Solving an unsolvable integral ?? [duplicate]
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This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
I recently stumbled upon an indefinite integral . sin(x)/x [ Another similar one is root (x) times sin x.
However if we substitute sin(x) in terms of x as Maclaurin series we could get a series of infinite yet integrable polynomials . What's the catch?
indefinite-integrals
asked Jan 11 at 15:43
Mikhail TalMikhail Tal
31
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marked as duplicate by user170231, Lord_Farin, amWhy, Did, Leucippus Jan 12 at 2:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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show 9 more comments
$begingroup$
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
I recently stumbled upon an indefinite integral . sin(x)/x [ Another similar one is root (x) times sin x.
However if we substitute sin(x) in terms of x as Maclaurin series we could get a series of infinite yet integrable polynomials . What's the catch?
indefinite-integrals
asked Jan 11 at 15:43
Mikhail TalMikhail Tal
31
$endgroup$
marked as duplicate by user170231, Lord_Farin, amWhy, Did, Leucippus Jan 12 at 2:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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Then you're left with a sum of infinite terms that has no compact closed form. So you define one.
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– user170231
Jan 11 at 15:50
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It's not solvable in terms of elementary functions.
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– pie314271
Jan 11 at 15:51
$begingroup$
Okay I get that fact .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
Ok I get it ....but the thing is it's mentioned as "unintegrable" .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
And what about the root(x) sin(x) one?
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
|
show 9 more comments
$begingroup$
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
I recently stumbled upon an indefinite integral . sin(x)/x [ Another similar one is root (x) times sin x.
However if we substitute sin(x) in terms of x as Maclaurin series we could get a series of infinite yet integrable polynomials . What's the catch?
indefinite-integrals
asked Jan 11 at 15:43
Mikhail TalMikhail Tal
31
$endgroup$
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
I recently stumbled upon an indefinite integral . sin(x)/x [ Another similar one is root (x) times sin x.
However if we substitute sin(x) in terms of x as Maclaurin series we could get a series of infinite yet integrable polynomials . What's the catch?
This question already has an answer here:
How can you prove that a function has no closed form integral?
7 answers
indefinite-integrals
indefinite-integrals
asked Jan 11 at 15:43
Mikhail TalMikhail Tal
31
asked Jan 11 at 15:43
Mikhail TalMikhail Tal
31
asked Jan 11 at 15:43
Mikhail TalMikhail Tal
31
asked Jan 11 at 15:43
Mikhail TalMikhail Tal
31
asked Jan 11 at 15:43
Mikhail TalMikhail Tal
31
31
marked as duplicate by user170231, Lord_Farin, amWhy, Did, Leucippus Jan 12 at 2:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by user170231, Lord_Farin, amWhy, Did, Leucippus Jan 12 at 2:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Then you're left with a sum of infinite terms that has no compact closed form. So you define one.
$endgroup$
– user170231
Jan 11 at 15:50
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It's not solvable in terms of elementary functions.
$endgroup$
– pie314271
Jan 11 at 15:51
$begingroup$
Okay I get that fact .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
Ok I get it ....but the thing is it's mentioned as "unintegrable" .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
And what about the root(x) sin(x) one?
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
|
show 9 more comments
1
$begingroup$
Then you're left with a sum of infinite terms that has no compact closed form. So you define one.
$endgroup$
– user170231
Jan 11 at 15:50
$begingroup$
It's not solvable in terms of elementary functions.
$endgroup$
– pie314271
Jan 11 at 15:51
$begingroup$
Okay I get that fact .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
Ok I get it ....but the thing is it's mentioned as "unintegrable" .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
And what about the root(x) sin(x) one?
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
1
1
$begingroup$
Then you're left with a sum of infinite terms that has no compact closed form. So you define one.
$endgroup$
– user170231
Jan 11 at 15:50
$begingroup$
Then you're left with a sum of infinite terms that has no compact closed form. So you define one.
$endgroup$
– user170231
Jan 11 at 15:50
$begingroup$
It's not solvable in terms of elementary functions.
$endgroup$
– pie314271
Jan 11 at 15:51
$begingroup$
It's not solvable in terms of elementary functions.
$endgroup$
– pie314271
Jan 11 at 15:51
$begingroup$
Okay I get that fact .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
Okay I get that fact .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
Ok I get it ....but the thing is it's mentioned as "unintegrable" .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
Ok I get it ....but the thing is it's mentioned as "unintegrable" .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
And what about the root(x) sin(x) one?
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
And what about the root(x) sin(x) one?
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
|
show 9 more comments
1 Answer
1
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The quote you're referring to is "It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for $int sqrt{x} sin; x ; dx$.
The reason is that there does not exist any function whose derivative is
$sqrt{x} sin ; x$."
As stated, this is simply wrong. A correct statement would be that there is not an elementary function whose derivative is $sqrt{x} sin; x$. Integration by parts "works" on $sqrt{x} sin; x$, but it doesn't give you a closed form for the antiderivative. It will say, e.g., that
$$ int sqrt{x} sin ; x ; dx = frac{2}{3} x^{3/2} sin; x - frac{2}{3} int x^{3/2} cos; x; dx $$
which is correct, but not really helpful: it leaves you no closer to finding a formula than before.
answered Jan 11 at 18:35
Robert IsraelRobert Israel
321k23210462
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The quote you're referring to is "It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for $int sqrt{x} sin; x ; dx$.
The reason is that there does not exist any function whose derivative is
$sqrt{x} sin ; x$."
As stated, this is simply wrong. A correct statement would be that there is not an elementary function whose derivative is $sqrt{x} sin; x$. Integration by parts "works" on $sqrt{x} sin; x$, but it doesn't give you a closed form for the antiderivative. It will say, e.g., that
$$ int sqrt{x} sin ; x ; dx = frac{2}{3} x^{3/2} sin; x - frac{2}{3} int x^{3/2} cos; x; dx $$
which is correct, but not really helpful: it leaves you no closer to finding a formula than before.
answered Jan 11 at 18:35
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
The quote you're referring to is "It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for $int sqrt{x} sin; x ; dx$.
The reason is that there does not exist any function whose derivative is
$sqrt{x} sin ; x$."
As stated, this is simply wrong. A correct statement would be that there is not an elementary function whose derivative is $sqrt{x} sin; x$. Integration by parts "works" on $sqrt{x} sin; x$, but it doesn't give you a closed form for the antiderivative. It will say, e.g., that
$$ int sqrt{x} sin ; x ; dx = frac{2}{3} x^{3/2} sin; x - frac{2}{3} int x^{3/2} cos; x; dx $$
which is correct, but not really helpful: it leaves you no closer to finding a formula than before.
answered Jan 11 at 18:35
Robert IsraelRobert Israel
321k23210462
$endgroup$
add a comment |
$begingroup$
The quote you're referring to is "It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for $int sqrt{x} sin; x ; dx$.
The reason is that there does not exist any function whose derivative is
$sqrt{x} sin ; x$."
As stated, this is simply wrong. A correct statement would be that there is not an elementary function whose derivative is $sqrt{x} sin; x$. Integration by parts "works" on $sqrt{x} sin; x$, but it doesn't give you a closed form for the antiderivative. It will say, e.g., that
$$ int sqrt{x} sin ; x ; dx = frac{2}{3} x^{3/2} sin; x - frac{2}{3} int x^{3/2} cos; x; dx $$
which is correct, but not really helpful: it leaves you no closer to finding a formula than before.
answered Jan 11 at 18:35
Robert IsraelRobert Israel
321k23210462
$endgroup$
The quote you're referring to is "It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for $int sqrt{x} sin; x ; dx$.
The reason is that there does not exist any function whose derivative is
$sqrt{x} sin ; x$."
As stated, this is simply wrong. A correct statement would be that there is not an elementary function whose derivative is $sqrt{x} sin; x$. Integration by parts "works" on $sqrt{x} sin; x$, but it doesn't give you a closed form for the antiderivative. It will say, e.g., that
$$ int sqrt{x} sin ; x ; dx = frac{2}{3} x^{3/2} sin; x - frac{2}{3} int x^{3/2} cos; x; dx $$
which is correct, but not really helpful: it leaves you no closer to finding a formula than before.
answered Jan 11 at 18:35
Robert IsraelRobert Israel
321k23210462
answered Jan 11 at 18:35
Robert IsraelRobert Israel
321k23210462
answered Jan 11 at 18:35
Robert IsraelRobert Israel
321k23210462
answered Jan 11 at 18:35
Robert IsraelRobert Israel
321k23210462
321k23210462
add a comment |
add a comment |
1
$begingroup$
Then you're left with a sum of infinite terms that has no compact closed form. So you define one.
$endgroup$
– user170231
Jan 11 at 15:50
$begingroup$
It's not solvable in terms of elementary functions.
$endgroup$
– pie314271
Jan 11 at 15:51
$begingroup$
Okay I get that fact .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
Ok I get it ....but the thing is it's mentioned as "unintegrable" .
$endgroup$
– Mikhail Tal
Jan 11 at 15:59
$begingroup$
And what about the root(x) sin(x) one?
$endgroup$
– Mikhail Tal
Jan 11 at 15:59