Solving $\displaystyle \int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}\log(2\cos \theta)d\theta$

Question

$$\int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}\int_{\frac{1}{a}}^{2\cos\theta}\frac{1}{x}dxd\theta=\int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}(\log(2\cos \theta)\,\mathrm +\log(a))d\theta=$$ $$=2\log(a)(\tan^{-1}\sqrt{4a^2-1})+\int_{-\tan^{-1}\sqrt{4a^2-1}}^{\tan^{-1}\sqrt{4a^2-1}}\log(2\cos \theta)d\theta$$

I want to to know if when $a\rightarrow \infty$ the original integral converges and I don't know how to solve the last integral and it seems its difficult (maybe its necesary use further techniques).

So maybe a better way its use inequalites and find an easier integral but I cant find it, so can you help me please?

Answer 1

Note that, for $a\to \infty$, $\tan^{-1}\sqrt{2a^2-1} \to \frac\pi2$. Then, the Integral

$$\int_{-\frac\pi2}^{\frac\pi2}\log(2\cos \theta)d\theta=0$$

However, the first term $2\log(a)(\tan^{-1}\sqrt{4a^2-1})$ diverges; so does the original integral.

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Edit:

\begin{align} I&=\int_0^{\frac{\pi}{2}}\log(2\cos x)dx=\int_0^{\frac{\pi}{2}}\log(2\sin x)dx\\ &=\int_0^{\frac{\pi}{2}}\log(2\cos\frac{x}{2})dx+\int_0^{\frac{\pi}{2}}\log(2\sin\frac{x}{2})dx\\ &=2\int_0^{\frac{\pi}{4}}\log(2\cos x)dx+2\int_0^{\frac{\pi}{4}}\log(2\sin x)dx\\ &=2\int_0^{\frac{\pi}{4}}\log(2\cos x)dx+2\int^{\frac\pi2}_{\frac{\pi}{4}}\log(2\cos x)dx\\ &=2\int_0^{\frac{\pi}{2}}\log(2\cos x)dx=2I \implies I=0 \end{align}