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Explore the continuity $F(y) = \int_{0}^{1} \arctan{\frac{x}{y}}dx$ on the set $Y = {\{y: y>0 }\}$ I have tried to explore uniform convergence of $F(y)$

$\arctan{\frac{x}{y}} \leq \frac{\pi}{2}$ hence $F(y)$ converges by Weierstrass and hence F(y) is continuous.

Am I right?

  • 0
    when you say $F(y)$ converges, what exactly do you mean? – 2017-02-24
  • 0
    that $\int_{0}^{1} \arctan{\frac{x}{y}}dx$ has uniform convergence – 2017-02-24

2 Answers 2

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The easiest way is to do the change of variables $x=t\,y$, leading to $$ F(y)=y\int_0^{1/y}\arctan t\,dt. $$

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By integration by parts $\int_{0}^{u}\arctan(t)\,dt = u\arctan(u)-\log\sqrt{1+u^2}$, hence your function equals $\frac{\pi}{2}-\arctan(y)-y\log\sqrt{1+\frac{1}{y^2}}$.