Are negative powers of cosine Lebesgue-integrable?

Question

I was trying to justify whether the integrals

$$\int_0^\frac{\pi}{2} \cos^\alpha x \,dx$$ are finite for negative $\alpha$, but I cannot come up with a proper inequality to apply comparison test nor an easy primitive.

Answer 1

The only issue with convergence will be at the right boundary. There, $(\cos x)^\alpha$ behaves like $(\frac \pi 2 - x)^\alpha$ near $x=\frac \pi 2$. This will be integrable iff $\alpha>-1$.

To give you a rigorous estimate: For $\alpha<0$ and $x\in(0, \frac \pi 2)$, we actually have $(\cos x)^\alpha >(\frac \pi 2 - x)^\alpha$. Since $\int_0^{\pi/2}(\frac \pi 2 - x)^\alpha\mathrm{d}x=\infty$ for $\alpha\le -1$, the original integral diverges as well.

Answer 2

Hint: Since $\sin (\frac {\pi} 2-x)=\cos x$, the integral is same as $\int_0^{\pi /2} (\sin x)^{\alpha} dx$. Since $\sin x \sim x$ near $0$ the integral is finite if and only if $\int_0^{\pi /2} x^{\alpha} dx$ is finite which means $\alpha > -1$.

Answer 3

The other answers gave the right argument.

If $\alpha>-1$ $$I(\alpha)=\int_0^\frac{\pi}{2} \cos^\alpha (x) \,dx=\frac{\sqrt{\pi }}{2}\,\,\frac{\Gamma \left(\frac{\alpha+1}{2}\right)}{\Gamma \left(\frac{\alpha+2}{2}\right)}$$

Expanded as a series around $\alpha=-1$ $$I(\alpha)=\frac{1}{\alpha+1}+\log (2)+\frac{12 \log ^2(2)-\pi^2}{24} (\alpha+1)+O\left((\alpha+1)^2\right)$$ which is a very good approximation.

For illustration, for $\alpha=-0.99$, the above approximation gives $$100+\frac{12 \log ^2(2)-\pi ^2}{2400}+\log (2)=\color{red}{ 100.6914}37$$