Limit of a function with a defined integral 3

Question

Let $F$ be function defined as an integral

$$F(x)=\int_{1}^{\infty}\dfrac{t^ke^{-xt}}{1+t^{5}}\textrm{d}t \quad \forall k\in \mathbb{N},\ x>0$$

Show that $\lim_{x\to \infty}F(x)=0$

Indeed,

i'm tried to find such $M \in \mathbb{R} \mid F(x)\leq M $ with limit of M equal zero when $x \to \infty$

note that $t\geq1 \implies t^5+1\geq 1 \implies \dfrac{1}{1+t^{5}}\leq 1 \implies \dfrac{e^{-xt}}{1+t^{5}}\leq e^{-xt} \implies \dfrac{t^k.e^{-xt}}{1+t^{5}}\leq t^k.e^{-xt} $

$$F(x)\leq \int_{1}^{\infty} t^k.e^{-xt}\textrm{d}t $$

• Am in the right way or please could you show me others ways

  Another way : note that $t^k=_{t\to \infty}o(e^{xt})$ then $\lim_{t \to \infty} t^{k}e^{-xt}=0$ then $\dfrac{t^{k}e^{-\frac{x}{2}t}}{1+t^5}\underset{t\to\infty}{\longrightarrow}\,0$ thus $\forall k>0,\quad \dfrac{t^{k}e^{-\frac{x}{2}t}}{1+t^5}\leq 1 \implies \dfrac{t^{k}e^{-xt}}{1+t^5}\leq e^{-\frac{x}{2}t}$ then \begin{align} F(x)&\leq \int_{1}^{\infty} e^{-\frac{x}{2}t}\textrm{d}t\\ &=\lim_{A\to \infty}\int_{1}^{A}e^{-\frac{x}{2}t}\textrm{d}t\\ &=\lim_{A\to \infty}[-\frac{2}{x}e^{-\frac{x}{2}t}]_{t=1}^{A}\\ &=\lim_{A\to \infty}[-\frac{2}{x}e^{-\frac{x}{2}t}]_{t=1}^{A}\\ &=\frac{2}{x}e^{-\frac{x}{2}}\\ 0 \leq &\lim_{x\to \infty }F(x)\leq \lim_{x\to \infty }\frac{2}{x}e^{-\frac{x}{2}}=0 \end{align} Then $$\lim_{x\to \infty}F(x)=0$$

• AM i right ??

• note that $ \dfrac{e^{-xt}}{1+t^5}\leq e^{-xt}$ can we say that we've this statement true for all $k \in \mathbb{N},\quad \int_{1}^{\infty}\dfrac{t^ke^{-xt}}{1+t^{5}}\textrm{d}t \leq \int_{1}^{\infty} e^{-xt}\textrm{d}t$

Answer 1

Since $F(x) \le \int_1^{\infty} t^ke^{-xt}\,dt$, you also have that $F(x) \le \int_0^{\infty} t^ke^{-xt}\,dt$. Do a change of variable and make use of a nice property of the $\Gamma$ function and you'll have your result.

Here's how I would continue. From my above argument,

$$F(x) \le \int_0^{\infty} t^k e^{-xt}\,dt.$$

Make a change of variable $\tau = xt$, then $d\tau = x\,dt$ and our inequality becomes

$$F(x) \le \int_0^{\infty} \left(\frac{\tau}{x}\right)^ke^{-\tau}\frac{1}{x}\,d\tau.$$

Doing a little regrouping gives

$$F(x)\le \frac{1}{x^{k+1}}\int_0^{\infty}\tau^ke^{-\tau}\,d\tau.$$

This integral has a nice closed-form solution and it is in fact nothing more than $k!\,$. (You can prove this to yourself by induction or doing a few examples.) Thus

$$F(x) \le \frac{k!}{x^{k+1}}.$$

Can you see how this gets you what you want?

Answer 2

For $x \geq 1$ we have

$$ \left|\frac{t^ke^{-xt}}{1+t^{5}}\right| \leq \frac{t^ke^{-t}}{1+t^{5}} \in L_1([1,\infty)). $$

Also, for every fixed $t \in [1,\infty)$ we have

$$ \lim_{x \to \infty} \frac{t^ke^{-xt}}{1+t^{5}} = 0. $$

By the dominated convergence theorem we therefore have

$$ \lim_{x \to \infty} \int_1^\infty \frac{t^ke^{-xt}}{1+t^{5}}\,dt = \int_1^\infty 0\,dt = 0. $$