Today I was surfing through YouTube and I cam across this unique and interesting integral on BriTheMathGuy YouTube channel. The integral is as follows :
$\text{Q. Compute}$ $$\int_{0}^{1} (x^x)^{(x^x)^{(x^x)^{\cdots}}}dx$$ I didn't watch the full video and tried to attempt it first on my own.
My Attempt : $$\text{Let}\space t = (x^x)^{(x^x)^{(x^x)^{\cdots}}}$$ $$\implies t = (x^x)^t$$ Taking the natural log of both sides, we get : $$\ln(t) = t\ln(x^x)$$ Transposing the $t$ to the LHS, we get : $$\ln(t)\space e^{-\ln(t)} = \ln(x^x) \tag{t = exp(ln(t))}$$ Now, after giving it some thoughts, I found out that the LHS is very close to the Lambert Function . So, I took the negative of both the sides and got : $$-\ln(t)\space e^{-\ln(t)} = -\ln(x^x)$$ Then taking the Lambert Function of both sides, we get : $$-\ln(t) = W(-\ln(x^x))$$ After all the necessary transposition, we get that : $$t = e^{-W(-\ln(x^x))}$$ After this I am completely lost and don't know what to do next. I watched the video but did not understand the steps mentioned. It would be a great help if anybody can help me with it. Any hint/answer is appreciated
Thanks in advance
This is not an asnwer. Just for the fun (I cannot watch the video)
I think that a series solution could be a way to go.
For level $n$, we have $$f_n=1+t+\frac {2n-1} {2!} t^2+\frac { 3n^2-2} {3!} t^3+\frac {4 n^3+12 n^2-24 n+9} {4!} t^4+$$ $$\frac {5 n^4+60 n^3-120 n^2+60 n-4} {5!} t^5+\frac {6 n^5+210 n^4-360 n^3+240 n-95} {6!} t^6+$$ $$\frac {7 n^6+630 n^5-420 n^4-2520 n^3+4200 n^2-2310 n+414} {7!} t^7+\cdots$$ with $t=x\log(x)$ and $$I_n=\int_0^1 \big[x \log(x)\big]^n\,dx=(-1)^n (n+1)^{-(n+1)}\, \Gamma (n+1)$$
Try to find the patterns for the next terms (I suppose that you will need many more).
In any manner, the result is going asymptotically to $1^-$ (quite slow convergence) $$\left( \begin{array}{cc} n & \text{approximation} \\ 1 & 0.7834282752 \\ 2 & 0.8319574446 \\ 3 & 0.8702455444 \\ 4 & 0.8999112011 \\ 5 & 0.9226399809 \\ 6 & 0.9399475265 \\ 7 & 0.9530953090 \\ 8 & 0.9630868363 \\ 9 & 0.9706985544 \\ 10 & 0.9765210224 \\ 20 & 0.9955618388 \\ 30 & 0.9983126577 \\ 40 & 0.9991074389 \\ 50 & 0.9994485136 \\ 60 & 0.9996236687 \\ 70 & 0.9997327583 \\ 80 & 0.9998003736 \\ 90 & 0.9998441300 \\ 100 & 0.9998503081 \end{array} \right)$$
Edit
Writing $$f_n=1+\sum_{k=1}^\infty \frac{P_k(n)} {k!} \big[x \log(x) \big]^k$$ $$\int_0^1 f_n\,dx=1+\sum_{k=1}^\infty (-1)^k \frac{P_k(n)}{(k+1)^{k+1} }$$