Is this a valid proof for showing $\int_{1}^{\infty}\frac{3e^{yt}}{y^4}\text{ d}y$ diverges?

I would like to show that for $t > 0$, $$\int_{1}^{\infty}\frac{3e^{yt}}{y^4}\text{ d}y$$ diverges. [This is equivalent to showing that the MGF for $Y$, with pdf $$f_{Y}(y) = \dfrac{3}{y^4}\text{, } y \in (1, \infty)$$ does not exist.]

My work:

\begin{equation*} \int_{1}^{\infty}e^{yt} \cdot \dfrac{3}{y^4}\text{ d}y = \lim_{u \to \infty}3\int_{1}^{u}\dfrac{e^{yt}}{y^4}\text{ d}y\text{.} \end{equation*} Thus, it suffices to show that $\int_{1}^{\infty}\dfrac{e^{yt}}{y^4}\text{ d}y = \infty$. Recall that since $t > 0$ \begin{equation*} \lim_{y \to \infty}\dfrac{e^{yt}}{y^4} = \infty\text{.} \end{equation*} This means, by definition, that $\forall \alpha \in \mathbb{R}$, $\exists K > 0$ such that $\forall y > K$, \begin{equation*} \dfrac{e^{yt}}{y^4} > \alpha\text{.} \end{equation*} Thus, $\forall y > K$, obviously $\dfrac{e^{yt}}{y^4} > 1$. We may assume $K > 1$. Therefore, $\forall u \geq K$, \begin{align*} \int_{1}^{u}\dfrac{e^{yt}}{y^4}\text{ d}y &= \int_{1}^{K}\dfrac{e^{yt}}{y^4}\text{ d}y + \int_{K}^{u}\dfrac{e^{yt}}{y^4}\text{ d}y \\ &\geq \int_{1}^{K}\dfrac{e^{yt}}{y^4}\text{ d}y + \int_{K}^{u}1\text{ d}y \\ &\geq 0 + u - K \\ &= u - K\text{.} \end{align*} As $u \to \infty$, obviously $u - K \to \infty$.

Hence, $\displaystyle\int_{1}^{u}\dfrac{e^{yt}}{y^4}\text{ d}y \to \infty$ as well, and the integral diverges.

I'm just nervous about the sudden $K > 1$ assumption. Is this valid? Obviously $K$ must be quite large.

• The integrand function is greater than $1$ for any $y$ large enough. – Jack D'Aurizio Oct 1 '16 at 15:07
• Since exponential growth always wins over polynomials, the integrand $\to \infty$ at $\infty.$ – zhw. Oct 1 '16 at 15:11

Yes it is valid to take $K>1$.
We assume that $t>0$. One may observe that, for $K>1$, $$\frac{3e^{yt}}{y^4}\ge \frac{3e^{yt}}{K^4}, \qquad y \in [1,K],$$ giving $$\int_1^K\frac{3e^{yt}}{y^4}\:dy\ge \frac{3}{K^4}\int_1^K e^{ty}\:dy=\frac{3}{K^4}\cdot \frac{e^{K t}-e^t}{t}$$ the latter expression tends to $\infty$ as $K \to \infty$.