Conditionally convergent integral
$begingroup$
It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.
My question is about the following integral:
$intlimits_{ - infty }^infty {{{sin left( x right)} over x}dx} $
It's known that this integral is conditionally convergent: it converges, but no absolutely.
Can one say that if ${a_n}buildrel {n to infty } over
longrightarrow infty $ and ${b_n}buildrel {n to infty } over
longrightarrow - infty $ that the value of
$$mathop {lim }limits_{n to infty } intlimits_{{b_n}}^{{a_n}} {{{sin left( x right)} over x}dx} $$
depends on the choice of the sequences ${a_n},{b_n}$ ?
I have experimented a bit in wolfram alpha but no matter what I tried, the result was $pi $.
I do remember hearing and in fact I've read also here:
Is there a rearrangement theorem for conditionally convergent improper integrals?
That the value of the conditionally convergent integral depends on the rearrangement.
improper-integrals
asked Dec 7 '18 at 8:52
zokomokozokomoko
157214
$endgroup$
add a comment |
$begingroup$
It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.
My question is about the following integral:
$intlimits_{ - infty }^infty {{{sin left( x right)} over x}dx} $
It's known that this integral is conditionally convergent: it converges, but no absolutely.
Can one say that if ${a_n}buildrel {n to infty } over
longrightarrow infty $ and ${b_n}buildrel {n to infty } over
longrightarrow - infty $ that the value of
$$mathop {lim }limits_{n to infty } intlimits_{{b_n}}^{{a_n}} {{{sin left( x right)} over x}dx} $$
depends on the choice of the sequences ${a_n},{b_n}$ ?
I have experimented a bit in wolfram alpha but no matter what I tried, the result was $pi $.
I do remember hearing and in fact I've read also here:
Is there a rearrangement theorem for conditionally convergent improper integrals?
That the value of the conditionally convergent integral depends on the rearrangement.
improper-integrals
asked Dec 7 '18 at 8:52
zokomokozokomoko
157214
$endgroup$
add a comment |
$begingroup$
It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.
My question is about the following integral:
$intlimits_{ - infty }^infty {{{sin left( x right)} over x}dx} $
It's known that this integral is conditionally convergent: it converges, but no absolutely.
Can one say that if ${a_n}buildrel {n to infty } over
longrightarrow infty $ and ${b_n}buildrel {n to infty } over
longrightarrow - infty $ that the value of
$$mathop {lim }limits_{n to infty } intlimits_{{b_n}}^{{a_n}} {{{sin left( x right)} over x}dx} $$
depends on the choice of the sequences ${a_n},{b_n}$ ?
I have experimented a bit in wolfram alpha but no matter what I tried, the result was $pi $.
I do remember hearing and in fact I've read also here:
Is there a rearrangement theorem for conditionally convergent improper integrals?
That the value of the conditionally convergent integral depends on the rearrangement.
improper-integrals
asked Dec 7 '18 at 8:52
zokomokozokomoko
157214
$endgroup$
It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.
My question is about the following integral:
$intlimits_{ - infty }^infty {{{sin left( x right)} over x}dx} $
It's known that this integral is conditionally convergent: it converges, but no absolutely.
Can one say that if ${a_n}buildrel {n to infty } over
longrightarrow infty $ and ${b_n}buildrel {n to infty } over
longrightarrow - infty $ that the value of
$$mathop {lim }limits_{n to infty } intlimits_{{b_n}}^{{a_n}} {{{sin left( x right)} over x}dx} $$
depends on the choice of the sequences ${a_n},{b_n}$ ?
I have experimented a bit in wolfram alpha but no matter what I tried, the result was $pi $.
I do remember hearing and in fact I've read also here:
Is there a rearrangement theorem for conditionally convergent improper integrals?
That the value of the conditionally convergent integral depends on the rearrangement.
improper-integrals
improper-integrals
asked Dec 7 '18 at 8:52
zokomokozokomoko
157214
asked Dec 7 '18 at 8:52
zokomokozokomoko
157214
asked Dec 7 '18 at 8:52
zokomokozokomoko
157214
asked Dec 7 '18 at 8:52
zokomokozokomoko
157214
asked Dec 7 '18 at 8:52
zokomokozokomoko
157214
157214
add a comment |
add a comment |
1 Answer
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The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
add a comment |
$begingroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
add a comment |
$begingroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
$endgroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
53.5k32055
53.5k32055
add a comment |
add a comment |
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