Numerical approximation of an integral I read a problem to determine the integral $\int_1^{100}x^xdx$ with error at most 5% from the book "Which way did the bicycle go". I was a bit disappointed to read the solution which used computer or calculator. I was wondering whether there is a solution to the problem which does not use computers or calculators. In particular, is there way to prove that the solution given in the book has a mistake because it claims that
$$\frac{99^{99}-1}{1+\ln 99}+\frac{100^{100}-99^{99}}{1+\ln 100}\leq \int_1^{100}x^xdx$$
gives a bound $1.78408\cdot 10^{199}\leq \int_1^{100}x^xdx$ but I think the LHS should be $1.78407\cdot 10^{199}\leq \int_1^{100}x^xdx$? I checked this by Sage and Wolfram Alpha but I was unable to do it by pen and paper.

Because your integrand grows so fast the whole integral is dominates for $x\approx 100$. We can write $x^x = \exp[ x \ln(x)]$ and then expanding $x \ln(x) = 100 \ln(100) + [1+ \ln(100)] (x- 100) + \cdots$ around $x = 100$ (note that it is important to expand inside the exponent). The integral can therefore be estimated as
$$\int_1^{100} dx \, x^x \approx 100^{100} \int_{-\infty}^{100} dx\, e^{[1+ \ln(100)] (x- 100) }
= \frac{100^{100}}{1 + \ln (100)}.$$ Numerics shows that this result is off by $3\times 10^{-4}$.
