Classic Feynman's Tick - $\int_0^1 \frac{x^{2026}-1}{\ln x}\ \mathrm{d}x$

Problem: $\displaystyle \int_0^1 \frac{x^{2026}-1}{\ln x}\ \mathrm{d}x$ Solution: We consider the general integral parametrized in $\alpha\neq -1$: $ \displaystyle I(\alpha) = \int_0^1 \frac{x^{\alpha}-1}{\ln x}\ \mathrm{d}x $. Note that: \begin{itemize} \item $\displaystyle I(0) = 0$. \item the problem requires $I(2026)$. \end{itemize} We differentiate $I(\alpha)$ with respect to $\alpha$, and using Leibniz Theorem: $$ I'(\alpha) = \int_0^1 \frac{d}{d\alpha}\left(\frac{x^{\alpha}-1}{\ln x}\right)\ \mathrm{d}x = \int_0^1 \left(\frac{x^{\alpha}}{\ln x}\cdot \ln x\right)\ \mathrm{d}x = \int_0^1 x^\alpha \ \mathrm{d}x = \frac{x^{\alpha+1}}{\alpha +1} \Bigr|_0^1 = \frac{1}{\alpha+1} $$ It follows that $I'(\alpha) = \dfrac{1}{\alpha+1}$, and by integrating with respect to $\alpha$, we get that $I(\alpha) = \ln (\alpha + 1)$ + C. By using the initial condition $I(0) = 0$, we get that $C = -\ln 1 = 0$. Thus, $I(\alpha) = \ln (\alpha + 1)$. Therefore, $I(2026) = \displaystyle \int_0^1 \frac{x^{2026}-1}{\ln x}\ \mathrm{d}x = \ln(2026+1) = \boxed{\ln 2027}$.