The integral $\int\ln(x-\ln(x))~dx$

Question

The integral $f(y)=\int_0^y\ln(x-\ln(x))~dx$ is on my mind.

I'm not sure if this has a closed form? Maybe we need to use the lambert-W function to solve this one?

If it cannot be done in closed form, I wonder what a good asymptotic is.

I considered using Taylor series both for solving the integral in closed form and to find a good asymptotic. However $\ln(x)$ (or $\ln(x+1)$) has a limited radius of convergence.

So I'm stuck.

Answer 1

Maple seems to only have a difficulty with $$\int_0^1\ln(x-\ln x) dx = 0.3224577\ldots =: C$$ The taylor series about $y=1$ is the given by $$I(y+1) = C + \frac16 y^3 - \frac1{12} y^4 + \frac1{40} y^5 - \frac1{180} y^6 + \mathcal O(y^7)$$

Answer 2

First of all, I think it is better to put the lower bound of integration to be $1$, and assume $y>0$ so that we get a definite integral: $$f(y)=\int_1^y\ln(x-\ln(x))dx$$ Substituting $x = e^z$ we get: $$\begin{array}{ll}f(y) & =\int_0^{\ln y}\ln(e^z-z)e^zdz\\&=\int_0^{\ln y}\ln(e^z-z)(e^z-1)dz+\int_0^{\ln y}\ln(e^z-z)dz\end{array}$$ Notice that the integral on the left is of the form $\int g(f(z))f'(z)dz=G(f(z))$, where $G$ is a primitive of $g$. Thus with $g(z)=\ln(z)$ and $f(z)=e^z-z$ we get: $$\begin{array}{ll}f(y) & =\bigg[ (e^z-z)\ln(e^z-z) - (e^z-z)\bigg]_0^{\ln y}+\int_0^{\ln y}\ln(e^z-z)dz\\&=(y-\ln y)[\ln(y-\ln y)-1]+1+\int_0^{\ln y}\ln(e^z-z)dz\end{array}$$ We are left with the problem of computing the integral: $$h(y)=\int_0^{\ln y}\ln(e^z-z)dz.$$ (Edit: I deleted what followed here in my old answer, because it was false. I haven't found a solution yet, but I propose the following way to proceed.)

Define $$H(y,a) = \int_0^{\ln(y)}(e^z - z)^adz,$$ then $h(y) = \partial_aH(y,0)$, so the original question is reduced to finding a closed form for $H$ (or showing that it doesn't exist), which looks more manageable.

One possible way to proceed from here could be as follows: use the (generalized) binomial theorem to write $$(e^z-z)^a = \sum_{k=0}^\infty(-1)^k\binom{a}{k}e^{(a-k)z}z^k.$$ If we can exchange somehow the integral with the sum (and there's some work to do), then we would get $$H(y,a) = \sum_{k=0}^\infty(-1)^k\binom{a}{k}\int_0^{\ln(y)}e^{(a-k)z}z^kdz.$$ We can now do the integral integrating by parts. Let $$s_k = \int_0^{\ln(y)}e^{(a-k)z}z^kdz.$$ Then $$\begin{align} s_0 = & \int_0^{\ln(y)}e^{az}dz = \frac{y^a-1}{a}\\ s_{k>0} = & \left[\frac{e^{(a-k)z}}{a-k}z^k\right]_0^{\ln(y)} - \int_0^{\ln(y)}\frac{e^{(a-k)z}}{a-k}kz^{k-1}dz\\ = & \frac{y^{a-k}}{a-k}\ln(y)^k - \frac{k}{a-k}s_{k-1}. \end{align}$$ Now this looks somewhat ugly, but considering instead the whole term $t_k = (-1)^k\binom{a}{k}s_k$ and using $\binom{a}{k} = \tfrac{a-k}{k}\binom{a}{k-1}$, we obtain $$\begin{align} t_0 = & \frac{y^{a-1}}{a}\\ t_{k>0} = & \binom{a}{k-1}\frac{y^{a-k}\ln(y)^k}{k} + t_{k-1}. \end{align}$$ Therefore, $$t_k = \frac{y^a-1}{a} + \sum_{i=1}^k(-1)^i\binom{a}{i-1}\frac{y^{a-i}\ln(y)^i}{i}$$

Now we would like to sum over all $t_k$, and then try to differentiate the resulting expression. This again looks like an ugly beast (a priori, it looks as if it shouldn't converge, so I'm not really sure this approach will work). I will try to work on this in the next few days if I find the time, meanwhile any help, comments, hints, partial solutions or alternative methods would be greatly appreciated.