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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

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  • $\begingroup$ Note that $$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$ And $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$ $\endgroup$ – clathratus Feb 12 '19 at 17:44
  • $\begingroup$ yes that i know. i m left with I= 2cos I $\endgroup$ – maveric Feb 12 '19 at 17:57
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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 $$ g'(x)=1+\cos(g(x))g'(x)\Rightarrow g'(x)=\frac1{1-\cos(g(x))}>0. $$

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