Determine which $p \in (0,\infty)$ where $f \in \mathcal{L}^p(\lambda)$.

Question

Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ given by $$f(t) = \left\{ \begin{array}{ll} tan(t) \quad t \in (0,\frac{\pi}{2}) \\ 0 \quad \text{else} \end{array} \right.$$

And for a measureable space $(X, \mathcal{E},\mu)$, let $$\mathcal{L}^p(\mu)=\bigg\{ f \in \mathcal{M}(\mathcal{E}):\int_X \mid{f}\mid^p d\mu < \infty \bigg\}$$

Determine which $p \in (0,\infty)$ where $f \in \mathcal{L}^p(\lambda)$.

---

My thoughts: So I know I have to compute the following lebesgue integral and I think I have reached these steps:

$$\int_\mathbb{R} \mid{f}\mid^p d\lambda=\int_0^{\pi/2} \mid{tan(t)}\mid^p \lambda(dt)+0=\int_0^{\pi/2} tan(t)^p \lambda(dt)=\int_0^{\pi/2} tan(t)^p dt$$ where the last integral is a Riemann integral. However I am having trouble computing this integral using the fundemental theorem of analysis.

Any help would be much appreciated! And is what I have done so far correct?

Answer 1

You're not going to be able to evaluate that integral for arbitrary $p > 0$ by the fundamental theorem of calculus. Show that $$\lim_{x \to \pi/2} (x - \pi/2)\tan x= -1.$$ Then we have $$\frac{1}{2}\vert x - \pi/2 \vert^{-p} \leq \vert \tan x \vert^p \leq 2\vert x - \pi/2 \vert^{-p}$$ for $x$ sufficiently close to $\pi/2$. If you cut off the support of your function $f$ so as to avoid a neighborhood of $\pi/2$, then you obtain a bounded function with compact support, hence in $L^p(\lambda)$ for all $p > 0$. The above estimates thus show that $f \in L^p(\lambda)$ if and only if $$\int_0^{\pi/2} \vert x - \pi/2 \vert^{-p} \hspace{.05cm} d\lambda(x) < \infty,$$ which occurs if and only if $0 < p < 1$.

Answer 2

Perform the transformation $y=\pi/2-x$ to the integral $\displaystyle\int_{0}^{\pi/2}\tan^{p}xdx$ we get $\displaystyle\int_{0}^{\pi/2}\cot^{p}xdx$. Since $\cos x\approx 1$ near $0$, we are to look at $\displaystyle\int_{0}^{\pi/2}\dfrac{1}{\sin^{p}x}dx$. Since $x/\sin x\approx 1$ near $0$, so we can look at $\displaystyle\int_{0}^{\pi/2}\dfrac{1}{x^{p}}dx$, and this is convergent whenever $0<p<1$.

To be precise, since $\cos x\rightarrow 1$ and $x/\sin x\rightarrow 1$ as $x\rightarrow 0^{+}$, we have some $\delta>0$ such that \begin{align*} \dfrac{1}{2}<\cos x,\dfrac{x}{\sin x}<\dfrac{3}{2},~~~~0<x<\delta. \end{align*} We can just look at $\displaystyle\int_{0}^{\delta}\cot^{p}(x)dx$ since the integrand in $\displaystyle\int_{\delta}^{1}\cot^{p}(x)dx$ has no singularity at all.

As a result, \begin{align*} \left(\dfrac{1}{2}\right)^{2p}\int_{0}^{\delta}\dfrac{1}{x^{p}}dx\leq\int_{0}^{\delta}\cot^{p}xdx\leq\left(\dfrac{3}{2}\right)^{2p}\int_{0}^{\delta}\dfrac{1}{x^{p}}dx, \end{align*} so the convergence of $\displaystyle\int_{0}^{\delta}\dfrac{1}{x^{p}}dx$ says it all.