A better lower bound of the definite integral $\int_0^1\frac{\ln^2x}{e^{2x}}\,dx$

Question

In this post of my blog, I proved that $$\int_0^1\frac{(x^2-3x+1)\ln x}{e^x}\,dx=-\frac1e.$$ Now if we apply the Cauchy-Schwarz inequality for integrals, we get $$\frac1{e^2}\le\int_0^1(x^2-3x+1)^2\,dx\int_0^1\frac{\ln^2x}{e^{2x}}\,dx.$$ The first integral is easy to evaluate, so we arrive at $$\int_0^1\frac{\ln^2x}{e^{2x}}\,dx\ge\frac{30}{11e^2}\approx0.36909\cdots$$ If I plug the integral into WA, the actual value is $2\,_3F_3(1,1,1;2,2,2;-2)\approx1.61511$ which involves hypergeometric functions. We can see that my approximation is not very useful.

So could a closer lower bound be found using 'simple' methods (e.g. not requiring the use of hypergeometric or gamma functions)?

P.S. An analytical method would be best.

Answer 1

You can just truncate the exponential series at an odd index: For $n \in \mathbb{N}$ we have $$ I \equiv \int \limits_0^1 \ln^2(x) \mathrm{e}^{-2 x} \, \mathrm{d} x \geq \sum \limits_{k=0}^{2n-1} \frac{(-2)^k}{k!} \int \limits_0^1 \ln^2 (x) x^k \, \mathrm{d}x = \sum \limits_{k=0}^{2n-1} \frac{(-1)^k 2^{k+1}}{k! (k+1)^3} \equiv I_n \, . $$ Then $I_1 = \frac{3}{2} = 1.5$, $I_2 = \frac{347}{216} \approx 1.60648$, $I_3 = \frac{130789}{81000} \approx 1.61468$ and so on.

Answer 2

It's trivial to get a better lower bound using a trivial method:

$$\frac{1}{2} \cdot 0.3 \cdot 5 = 0.75$$