# Suggestions for $\lim_{(x,y)\to (0,0)}\int_x^y \frac{\arctan{(t^3)}}{x+y} dt\$?

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:

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?

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.
• Yes, but he had $x<0<y$, which is a case I think you have to exclude. Apr 5, 2020 at 5:00
• @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.. Apr 5, 2020 at 8:30
• 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? Apr 5, 2020 at 20:08