Proof $f:\frac{(\ln(\frac{\pi}{2x}))^\gamma}{(\cos x)^\beta (\sin x)^\alpha}$ is Lebesgue integrable.

Question

Values of parameters such that $f:(0,\frac{\pi}{2}) \rightarrow \mathbb{R}: x \rightarrow \frac{(\ln(\frac{\pi}{2x}))^\gamma}{(\cos x)^\beta (\sin x)^\alpha}$ is Lebesgue integrable.

When $x\rightarrow 0$:

I already found that $\alpha < 1$ is needed because $(\frac{1}{\sin x})^\alpha = o(\frac{1}{x^\alpha})$ if $\alpha<1$. But I can not find how I can prove that the function $\ln$ is Lebesgue integrable.

When $x\rightarrow \frac{\pi}{2}$:

In this case the function $\ln$ and the cosine function equal zero. I wanted to try to figure this out with the rule of l'Hôpital but I could not figure this out...

Can anybody give me some help to finish this proof?

Answer 1

This is how I would start. Recall that $x\mapsto x^a(-{\ln x})^b$ is integrable at $0$ if and only if $$a>-1\quad\text{or}\quad(a=-1\enspace\text{and}\enspace b<-1).$$

• As $x\to0^+$, we have $\bigl(\ln(\frac\pi{2x})\bigr)^\gamma\sim(-{\ln x})^\gamma$ and $(\cos x)^\beta(\sin x)^\alpha\sim x^\alpha$, so $$f(x)\underset{x\to0^+}\sim(-{\ln x})^\gamma x^{-\alpha}.$$ Therefore $f$ is integrable at $0^+$ if and only if $\alpha<1$ or ($\alpha=1$ and $\gamma<-1$).

• As $x\to\frac\pi2^-$, we have $\bigl(\ln(\frac\pi{2x})\bigr)^\gamma\sim(\frac2\pi)^\gamma(\frac\pi2-x)^\gamma$ and $(\cos x)^\beta(\sin x)^\alpha\sim(\frac\pi2-x)^\beta$, so $$f(x)\underset{x\to\frac\pi2^-}\sim\left(\frac2\pi\right)^\gamma\left(\frac\pi2-x\right)^{\gamma-\beta}.$$ Hence $f$ is integrable at $\frac\pi2^-$ if and only if $\gamma-\beta>-1$.

• It now suffices to take the intersection of the two conditions to get the integrability of $f$ on $(0,\frac\pi2)$.