Prove that $-\int_0^1 (2(1-2x)^{2n} - 2)dx - \sum_{k=1}^n \int_0^1 {n\choose k} \dfrac{(-4x(1-x))^k}k dx = 2H_{2n} - H_n$

Question

Prove that $-\int_0^1 (2(1-2x)^{2n} - 2)dx - \sum_{k=1}^n \int_0^1 {n\choose k} \dfrac{(-4x(1-x))^k}k dx = 2H_{2n} - H_n$ where $H_n$ is the nth Harmonic number.

For the first part, I think one can use induction on $N$ with the base cases $N=0,1$ being trivial. Though I'm not sure how to prove the inductive step. I know we can ignore the case $K=0$ since the result holds in that case. Using the Binomial theorem to directly expand the expression $(1-x)^{N-K}$ inside the integral doesn't seem to help much.

One can check by standard methods that $-\int_0^1 (2(1-2x)^{2n} - 1)dx = \dfrac{4n}{2n+1}$ (i.e. $\int 2(1-2x)^{2n}dx = -\dfrac{(1-2x)^{2n+1}}{2n+1} $ so the desired integral equals $2 + (\dfrac{(1-2x)^{2n+1}}{2n+1})_0^{1} = \dfrac{4n}{2n+1}$). We have $\int \sum_{k=1}^n {n\choose k} (-z)^{k-1}dz = \int \dfrac{(1-z)^n-1}{(1-z)-1}dz = \int \sum_{i=0}^{n-1} (1-z)^i dz = - \sum_{k=1}^n \dfrac{(1-z)^k}{k}\tag{1}$ I think there might be some typos in the answer I linked at the bottom. I think (1) might be useful, but unlike the answer suggests it doesn't seem like we can directly substitute it into the expression $\sum_{k=1}^n \int_0^1 {n\choose k} \dfrac{(-4x(1-x))^k}k dx$.

Source: The answer by ZA here.

This post seems to have some useful proofs as well.

Answer 1

Given $$I = -\int_{0}^{1} (2 \, (1-2 x)^{2 n} - 2) \, dx - \sum_{k=1}^{n} \int_{0}^{1} {n\choose k} \dfrac{(-4 \, x (1-x))^k}{k} \, dx $$ then, by use of the Beta function, \begin{align} I &= 2 \, \int_{0}^{1} dx - 2 \, \int_{0}^{1} (1 - 2 x)^{2 n} \, dx - \sum_{k=1}^{n} \binom{n}{k} \, \frac{(-4)^{k}}{k} \, \int_{0}^{1} x^k \, (1-x)^k \, dx \\ &= 2 + 2 \, \left[ \frac{(1- 2 x)^{2 n + 1}}{2 \, (2 n + 1)} \right]_{0}^{1} - \sum_{k=1}^{n} \binom{n}{k} \frac{(-4)^k}{k} \, B(k+1, k+1) \\ &= 2 - \frac{2}{2 n + 1} - \sum_{k=1}^{n} \binom{n}{k} \frac{(-4)^k \, (k-1)! \, k!}{(2 k + 1)!} \\ &= 2 - \frac{2}{2 n + 1} + \psi\left(n + \frac{3}{2}\right) - \psi\left( \frac{3}{2} \right) \\ &= \psi\left(n + \frac{1}{2}\right) - \psi\left(\frac{1}{2}\right), \end{align} where $$ \psi(x+1) = \psi(x) + \frac{1}{x} $$ was used. This can also be given as $$ I = H_{n - \frac{1}{2}} + 2 \, \ln 2,$$ where $H_{n}$ is the $n^{th}$ harmonic number.