Find the Inverse operator Find $A^{-1}$ if $Af(x)=f(x)+\int\limits_{0}^{1}\cos(x+y)f(y)dy$.
Any ideas? I am not sure how to approach it...
 A: Hints:


*

*Define notation $c:=\cos$ and $s:=\sin$. 

*Define inner product $\langle f, g \rangle := \int_0^1 \overline{f(x)} g(x) dx $.

*Then $(Af)(x)~=~f(x) + c(x) \langle c, f \rangle  - s(x)\langle s, f \rangle .$

*Guess that the inverse is of the form 
$$(A^{-1}f)(x)~=~f(x) +  [c(x)~s(x)] M
 \left[\begin{array}{c}  \langle c, f \rangle \cr \langle s, f \rangle \end{array}\right] ,$$
where $M$ is a constant $2\times 2$ matrix independent of the function $f$ and the argument $x$. 

*Determine $M$ by checking either $A^{-1}A=id$ or $AA^{-1}=id$.
A: Thank you all, I solved it myself last night. Here is my solution:
$g(x)=f(x)-\int\limits_{0}^{1}\cos(x+y)f(y)dy=f(x)-\cos{x}\int\limits_{0}^{1}\cos{y}f(y)dy+\sin{x}\int\limits_{0}^{1}\sin{y}f(y)dy=\\$
$f(x)-\cos{x}c_f+\sin{x}d_f$, where $c_f=\int\limits_{0}^{1}\cos{y}f(y)dy$ and $d_f=\int\limits_{0}^{1}\sin{y}f(y)dy$. Now:
$f(x)=g(x)+\cos{x}c_f-\sin{x}d_f$ and if we substitute $f(x)$ in expressions for $c_f$ and $d_f$ we get:
$c_f=\int\limits_{0}^{1}\cos{x}f(x)dx=\int\limits_{0}^{1}\cos{x}(g(x)+\cos{x}c_f-\sin{x}d_f)=c_g+c_f\int\limits_{0}^{1}\cos^2{x}-d_f\int\limits_{0}^{1}\sin{x}\cos{x}dx.$
We can easily calculate these integrals and obtain first equation (we get another equation by doing the same thing, only starting with $d_f$). At the end we get a 2x2 system, where $c_g$ and $d_g$ are "constants" and $c_f$ and $d_f$ are unknown. By doing this we get $f(x)=g(x)+$something that depends only of $g(x)$ and some trigonometric functions and constants.
A: $\def\l{\lambda}$
We assume the integral is over $[0,\pi]$ and not $[0,1]$. 
If $[0,1]$ is the intended interval the method below will work but the eigenfunctions and eigevalues will be different.
We use the standard shorthand
$(K f)(x) = \int_0^\pi dy\, K(x,y) f(y)$ and 
$\langle f,g\rangle = \int_0^\pi dx f(x)^* g(x)$. 
The norm of $f$ is $\sqrt{\langle f,f\rangle}$.
Let $A f = g$. 
We wish to solve $g = f + K f$, for $f$, that is, 
to solve the inhomogeneous integral equation 
$$\begin{equation*}
f = g + \l K f,\tag{1}
\end{equation*}$$
where $\l = -1$. 
The kernel $K(x,y) = \cos(x+y)$ is that of a degenerate Hilbert-Schmidt integral operator.
We expect there to be a finite number of orthogonal eigenfunctions.
A standard technique involves first examining the homogeneous eigenvalue equation
$$u_i = \l_i K u_i.$$
This is straightforward to solve. 
In detail 
$$\begin{eqnarray*}
u_i(x) &=& \l_i \int_0^\pi dy\, \cos(x+y) u_i(y) \\
&=& \l_i \left( \cos x \int_0^\pi dy\, \cos(y) u_i(y)
-\sin x \int_0^\pi dy\, \sin(y) u_i(y)\right).
\end{eqnarray*}$$
The eigenfunctions must be of the form $A \cos x + B \sin x$. 
Plugging this into the equation above allows us to find the eigenfunctions 
(up to an overall constant) and eigenvalues.
We find the normalized eigenfunctions are 
$\sqrt{\frac{2}{\pi}}\cos x$ and 
$\sqrt{\frac{2}{\pi}}\sin x$, 
with eigenvalues $2/\pi$ and $-2/\pi$, respectively. 
Having the eigenfunctions, one can show that the solution to (1) is 
$$f = g + \l \sum_i \frac{u_i}{\l_i-\l} \langle u_i,g\rangle,$$
that is, 
$$\begin{equation*}
A^{-1}f = f + \l \sum_i \frac{u_i}{\l_i-\l} \langle u_i,f\rangle.\tag{2}
\end{equation*}$$
Addendum: 
For the interval $[0,1]$, the eigenfunctions are 
$$\begin{eqnarray*}
u_1(x) &=& c_1(\cos x - \alpha \sin x) \\
u_2(x) &=& c_2(\sin x - \alpha \cos x), 
\end{eqnarray*}$$
where 
$\alpha = \frac{1}{2} \left(2-\sqrt{6-2\cos2} \, \cos1\right)\csc^2 1$.
It it straightforward to verify the $u_i$ are orthogonal.
The eigenvalues are 
$$\begin{eqnarray*}
\lambda_1 &=& (\sqrt{6-2\cos 2} - 2\sin 1)\sec 1\\
\lambda_2 &=& -(\sqrt{6-2\cos 2} + 2\sin 1)\sec 1.
\end{eqnarray*}$$
The $c_i$ are got by imposing $\langle u_i,u_i\rangle = 1$,
$$\begin{eqnarray*}
c_1 &=& \frac{2}{\sqrt{(2-\sin 2) \alpha^2-4 \alpha\sin ^2 1 +\sin 2+2}} \\
c_2 &=& \frac{2}{\sqrt{(2+\sin 2) \alpha^2-4 \alpha\sin ^2 1 -\sin 2+2}}.
\end{eqnarray*}$$
