# Periodic functions for the definite integral

Considering this question where there is this integral:

$$\int_{0}^{4\pi} \ln|13\sin x+3\sqrt3 \cos x|\mathrm dx \tag 1$$

Easily all the periodic function $$a'\sin(x)+b\cos(x)+c=0 \tag 2$$ can be written as:

$$A\sin(x+\phi)+c=0, \ A=\sqrt{a'^2+b^2}\quad \text{ or }\quad A\cos(x+\varphi)+c=0\tag 3$$ where $$\phi, \varphi=\arctan \ldots$$ are angles definited in radians, hence $$\in\Bbb R$$. Reading the comments of the user @Sangchul Lee, I think that $$|\sin(x)|$$ is a even-function and $$\pi-$$periodic,

$$\int_{0}^{4\pi} \ln|13\sin x+3\sqrt3 \cos x|\,\mathrm{d}x=4\int_{0}^{\pi}\ln| A\sin(x+\phi)|\,\mathrm{d}x=4\int_{0}^{\pi}\ln(A| \sin(x+\phi)|)\,\mathrm{d}x$$

1. Why $$\phi$$ vanished? It is true if $$\phi=K\pi$$, with $$K\in\Bbb Z$$. I not remember this now.
2. Considering the comment "Let $$f:\mathbb R→\mathbb R$$ be $$T$$-periodic and integrable on any finite interval then $$∫_0^Tf(x)dx=∫_0^Tf(x+a)dx$$" when is it useful, for a periodic function,

$$\int_{0}^{T}f(x)\,\mathrm{d}x=\int_{0}^{T}f(x+a)\,\mathrm{d}x=\int_{\color{red}{-a}}^{\color{red}{T-a}}f(x+a)\,\mathrm{d}x$$

and if are there general rules (or what is it happen) for the limits of the integral of a generic periodic function?

$$\int_{\color{blue}{\lambda}}^{\color{blue}{\mu}}f(x+a)\,\mathrm{d}x=\int_{\color{blue}{\lambda}}^{\color{blue}{\mu}}f(x)\,\mathrm{d}x=C\int_{\color{magenta}{\cdots}}^{\color{magenta}{\cdots}}f(x)\,\mathrm{d}x$$ where $$C=C(\lambda)$$ (upper bound) or $$C=C(\mu)$$ (lower bound) is a real constant.

• @Notsredt even function ("funzione pari"); sometimes I done a mistake odd/even in English language. Commented Jul 22, 2020 at 10:23

We have $$13^2+27=14^2$$, so $$13\sin(x)+3\sqrt{3}\cos(x) = 14\sin(x+\varphi),\qquad \varphi=\arctan\frac{3\sqrt{3}}{13}\not\in\pi\mathbb{Z}$$ and $$\begin{eqnarray*}\int_{0}^{4\pi}\log\left|13\sin(x)+3\sqrt{3}\cos(x)\right|\,dx &=& 4\pi\log(14)+\int_{0}^{4\pi}\log\left|\sin x\right|\,dx\\&=&4\pi\log(14)+4\int_{0}^{\pi}\log\sin(x)\,dx\end{eqnarray*}$$ where $$\begin{eqnarray*} I=\int_{0}^{\pi}\log\sin(x)\,dx &=& 2\int_{0}^{\pi/2}\log\sin(2z)\,dz\\&=&\pi\log(2)+2\int_{0}^{\pi/2}\log\sin(z)\,dz+2\int_{0}^{\pi/2}\log\cos(z)\,dz\\&=&\pi\log(2)+2I\end{eqnarray*}$$ leads to $$\int_{0}^{4\pi}\log\left|13\sin(x)+3\sqrt{3}\cos(x)\right|\,dx =4\pi\log(14)-4\pi\log(2) = \color{red}{4\pi\log(7)}.$$
• Always thank you and I ask always excuse me for my low level in English language. I have always a doubt.ì if exist a particular value of upper bound (low) where I have it is not possible to write $B\sin(x+\varphi)=B\sin x$, when $\varphi\notin k\Bbb Z$. Commented Jul 22, 2020 at 10:44
• @Sebastiano: the point is that we don't need $\varphi\in \pi\mathbb{Z}$: since $\sin$ is $2\pi$-periodic, $$\int_{0}^{4\pi} g(\sin(x+\varphi))\,dx = \int_{0}^{4\pi}g(\sin x)\,dx$$ for any $\varphi$. Commented Jul 22, 2020 at 10:50
• Yes, of course. :-)....$4\pi=2(2\pi)$. But for my 2nd request? :-) "Grazie assai". Commented Jul 22, 2020 at 10:52
• @Sebastiano: if the length of the integration range is not a multiple of the period the identity $$\int_{\mu}^{\lambda}f(x+a)\,dx = \int_{\mu}^{\lambda}f(x)\,dx$$ is not granted to hold. Prego ;) Commented Jul 22, 2020 at 10:54
• Easily done: consider $f(x)=\sin(x)$, $\mu=\pi/4, \lambda=\pi/3$ and $a=\pi/2$. Commented Jul 22, 2020 at 10:57