Show that the function $u(x)=\arctan(\frac{1}{x})$ is in $L^p(0,+\infty)$ $ \ \forall p>1$ I need to show that $||u(x)||_{L^p}$ is finite.
$$||u(x)||_{L^p}^p=\int_0^{+\infty}|u(x)|^pdx=\int_0^{+\infty}|\arctan\bigg(\frac{1}{x}\bigg)|^pdx=\int_0^{+\infty}\arctan^p\bigg(\frac{1}{x}\bigg)dx$$
At this point I thought a substitution: $t=\frac{1}{x}$, so $dx=-\frac{1}{t^2}dt$ and the integral become 
$$-\int_{+\infty}^0\bigg(\frac{\arctan(t)}{t^2}\bigg)^pdt=\int_0^{+\infty}\bigg(\frac{\arctan(t)}{t^2}\bigg)^pdt\le\bigg(\frac{\pi}{2}\bigg)^p\int_0^{+\infty}\frac{dt}{t^2}=\bigg(\frac{\pi}{2}\bigg)^p\bigg[-\frac{1}{t}\bigg]_0^{+\infty}$$
but it diverges, I can't understand how to bound above this thing. Someone could help me?

$$
\text{arctan}^p\left(\frac{1}{x}\right) \underset{(+\infty)}{\sim}\frac{1}{x^p}
$$
and the function $\displaystyle x \mapsto \frac{1}{x^p}$ is integrable on $\left[1,+\infty\right[$. Then
$$
\text{arctan}^p\left(\frac{1}{x}\right) \underset{(0^{+})}{\sim}\left(\frac{\pi}{2}\right)^p
$$
So $\displaystyle x \mapsto \text{arctan}^p\left(\frac{1}{x}\right)$ is continuous on $\left[0,1\right]$, so it is integrable on $\left]0,1\right]$. 
Hence it is integrable on $\left]0,+\infty\right[$.
