Estimating the sign of an integral
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I would like to show that the expression below
$$c|t|^{alpha} bigg( int limits_{|u| < |t|} frac{e^{iu} -1 -iu}{|u|^{1 + alpha}} mbox{d}u + int limits_{|u| ge |t|} frac{e^{iu} -1}{|u|^{1 + alpha}} mbox{d}u bigg) $$
can be rewritten as
$$-b |t|^{alpha}$$
for some $b > 0$.
It's easy to show that both integrals converge. However there are two issues I don't know how to deal with. The first one is connected to the sign of the integral - how can it be shown that both the sum of integrals is negative? The second issue is linked to $|t|$ in the integration limits. Is that a problem? I suppose that the sum of the integrals in brackets should be a negative constant.
real-analysis
asked Jan 8 at 13:13
HendrraHendrra
1,200516
$endgroup$
|
show 1 more comment
$begingroup$
I would like to show that the expression below
$$c|t|^{alpha} bigg( int limits_{|u| < |t|} frac{e^{iu} -1 -iu}{|u|^{1 + alpha}} mbox{d}u + int limits_{|u| ge |t|} frac{e^{iu} -1}{|u|^{1 + alpha}} mbox{d}u bigg) $$
can be rewritten as
$$-b |t|^{alpha}$$
for some $b > 0$.
It's easy to show that both integrals converge. However there are two issues I don't know how to deal with. The first one is connected to the sign of the integral - how can it be shown that both the sum of integrals is negative? The second issue is linked to $|t|$ in the integration limits. Is that a problem? I suppose that the sum of the integrals in brackets should be a negative constant.
real-analysis
asked Jan 8 at 13:13
HendrraHendrra
1,200516
$endgroup$
$begingroup$
You assume that $a=0$, but I see no further mention of any $a$. Also, I see a lot of irrelevant information; you want to show that there exists some $b>0$ such that $$cint limits_{- infty}^{infty}frac{e^{itx} -1 - itx mathbb{1}_{(-1,1)}(x)}{|x|^{1+alpha}} mbox{d}x=-b|t|^{alpha}+2pi ki,$$ for some $kinBbb{Z}$, right?
$endgroup$
– Servaes
Jan 8 at 14:26
$begingroup$
@Servaes you're right! I've made a typo. Now it's fixed. And yes - basically I would like to show what you wrote in your comment
$endgroup$
– Hendrra
Jan 8 at 14:26
$begingroup$
@Servaes there was no $2 pi ki$ mentioned in the content of the task
$endgroup$
– Hendrra
Jan 8 at 14:28
$begingroup$
Consider what it means for the two powers of $e$ to be equal. Though for such $b$ and $k$ to exist for every $c>0$, you must clearly have $k=0$.
$endgroup$
– Servaes
Jan 8 at 14:28
1
$begingroup$
@Servaes I've made some edits. I think now my post is much better and contains less irrelevant information
$endgroup$
– Hendrra
Jan 9 at 14:29
|
show 1 more comment
$begingroup$
I would like to show that the expression below
$$c|t|^{alpha} bigg( int limits_{|u| < |t|} frac{e^{iu} -1 -iu}{|u|^{1 + alpha}} mbox{d}u + int limits_{|u| ge |t|} frac{e^{iu} -1}{|u|^{1 + alpha}} mbox{d}u bigg) $$
can be rewritten as
$$-b |t|^{alpha}$$
for some $b > 0$.
It's easy to show that both integrals converge. However there are two issues I don't know how to deal with. The first one is connected to the sign of the integral - how can it be shown that both the sum of integrals is negative? The second issue is linked to $|t|$ in the integration limits. Is that a problem? I suppose that the sum of the integrals in brackets should be a negative constant.
real-analysis
asked Jan 8 at 13:13
HendrraHendrra
1,200516
$endgroup$
I would like to show that the expression below
$$c|t|^{alpha} bigg( int limits_{|u| < |t|} frac{e^{iu} -1 -iu}{|u|^{1 + alpha}} mbox{d}u + int limits_{|u| ge |t|} frac{e^{iu} -1}{|u|^{1 + alpha}} mbox{d}u bigg) $$
can be rewritten as
$$-b |t|^{alpha}$$
for some $b > 0$.
It's easy to show that both integrals converge. However there are two issues I don't know how to deal with. The first one is connected to the sign of the integral - how can it be shown that both the sum of integrals is negative? The second issue is linked to $|t|$ in the integration limits. Is that a problem? I suppose that the sum of the integrals in brackets should be a negative constant.
real-analysis
real-analysis
asked Jan 8 at 13:13
HendrraHendrra
1,200516
asked Jan 8 at 13:13
HendrraHendrra
1,200516
edited Jan 9 at 14:25
Hendrra
asked Jan 8 at 13:13
HendrraHendrra
1,200516
asked Jan 8 at 13:13
HendrraHendrra
1,200516
asked Jan 8 at 13:13
HendrraHendrra
1,200516
1,200516
$begingroup$
You assume that $a=0$, but I see no further mention of any $a$. Also, I see a lot of irrelevant information; you want to show that there exists some $b>0$ such that $$cint limits_{- infty}^{infty}frac{e^{itx} -1 - itx mathbb{1}_{(-1,1)}(x)}{|x|^{1+alpha}} mbox{d}x=-b|t|^{alpha}+2pi ki,$$ for some $kinBbb{Z}$, right?
$endgroup$
– Servaes
Jan 8 at 14:26
$begingroup$
@Servaes you're right! I've made a typo. Now it's fixed. And yes - basically I would like to show what you wrote in your comment
$endgroup$
– Hendrra
Jan 8 at 14:26
$begingroup$
@Servaes there was no $2 pi ki$ mentioned in the content of the task
$endgroup$
– Hendrra
Jan 8 at 14:28
$begingroup$
Consider what it means for the two powers of $e$ to be equal. Though for such $b$ and $k$ to exist for every $c>0$, you must clearly have $k=0$.
$endgroup$
– Servaes
Jan 8 at 14:28
1
$begingroup$
@Servaes I've made some edits. I think now my post is much better and contains less irrelevant information
$endgroup$
– Hendrra
Jan 9 at 14:29
|
show 1 more comment
$begingroup$
You assume that $a=0$, but I see no further mention of any $a$. Also, I see a lot of irrelevant information; you want to show that there exists some $b>0$ such that $$cint limits_{- infty}^{infty}frac{e^{itx} -1 - itx mathbb{1}_{(-1,1)}(x)}{|x|^{1+alpha}} mbox{d}x=-b|t|^{alpha}+2pi ki,$$ for some $kinBbb{Z}$, right?
$endgroup$
– Servaes
Jan 8 at 14:26
$begingroup$
@Servaes you're right! I've made a typo. Now it's fixed. And yes - basically I would like to show what you wrote in your comment
$endgroup$
– Hendrra
Jan 8 at 14:26
$begingroup$
@Servaes there was no $2 pi ki$ mentioned in the content of the task
$endgroup$
– Hendrra
Jan 8 at 14:28
$begingroup$
Consider what it means for the two powers of $e$ to be equal. Though for such $b$ and $k$ to exist for every $c>0$, you must clearly have $k=0$.
$endgroup$
– Servaes
Jan 8 at 14:28
1
$begingroup$
@Servaes I've made some edits. I think now my post is much better and contains less irrelevant information
$endgroup$
– Hendrra
Jan 9 at 14:29
$begingroup$
You assume that $a=0$, but I see no further mention of any $a$. Also, I see a lot of irrelevant information; you want to show that there exists some $b>0$ such that $$cint limits_{- infty}^{infty}frac{e^{itx} -1 - itx mathbb{1}_{(-1,1)}(x)}{|x|^{1+alpha}} mbox{d}x=-b|t|^{alpha}+2pi ki,$$ for some $kinBbb{Z}$, right?
$endgroup$
– Servaes
Jan 8 at 14:26
$begingroup$
You assume that $a=0$, but I see no further mention of any $a$. Also, I see a lot of irrelevant information; you want to show that there exists some $b>0$ such that $$cint limits_{- infty}^{infty}frac{e^{itx} -1 - itx mathbb{1}_{(-1,1)}(x)}{|x|^{1+alpha}} mbox{d}x=-b|t|^{alpha}+2pi ki,$$ for some $kinBbb{Z}$, right?
$endgroup$
– Servaes
Jan 8 at 14:26
$begingroup$
@Servaes you're right! I've made a typo. Now it's fixed. And yes - basically I would like to show what you wrote in your comment
$endgroup$
– Hendrra
Jan 8 at 14:26
$begingroup$
@Servaes you're right! I've made a typo. Now it's fixed. And yes - basically I would like to show what you wrote in your comment
$endgroup$
– Hendrra
Jan 8 at 14:26
$begingroup$
@Servaes there was no $2 pi ki$ mentioned in the content of the task
$endgroup$
– Hendrra
Jan 8 at 14:28
$begingroup$
@Servaes there was no $2 pi ki$ mentioned in the content of the task
$endgroup$
– Hendrra
Jan 8 at 14:28
$begingroup$
Consider what it means for the two powers of $e$ to be equal. Though for such $b$ and $k$ to exist for every $c>0$, you must clearly have $k=0$.
$endgroup$
– Servaes
Jan 8 at 14:28
$begingroup$
Consider what it means for the two powers of $e$ to be equal. Though for such $b$ and $k$ to exist for every $c>0$, you must clearly have $k=0$.
$endgroup$
– Servaes
Jan 8 at 14:28
1
1
$begingroup$
@Servaes I've made some edits. I think now my post is much better and contains less irrelevant information
$endgroup$
– Hendrra
Jan 9 at 14:29
$begingroup$
@Servaes I've made some edits. I think now my post is much better and contains less irrelevant information
$endgroup$
– Hendrra
Jan 9 at 14:29
|
show 1 more comment
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$begingroup$
You assume that $a=0$, but I see no further mention of any $a$. Also, I see a lot of irrelevant information; you want to show that there exists some $b>0$ such that $$cint limits_{- infty}^{infty}frac{e^{itx} -1 - itx mathbb{1}_{(-1,1)}(x)}{|x|^{1+alpha}} mbox{d}x=-b|t|^{alpha}+2pi ki,$$ for some $kinBbb{Z}$, right?
$endgroup$
– Servaes
Jan 8 at 14:26
$begingroup$
@Servaes you're right! I've made a typo. Now it's fixed. And yes - basically I would like to show what you wrote in your comment
$endgroup$
– Hendrra
Jan 8 at 14:26
$begingroup$
@Servaes there was no $2 pi ki$ mentioned in the content of the task
$endgroup$
– Hendrra
Jan 8 at 14:28
$begingroup$
Consider what it means for the two powers of $e$ to be equal. Though for such $b$ and $k$ to exist for every $c>0$, you must clearly have $k=0$.
$endgroup$
– Servaes
Jan 8 at 14:28
1
$begingroup$
@Servaes I've made some edits. I think now my post is much better and contains less irrelevant information
$endgroup$
– Hendrra
Jan 9 at 14:29