Proving $\int_0^{2n}e^t t^{2n-1} dt > (2n-1)!$

Question

Let $n$ be a positive integer. Prove $\displaystyle\int_0^{2n}e^t t^{2n-1} dt > (2n-1)!$

Is there a way to prove this inequality ? Numerical experimentation suggests $\int_0^{2n}e^t t^{2n-1} dt$ grows much faster than $(2n-1)!$

I tried to relate the inequality to the Gamma function but the LHS is bugging me...

Answer 1

$e^t > 1$ for $t > 0$ and therefore $$ \int_0^{2n}e^t t^{2n-1} dt > \int_0^{2n} t^{2n-1} dt = \frac{(2n)^{2n}}{2n} = (2n)^{2n-1} > (2n-1)! $$

Answer 2

This can be proved by induction. Consider the base case $n = 1$:

$$\int_{0}^{2}x e^{x}\:\mathrm{d}x = 1 + e^{2} > 1$$

Now if we accept that the hypothesis holds true for $n$, then we can see that for $n + 1$:

$$\begin{align*}\int_{0}^{2n+2}x^{2n+1}e^{x}\:\mathrm{d}x &= \left.(2n+1)x^{2n}\right|_{x=0}^{2n+2}+\int_{0}^{2n}x^{2n} e^{x}\:\mathrm{d}x \\ &= \left.(2n+1)x^{2n}e^{x}\right|_{x=0}^{2n+2} + \left.2nx^{2n-1}e^{x}\right|_{x=0}^{2n+2} + 2n(2n+1)\int_{0}^{2n+2}x^{2n-1}e^{x}\:\mathrm{d}x\end{align*}$$

But we know that:

$$\int_{0}^{2n}x^{2n-1}e^{x}\:\mathrm{d}x > (2n-1)! \implies 2n(2n+1)\int_{0}^{2n+2}x^{2n-1}e^{x}\:\mathrm{d}x>2n(2n+1)(2n-1)!$$

Thus, as the other terms we introduced in the integration by parts are strictly positive, we have proved that:

$$\int_{0}^{2n}x^{2n-1}e^{x}\:\mathrm{d}x > (2n+1)!\qquad \forall n \in \mathbb{N}$$

Answer 3

I think induction on $n$ might work.

For $n-1$, we have $\int_0^2 e^tt dt = e^2+1>1!$ (by integration by parts)

Assuming that the result is true for $n$, i.e. for all natural numbers less than $(n+1)$, we assume

$\int_0^{2n}e^t t^{2n-1}dt>(2n-1)!$.

For $n+1$, we get

\begin{align} \int_0^{2n+2}e^tt^{2n+1}dt=[t^{2n+1}e^t-(2n+1)\int_0^{2n+2} t^{2n}e^t]_0^{2n+2}\\ = [t^{2n+1}e^t-(2n+1)(t^{2n}e^t-2n\int_0^{2n+2}t^{2n-1}e^t)]_0^{2n+2}> \\ \textit{a positive quantity} + (2n+1)(2n)\int_0^{2n+2}t^{2n-1}e^t> (2n+1)(2n)(2n-1)!=(2n+1)! \end{align}

where I have used the induction hypothesis in the last step and the inequality $\int_0^{2n+2}t^{2n-1}e^tdt>\int_0^{2n}t^{2n-1}e^tdt$.

Hence the inequality holds by induction.

I hope this is correct. Tell me if you find any objectionable statement.