# Integration of looped sin function At infinity, i took limits to be L. so L= sin (x+L)

then after integration i am left withL= 2 cosL. how to get value of L

• Note that $$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$ And $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$ – clathratus Feb 12 '19 at 17:44
• yes that i know. i m left with I= 2cos I – maveric Feb 12 '19 at 17:57

Hint:

Assuming $$\lim\limits_{n\to\infty}f_n(x)=f(x)$$ the functional equation for $$f(x)$$ reads: $$f(x)=\sin(x+f(x)).\tag1$$

Instead of trying to resolve the equation (1) wrt $$f$$ one can easily resolve it (taking into account multivaluedness of $$\arcsin$$ function) wrt $$x$$: $$x=\begin{cases} X_1(f)\equiv\arcsin(f)-f,& 0\le x\le\frac\pi2-1\\ X_2(f)\equiv\pi-\arcsin(f)-f,& \frac\pi2-1\le x\le\pi \end{cases}.\tag2$$ and compute the integral in question as $$\int_0^\pi f(x)dx=\int_0^1[X_2(f)-X_1(f)] df=\int_0^1[\pi-2\arcsin(f)] df=\color{red}2.$$

Extended hint:

To justify the equation $$(2)$$ it is necessary that the function $$f(x)$$ consists of exactly two monotonic branches: one increasing from $$0$$ to $$1$$ followed by the other decreasing from $$1$$ to $$0$$. The properties of $$f(x)$$ follow immediately from the equation $$(1)$$, provided that $$g(x)\equiv x+f(x)$$ is a monotonic function mapping $$[0,\pi]\mapsto[0,\pi]$$, and this is easy to demonstrate rewriting the equation $$(1)$$ as: $$g(x)=x+\sin(g(x)).$$

Indeed, $$g(0)=0$$ and $$g(\pi)=\pi$$ are the only possible solutions of

$$g(0)=\sin(g(0)),\quad\text{and}\quad g(\pi)=\pi+\sin(g(\pi)),$$
respectively, and $$g'(x)=1+\cos(g(x))g'(x)\Rightarrow g'(x)=\frac1{1-\cos(g(x))}>0.$$