So I was browsing the homepage of Youtube to see if there were any math equations that I thought that I might be able to solve when I came across this video by Dr. Trevor Bazett saying that the Bernoulli integral (this dude $\longrightarrow$ $\int_0^1 x^xdx$) is ridiculous, which I thought that I might be able to do. Here are my attempts at solving the Bernoulli integral without a calculator:
$$\mathbf{\text{My first attempt}}$$
The problem is, I really didn't know where to start because unlike my question where it turned out that I just had to remember whether I should differentiate with respect to $t$ or to $x$, however, since I had an $x^x$ before the $dx$, I was unable to continue from there. This concludes my first attempt on solving $\int_0^1 x^xdx$ without a calculator.
$$\mathbf{\text{My second attempt}}$$
Looking back at my post, I actually realized I could write it as $$\int_0^1 e^{\ln(x)^x}dx$$Or$$\int_0^1 e^{x\ln(x)}dx$$
Now, knowing that the Taylor Series is defined as $$e^{\color{red}{x}}=\sum_{n=0}^\infty \frac{\color{red}{x}^n}{n!}$$
I can rewrite it as a sum of integrals like$$\int_0^1 \sum_{n=0}^\infty \frac{(x\ln(x))^n}{n!}dx$$Or$$\sum_{n=0}^\infty \int_0^1 \frac{(x\ln(x))^n}{n!}dx$$
Which can be further simplified as$$\sum_{n=0}^\infty \frac{1}{n!}\int_0^1 x^n(\ln(x))^ndx$$Since I know that the Gamma function (factorial function) is$$\Gamma(n)=\int_0^\infty x^{\color{red}{n}}e^{-x}dx$$
However, I was having a hard time rewriting my current integral when I realized I could substitute $u$ for $-\ln(x)$, $du$ for $-\frac{1}{x}$ (since the derivative of $\ln(x)=\frac{1}{x})$ so I can rewrite it as$$(-1)^n\int_0^\infty e^{-u(n+1)}u^ndu$$
Which substituting $v=(n+1)u$, I can rewrite it as$$\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty e^{-v}v^ndv$$Which now I can rewrite it as$$\frac{(-1)^n}{(n+1)^{n+1}}\cdot\Gamma(n)$$
And now when I plug this into the integral/sum that we had before we started simplifying, we get$$\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^{n+1}}$$since we had a $\frac{1}{n!}$ which cancelled that and the $n!$ out. This is approximately$$0.783440501071313$$
$$\mathbf{\text{My question}}$$
Is my solution correct, and if it is not, what could I do to attain the correct solution or attain it more easily?
$$\mathbf{\text{Mistakes I might have made}}$$
- Rewriting the integrals (I have to have gone wrong somewhere, since there's no way I did this all correctly without a calculator)
- Getting the final answer as a number.
A trick to solve this integral without using Gamma function. For a more general case, we can solve
$$\int_0^1 x^s\ln^n(x)dx,~~~s\in\mathbb{R},~n\in \mathbb{N}$$
Consider:
$$f(a)=\int_0^1 x^a ~dx=\frac{1}{1+a}$$ Therefore, $$\int_0^1 x^s\ln^n(x) ~dx=f^{(n)}(a)\big|_{a=s}=\frac{(-1)^nn!}{(1+s)^{n+1}}$$