Extending Sophomore's Dream to include a constant in the exponent. I am trying to extend the Sophomore's Dream by proving the formula: 
$$\int _0^1\limits  x^{a-x}\text d x=\sum _{n=1}^\infty \limits \frac{1}{(a+n)^n}$$
I have tried induction, integration by parts, and using the same proof as here and substituting $a-x$, but haven't had much luck.
Thanks in advance!

$$
x^{a-x}=x^a e^{-x\ln x}=x^a\left(\sum_{n=0}^{\infty}(-1)^n\frac{x^n\ln^n x}{n!}\right)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+a}\ln^n x}{n!}.
$$
Intergrate by parts
$$
\int_0^1 (-1)^n\frac{x^{n+a}\ln^n x}{n!}\mathrm{d}x=(-1)^{n}\frac{x^{n+a+1}\ln^n x}{(n+a+1)n!}|_{x=0}^{x=1}+(-1)^{n-1}\int_0^1 \frac{x^{n+a}\ln^{n-1}x}{(n+a+1)(n-1)!}\mathrm{d}x
$$
$$
=\cdots=\frac{1}{(n+a+1)^{n+1}}.
$$
Hence we have
$$
\int _0^1\limits  x^{a-x}\text d x=\sum _{n=1}^\infty \limits \frac{1}{(a+n)^n}.
$$ 
