Prove of refute uniform convergence of integral.
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I can't prove (or refute) uniform convergence of
$$
int_{1}^{+infty} arctanfrac{2y}{x^2+y^2}dx,;yinmathbb{R}
$$
I tried Weierstrass M-test, but failed. Can't find the solution to this.
real-analysis integration convergence
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add a comment |
$begingroup$
I can't prove (or refute) uniform convergence of
$$
int_{1}^{+infty} arctanfrac{2y}{x^2+y^2}dx,;yinmathbb{R}
$$
I tried Weierstrass M-test, but failed. Can't find the solution to this.
real-analysis integration convergence
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This is an interesting question. Since you are new you should be made aware that the more effort you show the better received the question. Read the FAQ on how to ask a good question on this site.
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– RRL
Dec 1 '18 at 6:47
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For example, can you show some of your effort in applying the Weierstrass test?
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– RRL
Dec 1 '18 at 6:52
$begingroup$
Please see this on how to ask a good question on this site.
$endgroup$
– RRL
Dec 3 '18 at 0:50
add a comment |
$begingroup$
I can't prove (or refute) uniform convergence of
$$
int_{1}^{+infty} arctanfrac{2y}{x^2+y^2}dx,;yinmathbb{R}
$$
I tried Weierstrass M-test, but failed. Can't find the solution to this.
real-analysis integration convergence
$endgroup$
I can't prove (or refute) uniform convergence of
$$
int_{1}^{+infty} arctanfrac{2y}{x^2+y^2}dx,;yinmathbb{R}
$$
I tried Weierstrass M-test, but failed. Can't find the solution to this.
real-analysis integration convergence
real-analysis integration convergence
edited Dec 1 '18 at 6:36
Chinnapparaj R
5,3131828
5,3131828
asked Dec 1 '18 at 6:11
guarandooguarandoo
133
133
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This is an interesting question. Since you are new you should be made aware that the more effort you show the better received the question. Read the FAQ on how to ask a good question on this site.
$endgroup$
– RRL
Dec 1 '18 at 6:47
$begingroup$
For example, can you show some of your effort in applying the Weierstrass test?
$endgroup$
– RRL
Dec 1 '18 at 6:52
$begingroup$
Please see this on how to ask a good question on this site.
$endgroup$
– RRL
Dec 3 '18 at 0:50
add a comment |
$begingroup$
This is an interesting question. Since you are new you should be made aware that the more effort you show the better received the question. Read the FAQ on how to ask a good question on this site.
$endgroup$
– RRL
Dec 1 '18 at 6:47
$begingroup$
For example, can you show some of your effort in applying the Weierstrass test?
$endgroup$
– RRL
Dec 1 '18 at 6:52
$begingroup$
Please see this on how to ask a good question on this site.
$endgroup$
– RRL
Dec 3 '18 at 0:50
$begingroup$
This is an interesting question. Since you are new you should be made aware that the more effort you show the better received the question. Read the FAQ on how to ask a good question on this site.
$endgroup$
– RRL
Dec 1 '18 at 6:47
$begingroup$
This is an interesting question. Since you are new you should be made aware that the more effort you show the better received the question. Read the FAQ on how to ask a good question on this site.
$endgroup$
– RRL
Dec 1 '18 at 6:47
$begingroup$
For example, can you show some of your effort in applying the Weierstrass test?
$endgroup$
– RRL
Dec 1 '18 at 6:52
$begingroup$
For example, can you show some of your effort in applying the Weierstrass test?
$endgroup$
– RRL
Dec 1 '18 at 6:52
$begingroup$
Please see this on how to ask a good question on this site.
$endgroup$
– RRL
Dec 3 '18 at 0:50
$begingroup$
Please see this on how to ask a good question on this site.
$endgroup$
– RRL
Dec 3 '18 at 0:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $arctan(cdot)$ is increasing on $[0,1]$ and for $x geqslant 1$ and $y > 0$ we have
$$0 leqslant frac{2y}{x^2 + y^2} leqslant frac{2xy}{x^2 + y^2} leqslant 1$$
Thus,
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > sup_{y in mathbb{R^+}}int_n^{2n} arctanfrac{2y}{x^2 + y^2} , dx \ > sup_{y in mathbb{R^+}}left(narctanfrac{2y}{(2n)^2 + y^2}right) \ geqslant narctanfrac{2n}{4n^2 + n^2 } \= narctanfrac{2}{5n } $$
Using the Taylor expansion $arctan x = x + mathcal{O}(x^3)$ for $x in (-1,1]$ we have
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > nleft[frac{2}{5n} + mathcal{O}left(frac{1}{n^3} right)right] \ = frac{2}{5} + mathcal{O} left(frac{1}{n^2} right)$$
Since the RHS does not converge to $0$ as $n to infty$, the convergence of the improper integral is not uniform for $y in [0,infty)$ and, consequently, for $y in mathbb{R}$.
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$begingroup$
The proof is the same showing convergence is not uniform on any interval where $y$ is unbounded to the right.
$endgroup$
– RRL
Dec 1 '18 at 6:45
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In $$arctanfrac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$$ $frac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$ is argument of $arctan$ in this case?
$endgroup$
– guarandoo
Dec 1 '18 at 20:34
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No I pulled out the $C= arctan( 2y/ ((2n)^2 +y ^2)$ (which does not depend on $x$) from inside the integral. It just means $int_n^{2n} C , dx = C(2n-n) = Cn$.
$endgroup$
– RRL
Dec 1 '18 at 20:38
$begingroup$
But why $arctanfrac{2yn}{(2n)^2+y^2}$ instead of $narctanfrac{2y}{(2n)^2+y^2}$?
$endgroup$
– guarandoo
Dec 1 '18 at 20:45
$begingroup$
You are right. That is a mistake from the earlier answer before you added arctan. I can fix it. Give me some time. Thanks
$endgroup$
– RRL
Dec 1 '18 at 20:52
|
show 2 more comments
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$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
and $f(A,0) = 0$. So the limit function $g(y) = 2cdot 1_{mathbb R setminus {0}} (y)(pi sgn y - arctan (1/y))$.
Now consider
$$
sup_{y in mathbb R^*} vert f(A,y) - g(y) vert = sup_{mathbb R^*} vert 2 arctan (A/y) - pi sgn yvert = sgn_{mathbb R^+} vert 2arctan (A/y) - pi vert geqslant vert 2arctan (A/A) -pi vert =fracpi 2,
$$
thus the integral does not converge uniformly.
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Ummm… question changed.
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– xbh
Dec 1 '18 at 7:00
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I'm sorry, it's my bad.
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– guarandoo
Dec 1 '18 at 7:25
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@guarandoo Don't mind. I forgot to refresh the page in time.
$endgroup$
– xbh
Dec 1 '18 at 7:26
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $arctan(cdot)$ is increasing on $[0,1]$ and for $x geqslant 1$ and $y > 0$ we have
$$0 leqslant frac{2y}{x^2 + y^2} leqslant frac{2xy}{x^2 + y^2} leqslant 1$$
Thus,
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > sup_{y in mathbb{R^+}}int_n^{2n} arctanfrac{2y}{x^2 + y^2} , dx \ > sup_{y in mathbb{R^+}}left(narctanfrac{2y}{(2n)^2 + y^2}right) \ geqslant narctanfrac{2n}{4n^2 + n^2 } \= narctanfrac{2}{5n } $$
Using the Taylor expansion $arctan x = x + mathcal{O}(x^3)$ for $x in (-1,1]$ we have
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > nleft[frac{2}{5n} + mathcal{O}left(frac{1}{n^3} right)right] \ = frac{2}{5} + mathcal{O} left(frac{1}{n^2} right)$$
Since the RHS does not converge to $0$ as $n to infty$, the convergence of the improper integral is not uniform for $y in [0,infty)$ and, consequently, for $y in mathbb{R}$.
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$begingroup$
The proof is the same showing convergence is not uniform on any interval where $y$ is unbounded to the right.
$endgroup$
– RRL
Dec 1 '18 at 6:45
$begingroup$
In $$arctanfrac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$$ $frac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$ is argument of $arctan$ in this case?
$endgroup$
– guarandoo
Dec 1 '18 at 20:34
$begingroup$
No I pulled out the $C= arctan( 2y/ ((2n)^2 +y ^2)$ (which does not depend on $x$) from inside the integral. It just means $int_n^{2n} C , dx = C(2n-n) = Cn$.
$endgroup$
– RRL
Dec 1 '18 at 20:38
$begingroup$
But why $arctanfrac{2yn}{(2n)^2+y^2}$ instead of $narctanfrac{2y}{(2n)^2+y^2}$?
$endgroup$
– guarandoo
Dec 1 '18 at 20:45
$begingroup$
You are right. That is a mistake from the earlier answer before you added arctan. I can fix it. Give me some time. Thanks
$endgroup$
– RRL
Dec 1 '18 at 20:52
|
show 2 more comments
$begingroup$
Note that $arctan(cdot)$ is increasing on $[0,1]$ and for $x geqslant 1$ and $y > 0$ we have
$$0 leqslant frac{2y}{x^2 + y^2} leqslant frac{2xy}{x^2 + y^2} leqslant 1$$
Thus,
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > sup_{y in mathbb{R^+}}int_n^{2n} arctanfrac{2y}{x^2 + y^2} , dx \ > sup_{y in mathbb{R^+}}left(narctanfrac{2y}{(2n)^2 + y^2}right) \ geqslant narctanfrac{2n}{4n^2 + n^2 } \= narctanfrac{2}{5n } $$
Using the Taylor expansion $arctan x = x + mathcal{O}(x^3)$ for $x in (-1,1]$ we have
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > nleft[frac{2}{5n} + mathcal{O}left(frac{1}{n^3} right)right] \ = frac{2}{5} + mathcal{O} left(frac{1}{n^2} right)$$
Since the RHS does not converge to $0$ as $n to infty$, the convergence of the improper integral is not uniform for $y in [0,infty)$ and, consequently, for $y in mathbb{R}$.
$endgroup$
$begingroup$
The proof is the same showing convergence is not uniform on any interval where $y$ is unbounded to the right.
$endgroup$
– RRL
Dec 1 '18 at 6:45
$begingroup$
In $$arctanfrac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$$ $frac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$ is argument of $arctan$ in this case?
$endgroup$
– guarandoo
Dec 1 '18 at 20:34
$begingroup$
No I pulled out the $C= arctan( 2y/ ((2n)^2 +y ^2)$ (which does not depend on $x$) from inside the integral. It just means $int_n^{2n} C , dx = C(2n-n) = Cn$.
$endgroup$
– RRL
Dec 1 '18 at 20:38
$begingroup$
But why $arctanfrac{2yn}{(2n)^2+y^2}$ instead of $narctanfrac{2y}{(2n)^2+y^2}$?
$endgroup$
– guarandoo
Dec 1 '18 at 20:45
$begingroup$
You are right. That is a mistake from the earlier answer before you added arctan. I can fix it. Give me some time. Thanks
$endgroup$
– RRL
Dec 1 '18 at 20:52
|
show 2 more comments
$begingroup$
Note that $arctan(cdot)$ is increasing on $[0,1]$ and for $x geqslant 1$ and $y > 0$ we have
$$0 leqslant frac{2y}{x^2 + y^2} leqslant frac{2xy}{x^2 + y^2} leqslant 1$$
Thus,
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > sup_{y in mathbb{R^+}}int_n^{2n} arctanfrac{2y}{x^2 + y^2} , dx \ > sup_{y in mathbb{R^+}}left(narctanfrac{2y}{(2n)^2 + y^2}right) \ geqslant narctanfrac{2n}{4n^2 + n^2 } \= narctanfrac{2}{5n } $$
Using the Taylor expansion $arctan x = x + mathcal{O}(x^3)$ for $x in (-1,1]$ we have
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > nleft[frac{2}{5n} + mathcal{O}left(frac{1}{n^3} right)right] \ = frac{2}{5} + mathcal{O} left(frac{1}{n^2} right)$$
Since the RHS does not converge to $0$ as $n to infty$, the convergence of the improper integral is not uniform for $y in [0,infty)$ and, consequently, for $y in mathbb{R}$.
$endgroup$
Note that $arctan(cdot)$ is increasing on $[0,1]$ and for $x geqslant 1$ and $y > 0$ we have
$$0 leqslant frac{2y}{x^2 + y^2} leqslant frac{2xy}{x^2 + y^2} leqslant 1$$
Thus,
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > sup_{y in mathbb{R^+}}int_n^{2n} arctanfrac{2y}{x^2 + y^2} , dx \ > sup_{y in mathbb{R^+}}left(narctanfrac{2y}{(2n)^2 + y^2}right) \ geqslant narctanfrac{2n}{4n^2 + n^2 } \= narctanfrac{2}{5n } $$
Using the Taylor expansion $arctan x = x + mathcal{O}(x^3)$ for $x in (-1,1]$ we have
$$sup_{y in mathbb{R^+}}int_n^infty arctanfrac{2y}{x^2 + y^2} , dx > nleft[frac{2}{5n} + mathcal{O}left(frac{1}{n^3} right)right] \ = frac{2}{5} + mathcal{O} left(frac{1}{n^2} right)$$
Since the RHS does not converge to $0$ as $n to infty$, the convergence of the improper integral is not uniform for $y in [0,infty)$ and, consequently, for $y in mathbb{R}$.
edited Dec 1 '18 at 21:36
answered Dec 1 '18 at 6:28
RRLRRL
49.4k42573
49.4k42573
$begingroup$
The proof is the same showing convergence is not uniform on any interval where $y$ is unbounded to the right.
$endgroup$
– RRL
Dec 1 '18 at 6:45
$begingroup$
In $$arctanfrac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$$ $frac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$ is argument of $arctan$ in this case?
$endgroup$
– guarandoo
Dec 1 '18 at 20:34
$begingroup$
No I pulled out the $C= arctan( 2y/ ((2n)^2 +y ^2)$ (which does not depend on $x$) from inside the integral. It just means $int_n^{2n} C , dx = C(2n-n) = Cn$.
$endgroup$
– RRL
Dec 1 '18 at 20:38
$begingroup$
But why $arctanfrac{2yn}{(2n)^2+y^2}$ instead of $narctanfrac{2y}{(2n)^2+y^2}$?
$endgroup$
– guarandoo
Dec 1 '18 at 20:45
$begingroup$
You are right. That is a mistake from the earlier answer before you added arctan. I can fix it. Give me some time. Thanks
$endgroup$
– RRL
Dec 1 '18 at 20:52
|
show 2 more comments
$begingroup$
The proof is the same showing convergence is not uniform on any interval where $y$ is unbounded to the right.
$endgroup$
– RRL
Dec 1 '18 at 6:45
$begingroup$
In $$arctanfrac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$$ $frac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$ is argument of $arctan$ in this case?
$endgroup$
– guarandoo
Dec 1 '18 at 20:34
$begingroup$
No I pulled out the $C= arctan( 2y/ ((2n)^2 +y ^2)$ (which does not depend on $x$) from inside the integral. It just means $int_n^{2n} C , dx = C(2n-n) = Cn$.
$endgroup$
– RRL
Dec 1 '18 at 20:38
$begingroup$
But why $arctanfrac{2yn}{(2n)^2+y^2}$ instead of $narctanfrac{2y}{(2n)^2+y^2}$?
$endgroup$
– guarandoo
Dec 1 '18 at 20:45
$begingroup$
You are right. That is a mistake from the earlier answer before you added arctan. I can fix it. Give me some time. Thanks
$endgroup$
– RRL
Dec 1 '18 at 20:52
$begingroup$
The proof is the same showing convergence is not uniform on any interval where $y$ is unbounded to the right.
$endgroup$
– RRL
Dec 1 '18 at 6:45
$begingroup$
The proof is the same showing convergence is not uniform on any interval where $y$ is unbounded to the right.
$endgroup$
– RRL
Dec 1 '18 at 6:45
$begingroup$
In $$arctanfrac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$$ $frac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$ is argument of $arctan$ in this case?
$endgroup$
– guarandoo
Dec 1 '18 at 20:34
$begingroup$
In $$arctanfrac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$$ $frac{2y}{(2n)^2+y^2}int_{n}^{2n}1;dx$ is argument of $arctan$ in this case?
$endgroup$
– guarandoo
Dec 1 '18 at 20:34
$begingroup$
No I pulled out the $C= arctan( 2y/ ((2n)^2 +y ^2)$ (which does not depend on $x$) from inside the integral. It just means $int_n^{2n} C , dx = C(2n-n) = Cn$.
$endgroup$
– RRL
Dec 1 '18 at 20:38
$begingroup$
No I pulled out the $C= arctan( 2y/ ((2n)^2 +y ^2)$ (which does not depend on $x$) from inside the integral. It just means $int_n^{2n} C , dx = C(2n-n) = Cn$.
$endgroup$
– RRL
Dec 1 '18 at 20:38
$begingroup$
But why $arctanfrac{2yn}{(2n)^2+y^2}$ instead of $narctanfrac{2y}{(2n)^2+y^2}$?
$endgroup$
– guarandoo
Dec 1 '18 at 20:45
$begingroup$
But why $arctanfrac{2yn}{(2n)^2+y^2}$ instead of $narctanfrac{2y}{(2n)^2+y^2}$?
$endgroup$
– guarandoo
Dec 1 '18 at 20:45
$begingroup$
You are right. That is a mistake from the earlier answer before you added arctan. I can fix it. Give me some time. Thanks
$endgroup$
– RRL
Dec 1 '18 at 20:52
$begingroup$
You are right. That is a mistake from the earlier answer before you added arctan. I can fix it. Give me some time. Thanks
$endgroup$
– RRL
Dec 1 '18 at 20:52
|
show 2 more comments
$begingroup$
$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
and $f(A,0) = 0$. So the limit function $g(y) = 2cdot 1_{mathbb R setminus {0}} (y)(pi sgn y - arctan (1/y))$.
Now consider
$$
sup_{y in mathbb R^*} vert f(A,y) - g(y) vert = sup_{mathbb R^*} vert 2 arctan (A/y) - pi sgn yvert = sgn_{mathbb R^+} vert 2arctan (A/y) - pi vert geqslant vert 2arctan (A/A) -pi vert =fracpi 2,
$$
thus the integral does not converge uniformly.
$endgroup$
$begingroup$
Ummm… question changed.
$endgroup$
– xbh
Dec 1 '18 at 7:00
$begingroup$
I'm sorry, it's my bad.
$endgroup$
– guarandoo
Dec 1 '18 at 7:25
$begingroup$
@guarandoo Don't mind. I forgot to refresh the page in time.
$endgroup$
– xbh
Dec 1 '18 at 7:26
add a comment |
$begingroup$
$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
and $f(A,0) = 0$. So the limit function $g(y) = 2cdot 1_{mathbb R setminus {0}} (y)(pi sgn y - arctan (1/y))$.
Now consider
$$
sup_{y in mathbb R^*} vert f(A,y) - g(y) vert = sup_{mathbb R^*} vert 2 arctan (A/y) - pi sgn yvert = sgn_{mathbb R^+} vert 2arctan (A/y) - pi vert geqslant vert 2arctan (A/A) -pi vert =fracpi 2,
$$
thus the integral does not converge uniformly.
$endgroup$
$begingroup$
Ummm… question changed.
$endgroup$
– xbh
Dec 1 '18 at 7:00
$begingroup$
I'm sorry, it's my bad.
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– guarandoo
Dec 1 '18 at 7:25
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@guarandoo Don't mind. I forgot to refresh the page in time.
$endgroup$
– xbh
Dec 1 '18 at 7:26
add a comment |
$begingroup$
$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
and $f(A,0) = 0$. So the limit function $g(y) = 2cdot 1_{mathbb R setminus {0}} (y)(pi sgn y - arctan (1/y))$.
Now consider
$$
sup_{y in mathbb R^*} vert f(A,y) - g(y) vert = sup_{mathbb R^*} vert 2 arctan (A/y) - pi sgn yvert = sgn_{mathbb R^+} vert 2arctan (A/y) - pi vert geqslant vert 2arctan (A/A) -pi vert =fracpi 2,
$$
thus the integral does not converge uniformly.
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$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
and $f(A,0) = 0$. So the limit function $g(y) = 2cdot 1_{mathbb R setminus {0}} (y)(pi sgn y - arctan (1/y))$.
Now consider
$$
sup_{y in mathbb R^*} vert f(A,y) - g(y) vert = sup_{mathbb R^*} vert 2 arctan (A/y) - pi sgn yvert = sgn_{mathbb R^+} vert 2arctan (A/y) - pi vert geqslant vert 2arctan (A/A) -pi vert =fracpi 2,
$$
thus the integral does not converge uniformly.
answered Dec 1 '18 at 6:57
xbhxbh
5,9681522
5,9681522
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Ummm… question changed.
$endgroup$
– xbh
Dec 1 '18 at 7:00
$begingroup$
I'm sorry, it's my bad.
$endgroup$
– guarandoo
Dec 1 '18 at 7:25
$begingroup$
@guarandoo Don't mind. I forgot to refresh the page in time.
$endgroup$
– xbh
Dec 1 '18 at 7:26
add a comment |
$begingroup$
Ummm… question changed.
$endgroup$
– xbh
Dec 1 '18 at 7:00
$begingroup$
I'm sorry, it's my bad.
$endgroup$
– guarandoo
Dec 1 '18 at 7:25
$begingroup$
@guarandoo Don't mind. I forgot to refresh the page in time.
$endgroup$
– xbh
Dec 1 '18 at 7:26
$begingroup$
Ummm… question changed.
$endgroup$
– xbh
Dec 1 '18 at 7:00
$begingroup$
Ummm… question changed.
$endgroup$
– xbh
Dec 1 '18 at 7:00
$begingroup$
I'm sorry, it's my bad.
$endgroup$
– guarandoo
Dec 1 '18 at 7:25
$begingroup$
I'm sorry, it's my bad.
$endgroup$
– guarandoo
Dec 1 '18 at 7:25
$begingroup$
@guarandoo Don't mind. I forgot to refresh the page in time.
$endgroup$
– xbh
Dec 1 '18 at 7:26
$begingroup$
@guarandoo Don't mind. I forgot to refresh the page in time.
$endgroup$
– xbh
Dec 1 '18 at 7:26
add a comment |
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