Proving $e$ is irrational using a Beukers-like integral

Question

We know the following, for some integers $a_n,b_n$ and $n$ where $n\geq0$: \begin{align} &I_n = \int_0^1 x^n(1-2x)^n e^x dx = a_ne+b_n \\ &|I_n|= \left\lvert \int_0^1 x^n(1-2x)^n e^x dx \right\rvert \leq \left( \frac{1}{8} \right)^n(e-1) \end{align}

therefore, as $n\to\infty$ the integral $I_n$ tends to zero. So if we assume $e=p/q$ then we would have a integer that tends to zero which is not possible, unless the integer is $0$ itself. So, if we prove that $I_n$ is not zero, then this concludes the proof that $e$ is irrational.

But how to prove that $I_n \neq0$ for all integer $n\geq0\:?$

PS: I know that if we changed the polynomial inside $I_n$ it would be easier to show that $I_n \neq 0$ for all $n$, but I'm interested in this case in particular.

EDIT: I believe the estimation for $I_n$ is wrong. I estimated it by the following way:

\begin{align} &x(1-2x)\leq\text{max}[x(1-2x)] = 1/8\\ &x^n(1-2x)^n\leq \left( 1/8 \right)^n\\ & x^n(1-2x)^n e^x \leq \left( 1/8 \right)^n e^x\\ & \int_0^1 x^n(1-2x)^n e^x dx\leq \left( 1/8 \right)^n \int_0^1 e^xdx\\ & \int_0^1 x^n(1-2x)^n e^x dx\leq \left( 1/8 \right)^n (e-1) \end{align}

Answer 1

$\begin{array}\\ I_n &= \int_0^1 x^n(1-2x)^n e^x dx\\ &= \int_0^{1/2} x^n(1-2x)^n e^x dx+\int_{1/2}^1 x^n(1-2x)^n e^x dx\\ \\ &= \int_0^{1/2} x^n(1-2x)^n e^x dx+\int_0^{1/2} (1-x)^n(1-2(1-x))^n e^{1-x} dx\\ \\ &= \int_0^{1/2} x^n(1-2x)^n e^x dx+\int_0^{1/2} (1-x)^n(2x-1)^n e^{1-x} dx\\ \\ &= \int_0^{1/2} x^n(1-2x)^n e^x dx+\int_0^{1/2} (1-x)^n(-1)^n(1-2x)^n e^{1-x} dx\\ &= \int_0^{1/2} (1-2x)^n (x^ne^x +(1-x)^n(-1)^n e^{1-x}) dx\\ \end{array} $

If $n$ is even, $x^ne^x +(1-x)^n(-1)^n e^{1-x} =x^ne^x +(1-x)^n e^{1-x} \gt 0 $.

If $n$ is odd, $x^ne^x +(1-x)^n(-1)^n e^{1-x} =x^ne^x -(1-x)^n e^{1-x} \lt 0 $ for $0 < x < 1/2$ since $x < 1-x$ and $e^x < e^{1-x}$.

Answer 2

The case is trivial when $n$ is even ($I_n$ always $>0$), so we just assume $n$ is odd. Write $$I_n=\underbrace{\int_0^{1/2} x^n(1-2x)^n e^x~\mathrm dx}_{J_1}+\underbrace{\int_{1/2}^1 t^n(1-2t)^n e^t~\mathrm dt}_{J_2}.$$ By considering the area under the curve, $J_1>0$ while $J_2<0$.

Claim: $-J_2> J_1$ when $n$ is odd ($\Leftrightarrow I_n=J_1+J_2<0$).

Proof: Substitute $x=t-\frac12$ in $-J_2$, \begin{align*} -J_2=-\int_0^{1/2}\left(x+\frac12\right)^n(-2x)^n e^{x+1/2}~\mathrm dx&=\underbrace{(-1)^{n+1}}_{=1}e^{1/2}\int_0^{1/2}\left(2x+1\right)^n x^n e^x~\mathrm dx\\ (\text{Using the fact $e>1$})\quad&>\int_0^{1/2}\left(2x+1\right)^n x^n e^x~\mathrm dx\\ (2x+1>1-2x\text{ when } x>0)\quad&>\underbrace{\int_0^{1/2}\left(1-2x\right)^n x^n e^x~\mathrm dx}_{J_1}\\ \end{align*} and we are done.