Prove of refute uniform convergence of integral.












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 6:36









Chinnapparaj R

5,3131828




5,3131828










asked Dec 1 '18 at 6:11









guarandooguarandoo

133




133












  • $begingroup$
    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










  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $y in mathbb{R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $y in mathbb{R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $y in mathbb{R}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $y in mathbb{R}$.






share|cite|improve this answer











$endgroup$



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 $y in mathbb{R}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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


















  • $begingroup$
    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











0












$begingroup$

$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
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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    Ummm… question changed.
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    Dec 1 '18 at 7:25










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0












$begingroup$

$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
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.






share|cite|improve this answer









$endgroup$



$$DeclareMathOperatorsgn{sgn}
f(A,y)=int_1^A frac {2y, mathrm dx}{x^2+y^2} = int_1^A frac {2ycdot y}{y^2} cdot frac {mathrm d(x/y)}{(x/y)^2 + 1} = 2int_{1/y}^{A/y} frac {mathrm d u}{1+u^2} = 2arctan (A/y) - 2arctan (1/y) to pi sgn y - 2arctan (1/y) [Ato +infty, y neq 0],
$$
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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 6:57









xbhxbh

5,9681522




5,9681522












  • $begingroup$
    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










  • $begingroup$
    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


















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