About the Fourier-Legendre series of $f(x)=e^{-x}$ So for the function $f(x) = \exp(-x)$ I have the formula for the coefficients of
$$f(x) = \sum_{n=0}^{\infty}a_n P_n(x)$$
which is(by using Rodrigues formula) 
$$a_n = \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^ndx$$
I think my next step is the integrate by parts, 
$$a_n = \frac{2n+1}{2} -[\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n]^{1}_{-1} + \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx = \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx  $$
Since the term in the brackets is zero.
It seems to me that if we do this $n$ times (further $n-1$ times). We obtain.
$$a_n = \frac{2n+1}{2}\int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}(x^2-1)^ndx$$
I think I have an idea on how to proceed.
EDIT:
Let
$$\alpha_n = \int_{-1}^{1}\exp(-x)(x^2-1)^ndx$$
Then  $$\alpha_n = 2n\int_{-1}^{1}\exp(-x)x(x^2-1)^{n-1}dx = $$ $$-2n(n-1)\alpha_{n-1} - 2n(n-1)\alpha_{n-2} - 2n(n-1)\int_{-1}^{1}\exp(-x)x(x^2-1)^{n-2}dx$$
I can not seem to rid the $x$ in the last integral.
I have integrated $xe^{-x}$ before integrating the second time then added and subtracted $\alpha_{n-2}$

You have noted that
\begin{align}
   a_n & = \frac{2n+1}{2}\int_{-1}^{1}e^{-x}\frac{1}{2^n n!}\frac{d^{n}}{dx^{n}}(x^2-1)^{n}dx \\
    & = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}\frac{d^{n}}{dx^{n}}(x^2-1)^ndx 
\end{align}
And,
\begin{align}
     \int fg^{(n)}dx&=fg^{(n-1)}-f^{(1)}f^{(n-2)}+f^{(2)}{g^{(n-3)}}-\cdots+(-1)^{n}\int f^{(n)}gdx
\end{align}
Because $(x^2-1)^n=(x-1)^n(x+1)^n$, then all derivatives of $(x^2-1)$ of order $k < n$ vanish at the endpoints of $[-1,1]$. So, using a standard binomial expansion of $(x^2-1)^n$,
\begin{align}
      a_n & = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}(x^2-1)^ndx \\
  & = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}\sum_{k=0}^{n}{{n}\choose{k}}(-1)^{k}x^{2k}dx \\
  & = \frac{2n+1}{2^{n+1}n!}\sum_{k=0}^{n}{{n}\choose{k}}\int_{-1}^{1}e^{-x}x^{2k}dx
\end{align}
Then
\begin{align}
    \int e^{-x}x^{2k}dx & = -e^{-x}x^{2k}+\int e^{-x}2kx^{2k-1}dx \\
      & = -e^{-x}x^{2k}-e^{-x}(2k)x^{2k-1}+\int e^{-x}(2k)(2k-1)x^{2k-2}dx \\
      & = -e^{-x}\{ x^{2k}+(2k)x^{2k-1}+(2k)(2k-1)x^{2k-2}+\cdots+(2k)! \}
\end{align}
It's ugly, but you do get the coefficients once you evaluate the above from $-1$ to $1$.
