How to check these terms in $\int{x^x dx}$ power series are correct?

Question

I've got an infinite power series for $\int{x^x}{dx}$ over $x>0$ and I suspect it might have this form:

$$\int{x^x}{dx} = \sum_{n=1}^{\infty}{\frac{x^n}{A(n)}}\sum_{k=0}^{n-1}{B(n,k) \log^k(x)}$$ $$A(n)=\mathrm{lcm}(n^n,n!)$$ $$B(n,k)=\begin{cases}(-1)^{1+n+k}\times\frac{n!\times n^k}{\gcd(n^n,n!)\times k!} & n > k\\0 & \mathrm{otherwise}\end{cases}$$

The first bunch of terms in the sum look like this: $$x +\frac{x^2}{4} (-1 + 2 \log(x)) +\frac{x^3}{54} (2 - 6 \log(x) + 9 \log^2(x)) +\frac{x^4}{768} (-3 + 12 \log(x) - 24 \log^2(x) + 32 \log^3(x)) +\frac{x^5}{75000}((24 - 120 \log(x) + 300 \log^2(x) - 500 \log^3(x) + 625 \log^4(x))) + \dots $$

I checked OEIS and the terms followed patterns related to A055774, A051696, and A095996.

Is anybody aware of any existing derivation of this series and if not, how can I go about proving the general formula is correct and converges to the integral over $x>0$? I've checked using Mathematica to produce the terms. It matches out to 20 terms so far but I'm not sure what to do next to verify it as $n\rightarrow\infty$.

denom[n_] := LCM[n!, n^n]
logterm[n_, k_] := 
 If[n > k, (-1)^(k + n + 1) Denominator[n^n/n!]*n^k/k!, 0]
myseries[x_, n_] := 
 Sum[x^i/denom[i] Sum[logterm[i, j]*Log[x]^j, {j, 0, i - 1}], {i, 1, 
   n}]
(* evaluate and compare *)
Normal[Series[Integrate[x^x, x], {x, 0, 20}]] == myseries[x,20]

A related question Finding $\int x^xdx$ explores this integral but it does not feature this particular power series expansion.

-- edit -- I realized the $\mathrm{lcm}(n^n,n!)$ in $A(n)$ can be pulled into the inner sum combining with the $\gcd(n^n,n!)$ in $B(n,k)$ as $n^n \times n!$ so this simplifies things a bit and the series is now: $$ \int{x^x}{dx} = \sum_{n=1}^{\infty}{x^n}\sum_{k=0}^{n-1}{B(n,k) \log^k(x)} \\ B(n,k)=\begin{cases}(-1)^{1+n+k}\times\frac{n^{k-n}}{k!} & n > k\\0 & \mathrm{otherwise}\end{cases} $$

Answer 1

I think the situation can be simplified. Start with $$ f := \sum_{n=1}^\infty x^n \sum_{k=0}^{n-1} B(n,k)\, y^k. \tag{1}$$ Then use the total derivative to get $$ D(n,k) := D[ x^n y^k ] = x^{n-1}y^{k-1}(k+n\,y) \tag{2} $$ after using $\, D[ y ]=1/x.\,$ The derivative splits into $$ D_1(n,k) := kx^{n-1}y^{k-1},\, D_2(n,k) := nx^{n-1}y^k. \tag{3} $$ Now a bit of algebra shows that $$ -B(n,k)D_1(n,k)=B(n,k-1)D_2(n,k-1). \tag{4} $$ The minus sign leads to cancellation in the derivative of the sum in equation $(1)$ and simplifies into $\,D[ f ] = \exp(x\,y).\,$ Replacing $\,y\,$ with $\,\log(x)\,$ gives $\,x^x.\,$

Answer 2

@Somos's answer is much more elegant, but I also went about it the long-winded mechanical way to differentiate my result and work back to $x^x$.

1. Differentiating: $$ \sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{\partial }{\partial x}\frac{x^n \log ^k(x) (-1)^{1+n+k}}{n^{n-k}\ k!}=\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\frac{(-1)^{1+k+n} x^{n-1} \log ^{k-1}(x) (k+n \log (x))}{n^{n-k}\ k!} $$
2. Logarithms combine (I leaned on Mathematica here to process the inner sum) $$ \\=\sum_{n=1}^{\infty}\frac{(-1)^{2 n}}{n!} n x^{n-1} \log ^{n-1}(x) $$
3. Noting any even power of -1 is 1 and rebasing the outer sum from $n=0$ $$ \sum_{n=1}^{\infty}\frac{n x^{n-1}}{n!} \log ^{n-1}(x)=\sum_{n=0}^{\infty}\frac{n+1}{(n+1)!} x^{n} \log ^{n}(x) $$
4. Using $a/a!=\frac{1}{(a-1)!}$ we get a recognizable series for $x^x$ $$ =\sum_{n=0}^{\infty}\frac{x^{n}\log^{n}(x)}{n!}=x^x \therefore \int{x^x}{dx}=\sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{x^n \log ^k(x) (-1)^{1+n+k}}{n^{n-k}\ k!} $$

Answer 3

(1). For $u,v\in \Bbb Z^+$ we have lcm$(u,v)\cdot$ gcd$(u,v)=ab.$ So when $n>k>0$ we have $\frac {|B(n,k)|}{A(n)}=\frac {n!n^k} {(n^n\cdot n!)k!}=\frac {1}{n^{n-k}\cdot k!}.$

(2). For $x\in \Bbb R^+$ we have $x^x=e^{x\ln x}=\sum_{n=0}^{\infty}(x\ln x)^n/n!.$ We can find $\int (x\ln x)^n dx$ using recursion, as follows: For integers $a,b$ with $a\ge b\ge 0$ let $$F(a,b)=\int x^a(\ln x)^bdx.$$ Integrating by parts, when $a\ge b>0$ we have $$F(a,b)=\int (\ln x)^b (a+1)^{-1}d(x^{a+1})=$$ $$=(a+1)^{-1}x^{a+1}(\ln x)^b-(a+1)^{-1}\int x^{a+1}d((\ln x)^b)=$$ $$=(a+1)^{-1}x^{a+1}(\ln x)^b-(a+1)^{-1}bF(a,b-1)$$ ...(because $d((\ln x)^b)=bx^{-1} (\ln x)^{b-1}dx$ when $b\in \Bbb z^+$)...

giving $F(a,b)$ in terms of $F(a,b-1).$

(3). Regarding convergence: For $x>0 $ and $m\in \Bbb Z^+$ let $H_m(x)=\sum_{n=0}^m(x\ln x)^n/n!,$ and let $H_m(0)=1.$ And let $x^x=1$ when $x=0.$

Once we have $\int H_m(x)=\sum_{n=1}^{m+1}x^nP_n(\ln x)$ for all $x>0,$ where each $P_n$ is a polynomial of degree $n-1,$ we have $$\int_0^yH_m(x)dx=\lim_{r\to 0^+}\int_r^y H_m(x)dx=\sum_{n=1}^{m+1}y^nP_n(\ln y)$$ because $\lim_{r\to 0^+}r^n(\ln r)^j=0$ when $n,j$ are integers with $n>j\ge 0.$ Now each $H_m$ is continuous, and $H_m\to x^x$ uniformly on any bounded subset of $[0,\infty),$ as $m\to \infty,$ so $\int_0^yH_m(x)dx$ converges uniformly to $\int_0^yx^xdx$ on any bounded subset of $[0,\infty).$