Which method solves this integral equation? $\int_{-1}^{1}w(x)\,e^{tx}\,dx=6\left(\frac{\cosh(t)}{t^2}-\frac{\sinh(t)}{t^3}\right)$ Today I encountered this integral equation wrt. $w(x)$:
$$\int_{-1}^1 w(x)\ e^{t x}dx = 6\left(\frac{\cosh(t)}{t^2}-\frac{\sinh(t)}{t^3}\right)$$
I never solved such equations, and when I tried to read about it, I ended up sticking to maples intsolve() function, which failed to find a solution. However when I replaced the integral by a quadrature rule and solved it numerically, the solution looked pretty much like $(3-3x^2)/2$ which does in fact solve the equation.
If I encounter another similar equation, is there any proper way of preceding that doesn't rely on guessing the solution?
 A: A possible approach may be the following one. 
Despite its appearance, the RHS is an entire function:
$$\begin{eqnarray*}6\left(\frac{\cosh(t)}{t^2}-\frac{\sinh(t)}{t^3}\right)&=&6\left(\sum_{n\geq 0}\frac{t^{2n-2}}{(2n)!}-\sum_{n\geq 0}\frac{t^{2n-2}}{(2n+1)!}\right)\\&=&6\sum_{n\geq 0}\frac{2n t^{2n-2}}{(2n+1)!}\\&=&12\sum_{n\geq 0}\frac{(n+1)\,t^{2n}}{(2n+3)!}\\&=&3\sum_{n\geq 0}\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\frac{t^{2n}}{(2n)!}\tag{1}\end{eqnarray*}$$
By differentiating both sides of the initial identity $m$ times with respect to $t$, then evaluating at $t=0$, we get $\int_{-1}^{1}x^{2n+1}w(x)\,dx=0$ for any $n\geq 0$, hence $w(x)$ is an even function.
Moreover, we get:
$$\forall{n\geq 0},\qquad  \int_{-1}^{1}x^{2n}w(x)\,dx = 3\cdot\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\tag{2}$$
hence we know how $w(x)$ is represented by Legendre polynomials, that give an orthogonal base of $L^2(-1,1)$ with respect to the usual inner product. Under such assumption, $w\in L^2(-1,1)$, we have that the solution is unique. On the other hand, it is straightforward to check that $w(x)=\frac{3}{2}(1-x^2)$ is an even function that fulfills $(2)$, hence it is the wanted solution.
