you don't allow for complex numbers, then the result will of course be undefined.
$endgroup$
– JMoravitz
Jan 11 at 4:25
$begingroup$
OK JM, thank you - Will throw an asterisk in there -->Diverges *(unless we allow for complex numbers...)
$endgroup$
– Michael Schuster
Jan 11 at 4:32
1
1
$begingroup$
If it makes you feel better, treat $intlimits_{-1}^2sqrt{x}~dx$ as $intlimits_{-1}^0sqrt{x}~dx~+intlimits_{0}^2sqrt{x}~dx$ and recognize that assuming you allow for complex numbers that $intlimits_{-1}^0 sqrt{x}~dx = intlimits_0^1icdotsqrt{x}~dx$. If you don't allow for complex numbers, then the result will of course be undefined.
$endgroup$
– JMoravitz
Jan 11 at 4:25
$begingroup$
If it makes you feel better, treat $intlimits_{-1}^2sqrt{x}~dx$ as $intlimits_{-1}^0sqrt{x}~dx~+intlimits_{0}^2sqrt{x}~dx$ and recognize that assuming you allow for complex numbers that $intlimits_{-1}^0 sqrt{x}~dx = intlimits_0^1icdotsqrt{x}~dx$. If you don't allow for complex numbers, then the result will of course be undefined.
$endgroup$
– JMoravitz
Jan 11 at 4:25
$begingroup$
OK JM, thank you - Will throw an asterisk in there -->Diverges *(unless we allow for complex numbers...)
$endgroup$
– Michael Schuster
Jan 11 at 4:32
$begingroup$
OK JM, thank you - Will throw an asterisk in there -->Diverges *(unless we allow for complex numbers...)
$endgroup$
– Michael Schuster
Jan 11 at 4:32
add a comment |
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1
$begingroup$
If it makes you feel better, treat $intlimits_{-1}^2sqrt{x}~dx$ as $intlimits_{-1}^0sqrt{x}~dx~+intlimits_{0}^2sqrt{x}~dx$ and recognize that assuming you allow for complex numbers that $intlimits_{-1}^0 sqrt{x}~dx = intlimits_0^1icdotsqrt{x}~dx$. If you don't allow for complex numbers, then the result will of course be undefined.
$endgroup$
– JMoravitz
Jan 11 at 4:25
$begingroup$
OK JM, thank you - Will throw an asterisk in there -->Diverges *(unless we allow for complex numbers...)
$endgroup$
– Michael Schuster
Jan 11 at 4:32