I am trying to analytically solve a simple looking integral equation: \begin{align*} \int_{0}^1 e^{(1-t)x} \varphi(t) dt = 1, \hspace{0.2cm} \forall x \in [-1,0], \end{align*} but could neither say whether it is solvable and what function $\varphi(t)$ solves the equation.
In the case that this is not solvable, I would like to consider approximate solutions, i.e. roughly speaking \begin{align*} \int_{0}^1 e^{(1-t)x} \varphi(t) dt \approx 1, \hspace{0.2cm} \forall x \in [-1,0], \end{align*} either in the sense of the $L^2$-norm or the uniform-norm. Whichever is easier!
Any help is greatly appreciated!
Suppose that there is an integrable function $\phi$ on $[0,1]$ such that $$\forall\,x\in[-1,0],\qquad \int_0^1e^{-xt}\phi(t)dt=e^{-x}\tag{1}$$ Taking successive derivatives and evaluating the result for $x=0$ we see that $$\forall, n\in\mathbb{N},\quad \int_0^1 t^n \phi(t)dt=1$$ This implies that for every polynomial $P(X)\in \mathbb{R}[X]$ we have $$\int_0^1 P(t) \phi(t)dt=P(1)$$ In particular, $$\forall\,Q(X)\in \mathbb{R}[X],\qquad \int_0^1 Q(t)(t-1) \phi(t)dt=0$$ By Weirstrass theorem every continuous function on $[0,1]$ is the uniform limit of a sequence of polynomials. Thus, we conclude from the above result that $$\forall\,f\in C([0,1]),\qquad \int_0^1f(t)(t-1)\phi(t) dt=0$$ But this means that $(t-1)\phi(t)=0~ a.e.$, and consequently $\phi(t)=0~ a.e.$ which is absurd because it contradicts $(1)$.
As a conclusion: there is no solution $\phi$ to $(1)$ in the space of integrable functions on $[0,1]$.