Limit of a definite nonelementary integral

Question

I have to prove that $\lim_{n\to \infty}\int_0^{1/2}x^ne^{2x-1}\;dx=0$ without calculating the primitive. I have already proved that it can be defined as a recurrence sequence: $$y_1=\frac{1}{4e},\quad y_n=\frac{1}{2^{n+1}}-\frac{n}{2}\cdot y_{n-1}$$ And now I'm trying to use the monotone convergence theorem, so I've proved that it is bounded below by $0$ $(x^ne^{2x-1}>0\quad \forall n\in\mathbb{N} )$ , but I don't know how to prove the sequence is decreasing. Thanks in advance.

Edit:
I managed to complete the proof. $$0 \leq y_n=\frac{1}{2^{n+1}}-\frac{n}{2}\cdot y_{n-1}\leq \frac{1}{2^{n+1}}$$ Then by the squeeze theorem $\lim_{n\to\infty}y_n=0$

Answer 1

Let $I_n:=\int_0^{1/2}x^ne^{2x-1}\;dx.$

By the mean value theorem for integrals there is $t_n \in [0,1/2]$ such that

$$I_n= \frac{1}{2}t_n^n e^{2t_n-1}.$$

Since $2t_n-1 \le 0$ we get

$$0 \le I_n \le \frac{1}{2}t_n^n,$$

thus

$$ 0 \le I_n \le \frac{1}{2^{n+1}},$$

therefore $I_n \to 0 $ as $ n \to \infty.$

Answer 2

You need to use the decreasing version of Monotone Convergence Theorem:

$\varphi_{n}\downarrow 0$ with $\varphi_{1}\in L^{1}$, then $\int\varphi_{n}\rightarrow 0$.

Just start with the increasing sequence $(\varphi_{1}-\varphi_{n})_{n}$, then $\int(\varphi_{1}-\varphi_{n})\rightarrow\int\varphi_{1}$.