# Finding the limit $\lim_{x\rightarrow 0^{+}}\frac{\int_{1}^{+\infty}\frac{e^{-xy}\quad-1}{y^3}dy}{\ln(1+x)}.$

Finding the following limit:$$\lim_{x\rightarrow 0^{+}}\frac{\int_{1}^{+\infty}\frac{e^{-xy}\quad-1}{y^3}dy}{\ln(1+x)}.$$

To my way of thinking,L’Hopital’s rule is useful to this question.Then $\frac{d}{dx}\int_{1}^{+\infty }\frac{e^{-xy}\quad-1}{y^3}dy=\int_{1}^{+\infty }\frac{\partial}{\partial x}(\frac{e^{-xy}\quad-1}{y^3})dy,\forall x\in [0,1],$ but it seems hopeless to get the answer.I need some help,thanks.

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A possible approach:

1). Use the Dominated Convergence Theorem (or a variant/theorem derived from it) to show that
$$\frac{1}{x}\int_{[1,\infty)} \frac{e^{-xy}-1}{y^3} dy \xrightarrow[x\to0]{} -\int_{[1,\infty)} \frac{dy}{y^2}$$

2). Use this (and the fact that $\frac{x}{\ln(1+x)} \xrightarrow[x\to0]{} 1$) to show that
$$\frac{1}{\ln(1+x)}\int_{[1,\infty)} \frac{e^{-xy}-1}{y^3} dy \xrightarrow[x\to0]{} -\int_{[1,\infty)} \frac{dy}{y^2}$$

To see the intuition behind the first step: $e^{-xy} \operatorname*{=}_{x\to 0} 1 – xy + o(x)$, for any fixed $y$, so that (wishfully) $$\int_{[1,\infty)}\frac{e^{-xy}-1}{y^3} dy \approx \int_{[1,\infty)}\frac{-xy}{y^3} dy = \int_{[1,\infty)}\frac{-x}{y^2} dy$$ when $x\to 0$.

Let $f(x)=\int_{1}^{\infty}\frac{e^{-xy}-1}{y^3}dy,\quad g(x)=\ln(1+x).$

Using L’Hopital’s rule ,we have $$\lim_{x\rightarrow0 ^+}{\frac{f(x)}{g(x)}}=\lim_{x\rightarrow0 ^+}{\frac{f'(x)}{g'(x)}}.\quad (1)$$ and $$f'(x)=\int_{1}^{\infty }\frac{\partial}{\partial x}(\frac{e^{-xy}\quad-1}{y^3})dy=\int_{1}^{\infty}\frac{e^{-xy}}{y^2}dy.\quad(2)$$ $$(1)+(2)\Longrightarrow\lim_{x\rightarrow0 ^+}{\frac{f(x)}{g(x)}}=\lim_{x\rightarrow 0^+}{\int_{1}^{\infty}\frac{e^{-xy}}{y^2}dy}=\int_{1}^{\infty}\lim_{x\rightarrow 0^+}\frac{e^{-xy}}{y^2}dy=-1.$$