How about this?

Let I(a,b)=òx^a ln(x)^b dx

(By the way all integrals here are with respect to x and between 0 and 1.)

Integration by parts reveals that:

I(a,b)=(x^a+1 ln(x)^b)/(a+1)-b/(a+1)I(a,b-1)

Provided a+1 ³ b the first term on the right hand side will be zero after the limits are applied, so we can then say:

I(a,b)=-b/(a+1)I(a,b-1)

By successive application of this formula we can evaluate I(n,n) (for +ve integral n) as:

I(n,n)=(-n)(-n+1)(-n+2)...(-2)(-1)/(n+1)ⁿ×I(n,0)

Clearly I(n,0)=1/(n+1)

So I(n,n)=(-1)ⁿ n!/(n+1)ⁿ⁺¹

Now back to the problem.

òx^-x dx can be written as òe^{-x ln x} dx which can now be expanded out using the infinite series. We get

ò[1-x ln x+(x ln x)²/2!-(x ln x)³/3!+...] dx

=I(0,0)-I(1,1)+I(2,2)/2!-I(3,3)/3!+...

Now the factorials will cancel with the factorials contained within the I(n,n)s. The (-1)ⁿ will disappear and we are left with: