# For which values of $p$ and $q$ does an improper integral converge

Determine all $p$ and $q$ for which the following improper integral converges. Justify your answer.

$$\int_0^{\pi/2}\frac{\mathrm{d}x}{(\sin^px)(\cos^qx)}$$

I think I have an idea of how to approach this problem, but I'm not certain it's correct. What I've done is to split the interval so we have:

$$\int_{0}^{\pi/4}\frac{\mathrm{d}x}{(\sin^px)(\cos^qx)}+ \int_{\pi/4}^{\pi/2}\frac{\mathrm{d}x}{(\sin^px)(\cos^qx)}$$

Then I state that since $\frac{\sqrt{2}}{2} \le \cos^qx \le 1$ in the first interval, I can use the comparison test:

$$0 \le \frac{1}{\sin^px} \le \frac{1}{(\sin^px)(\cos^qx)}$$

and find that the comparison function converges when $p\le0$

After that I would do something similar for the second interval in which the comparison function would involve $\cos^qx$ in place of $\sin^px$

Is what I've done so far correct? Thanks in advance.

• There is something wrong about your comparison. You want to find an upper bound for the integrand, if you want to prove convergence. – Julien Feb 21 '13 at 3:39
• @ellipsis318: Note that you can type \sin and \cos in $\LaTeX$ to achieve $\sin$ and $\cos$ rather than $sin$ and $cos$. – JavaMan Feb 21 '13 at 3:59
• @JavaMan: Thanks, I noticed your edits and it looks much clearer this way. I'll be sure to write them this way from now on. – ellipsis318 Feb 21 '13 at 4:04
• @ellipsis318: No problem. I'm glad to share it! – JavaMan Feb 21 '13 at 4:05

I would consider the behavior at the endpoints. Near $x=0$, $\sin{x} \sim x$, so that convergence of the integral requires that $p<1$. Similarly, near $x=\pi/2$, $\cos{x} \sim (\pi/2)-x$, so that $q<1$ for convergence. Therefore, $p$ and $q$ each must be less than $1$ for convergence.
• Yes. We are talking about where the integrand blows up sufficiently for the integral to also blow up. So near $x=0$, $1/(\sin^p{x} \cos^q{x}) \sim 1/x^p$. This means that $p<1$ if the integral is to converge. Similar reasoning holds at the other endpoint. – Ron Gordon Feb 21 '13 at 4:10
This function is known as the Beta function, $$B(x,y)=2\int_0^{\pi/2} \sin(t)^{2x-1}\cos(t)^{2y-1}\, \mathrm{d}t.$$ In particular, you are looking at $$\frac{1}{2}B\left(\frac{1}{2}-\frac{p}{2},\frac{1}{2}-\frac{q}{2}\right).$$ This converges as long as $p,q<1$.
For a direct proof, note that by substitution $$\int_0^{\pi/2} \sin(t)^{-p}\cos(t)^{-q}\, \mathrm{d}t=\int_0^\infty \frac{t^{-p/2-1/2}}{(1+t)^{1-p/2-q/2}}\, \mathrm{d}t.$$ For convergence around $0$, we must have that $\frac{p}{2}<\frac{1}{2}$, and for convergence near $\infty$, we must have $\frac{q}{2}<\frac{1}{2}$. This yields $p,q<1$.
• @ellipsis318: $u=\tan(t)$, – Eric Naslund Feb 21 '13 at 4:40