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I'm trying to solve this limit over "its natural domain": $\Bbb R^2 \cap (x+y \neq 0)$, I suppose. $$ \lim_{(x,y)\to (0,0)}\int_x^y \frac{\arctan{(t^3)}}{x+y} dt\ $$

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My attempt:
1. $ \lim_{(x,y)\to (0,0)} f(x,mx) \to 0$ with $(m \neq -1)$
2. $ \lim_{(x,y)\to (0,0)} f(x,x^2) \to 0$

I tried to evaluate the limt "near" the line $(-y,y)$ where there could be some problems:
3. $ \lim_{(x,y)\to (0,0)} f(-y\pm y^2,y) \to 0$ (if anybody wants to see steps ask me and I'll edit the post)

So I was thinking the limits could be $0$. I also tried to graph the function:
enter image description here

and I can't see any "strange curve" which I can exploit to disprove my conjecture.

So I tried to make such an estimation:
* $ |f(x,y| \le g(x,y) $
* $ \lim_{(x,y)\to (0,0)} g(x,y) \to 0$

For first I did: $$ \left|\int_x^y \frac{\arctan{(t^3)}}{x+y} dt\ \right| \le \frac{\pi}{2} \frac{|y-x|}{|x+y|} $$
But the last limit doesn't exist.

Then I tried the change of variables: $|x|=u^2$, $|y|=v^2$, separating the various cases; but I fail when I evaluate the cases $(x>0, y<0)$ and $(x<0, y>0)$. For example, the firts: $$ \left|\int_{u^2}^{-v^2} \frac{\arctan{(t^3)}}{u^2-v^2} dt\ \right| \le -\int_{u^2}^{-v^2} \left|\frac{\arctan{(t^3)}}{u^2-v^2}\right| dt\ \le -\frac{1}{|u^2-v^2|} \int_{u^2}^{-v^2} \left|t^3 \right| dt\ \le \frac{u^8+v^8}{|u^2-v^2|} $$ Again, I think the last limit doesn't exist.

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Has anybody some hints to solve the limit?

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1 Answer 1

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I think you probably need to assume $x,y > 0$, otherwise if $x=-y$ it is infinite. With that assumption, since the derivative of arctan is 1 at 0, you can bound $arctan(t^3)$ by a constant times $t^3$, then just do the integration, and go from there.

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  • $\begingroup$ Correct me if I am wrong, but he did do that in the 3rd step of the last integral attempt? $\endgroup$
    – Gareth Ma
    Apr 4, 2020 at 19:44
  • $\begingroup$ Yes, but he had $x<0<y$, which is a case I think you have to exclude. $\endgroup$ Apr 5, 2020 at 5:00
  • $\begingroup$ @Greg I don't think I can exclude other cases except for $x>0, y>0$; infact it is the second question of exercise. I have at least to show the limit doesn't exist at all.. $\endgroup$ Apr 5, 2020 at 8:30
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    $\begingroup$ Wait, I was wrong, if $x=-y$ then you get $0/0$, I overlooked the fact that the integral is $0$. But what about the following argument? The power series of $arctan(t^3)$ is $\sum_{n=0}^\infty \frac{(-1)^n t^{6n+3}}{2n}$. Integrate term by term to get $\sum_{n=0}^\infty \frac{(-1)^n(y^{6n+4}-x^{6n+4})}{2n(6n+4)(x+y)}$. Then you can factor $(y^{6n+4}-x^{6n+4})$ and cancel the $x+y$ in the denominator. Then you obtain $(x-y)(y^{6n+2} + y^6 x^2 + ... + x^{6n+2})$, and the modulus is $\leq |x-y| (3n+2) \max(|x|,|y|)^{6n+2}$. I think this limit should be $0$, no? $\endgroup$ Apr 5, 2020 at 20:08
  • $\begingroup$ At the end, I found another simpler solution. But, I really can't understand how you have factorized the numerator. Also, can you handle infinite sum into limits? $\endgroup$ Apr 6, 2020 at 8:12

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