Is the integral less than $\frac{n!}2?$

Question

Knowing that $$\int_0^\infty x^n\exp(-x)\,dx=n!,$$ can we prove the following inequality: $$\int_0^n x^n\exp(-x) \,dx< \frac{n!}2 \;\;?$$

Answer 1

Let's take a step back and think a bit more elementary.

Consider the graph of $f(x) = x^n \exp(-x)$. Flip it over the line $x=n$ to get $f(2n-x)$. It is enough to show that

$$\int_{n}^{2n} f(2n-x) \, dx < \int_{n}^{2n} f(x) \, dx \,.$$

I will now show that if $x \in (n, 2n)$, then $f(2n-x) < f(x)$, which is enough to prove the above. Rearranging,

$$ \begin{align} (2n-x)^n \exp(x-2n) &< x^n \exp(-x) \\ \left( \frac{2n-x}{x} \right) &< \exp(2(n-x)). \end{align} $$

Now, let's do a little shift to $x \in (0, n)$ by doing $x \mapsto x-n$. Then we need to prove that

$$\left(\frac{n-x}{n+x}\right)^n < \exp(-2x).$$

Take the logarithm of both sides(it is well-defined) to get

$$n(\ln(n-x) - \ln(n+x)) < -2x$$

To prove the above inequality, let $a(x) = n(\ln(n-x) - \ln(n+x))$. I will now show that $a'(x) < -2$ for $x \in (0,n)$ which is enough to prove the above. Computing the derivative gives

$$a'(x) = -2 \left(\frac{n^2}{n^2-x^2}\right) < -2 \left(\frac{n^2-x^2}{n^2-x^2}\right) = -2.$$