Extending Sophomore's Dream to include a constant in the exponent.

$$ x^{a-x}=x^a e^{-x\ln x}=x^a\left(\sum_{n=0}^{\infty}(-1)^n\frac{x^n\ln^n x}{n!}\right)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+a}\ln^n x}{n!}. $$ 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 $$ \int _0^1\limits x^{a-x}\text d x=\sum _{n=1}^\infty \limits \frac{1}{(a+n)^n}. $$

$$\int _0^1\limits x^{a-x}\text d x=\int_0^1e^{(a-x)\ln(x)}dx=\int_0^1\sum_{n=0}^\infty\frac{((a-x)\ln(x))^n}{n!}dx=\sum_{n=0}^\infty\frac{1}{n!}\int_0^1((a-x)\ln(x))^ndx$$ Let $u=-\ln(x)$ and $-e^udu=dx$

So $$\sum_{n=0}^\infty\frac{1}{n!}\int_0^1((a-x)\ln(x))^ndx=\sum_{n=0}^\infty\frac{1}{n!}\int_0^\infty(e^{-u}-a)^nu^ne^udu$$

Now recall that $n!=\int_0^\infty u^n e^{-u}du$

Can you continue from here?