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$$ \int_{-\frac\pi2}^\frac\pi2 \left(\cos(x) \right)^{\alpha}\, \mathrm{d}x $$

I'm supposed to prove for what values this improper integral converges.

My "feeling" is that it converges for $\alpha \ge 0$, and diverges else, but I don't really know how to prove it.

I know the behavior of cosine should be like this:

In that area $\left(\frac\pi2 \text{ and } -\frac\pi2\right)$ when $\alpha \ge 0$ it should be kind of like upside-down parabola, and as $\alpha$ gets bigger, the parabola gets closer to the $y$ line. When $\alpha <0$ we should get an unbounded function.

But I don't really know how to prove my intuition.

I believe I should probably use a comparison test, but I don't know with what function as for proving for when $\alpha <0$ I don't really have an idea.

Robert Lee
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ilsapo
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  • Could you edit your statement so that it's more clear what you're trying to prove? I understand what you're saying but don't really see a question being asked here. – O.S. Mar 17 '22 at 11:27
  • you right, sorry. I was too absorbed in thinking about the answer, that I forgot to cleary state the question. edited just now, thanks! – ilsapo Mar 17 '22 at 11:36

1 Answers1

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I'll give a closed form for the integral as well as arguing for convergence. The Beta function $$ B(a,b) = 2\int_{0}^{\frac\pi2} \sin^{2a-1}(x) \cos^{2b-1}(x) \, \mathrm{d}x $$ converges when $a, b>0$. Applying this to your case we see that \begin{align*} \int_{-\frac\pi2}^{-\frac\pi2}\cos^\alpha(x)\, \mathrm{d}x &\overset{\color{blue}{\text{Even}}}{=} 2\int_{0}^{\frac\pi2} \cos^\alpha(x)\, \mathrm{d}x\\ & = 2\int_{0}^{\frac\pi2} \sin^{2\left(\frac12\right)-1}(x) \cos^{2\left(\frac\alpha2 + \frac12\right) -1}\mathrm{d}x\\ & = B\left(\frac{1}{2}, \frac\alpha2 + \frac12\right)\\ & =\frac{2 \sqrt{\pi} \, \Gamma\left(\frac{\alpha+1}{2}\right)}{\alpha\Gamma\left( \frac{\alpha}{2}\right)} \end{align*} And since $\frac{\alpha}{2} + \frac12 > 0 \implies \alpha >-1$ we can conclude $$ \boxed{\int_{-\frac\pi2}^{-\frac\pi2}\cos^\alpha(x)\, \mathrm{d}x = \frac{2 \sqrt{\pi} \, \Gamma\left(\frac{\alpha+1}{2}\right)}{\alpha\Gamma\left( \frac{\alpha}{2}\right)} \qquad \text{for} \quad \alpha > -1} $$

If you just want the convergence part, you can just adapt the proof for the convergence of the Beta function in the linked answer above. Notice that the $2$ different integral representations are connected through the substitution $t = \sin^2(x)$, and thus $$ 2\int_{0}^\frac\pi2 \cos^\alpha(x)\mathrm{d}x = \int_0^1 (1-t)^{\frac{\alpha-1}{2}}t^{-\frac{1}{2} }\mathrm{d}t $$ from where you can directly apply the same argument for convergence.

Robert Lee
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