1
$\begingroup$

Let $g_+(x) = \sin x$ and $g_-(x) = \cos x$ and similarly let $f_+(x)=\sin x$ be $f_-(x)=\cos x$. I was trying to study the integrals

$$ I_{(\pm,\pm)}(a,n) = \int_0^{\pi/2} \log\bigl( g_{\pm}(ax) \bigr) f_{\pm}(n x) \,\mathrm{d}x $$

Where $a \in\mathbb{R}$ and $n \in \mathbb{Z}$. Really there is no reason to study $g_\pm(ax)$ since

$$ I_{(\pm,\pm)}(a,n) = \int_0^{\pi/2} \log\bigl( g_{\pm}(ax) \bigr) f_{\pm}(n x) \,\mathrm{d}x = I_{(\pm, \pm)}(1,n) + \log a \int_0^{\pi/2} f_{\pm}(n x)\,\mathrm{d}x $$

Where the last integral is trivial. To make things easier introduce $I_{\pm,\pm}(1,n) = I_{\pm, \pm}(n)$. By some symmetry arguments it seems one can say that

$$ \begin{align*} I_{(+,-)}(4n+0) & = I_{(-,-)}(4n+0) = - \frac{\pi}{8n} \\ I_{(+,+)}(4n+0) & = -I_{(-,+)}(4n+0) = \frac{g}{h} \\ I_{(+,+)}(4n+1) & = I_{(-,-)}(4n+1) = \frac{1}{4n+1}\log 2 - \frac{t}{w}\\ I_{(+,-)}(4n+1) & = I_{(-,+)}(4n+1) = \frac{r}{s}\\ I_{(+,-)}(4n+2) & = -I_{(-,-)}(4n+2) = \frac{\pi}{2(4n+2)}\\ I_{(+,+)}(4n+2) & = -I_{(-,+)}(4n+2) = \frac{u}{v}\\ I_{(+,+)}(4n+3) & = -I_{(-,-)}(4n+3) = \frac{1}{4n+3}\log(2) - \frac{p}{q}\\ I_{(+,-)}(4n+3) & = -I_{(-,+)}(4n+3) = \frac{c}{d}\\ \end{align*} $$ Sorry about the formating, not quite sure how to make it more readable. Seems like there are three cases. 1.) $I_n = (-1)^{2n} \pi/n$, $I_n = q_n/p_n$ and lastly $I_n = \frac{1}{n}\log 2 - \frac{a_n}{b_n}$.

Are there any simple closed forms for $a_n, b_n$ or $q_n, p_n$? I guess the $4n$ and $4n+2$ cases with $\pi/n$ are easy to prove. But I did not quite suceed. Any help or suggestions are welcome =)

$\endgroup$
  • $\begingroup$ Fourier Series of $\log 2\sin x$ or $\log 2\cos x$ and the multiple angle formulae? :-) $\endgroup$ – r9m Dec 4 '15 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.