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

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?

• Examine the local behaviour at the endpoints using Taylor polynomials.
– Gary
Aug 13 at 17:44

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 $$(0,\frac\pi2)$$.