I have been asked to find the asymptotic approximation to the integral
$$E_n(x) = \int_1^\infty t^{-n}e^{-xt}\,{\rm d}t$$
where $n={\rm ord}(1)$ and $x\to\infty$ and then estimate the remainder.
I don't really have any clue how to do this however I have attempted something. Write
$$e^{-xt} = \sum_{k=0}^\infty \frac{(-xt)^k}{k!}$$
then
$$E_n(x) = \sum_{k=0}^\infty \frac{(-x)^k}{k!}\int_1^\infty t^{k-n}\,{\rm d}t$$
However I'm not sure how to evaluate that integral. Could someone help?
On
$$E_1(x) = e^{-x}\int_{0}^{+\infty}\frac{e^{-xt}}{t+1}\,dt = \frac{1}{x e^x}\int_{0}^{+\infty}\frac{e^{-u}}{1+\frac{u}{x}}\,du $$ hence for large $x$, $E_1(x)$ is expected to behave like $\frac{1}{xe^x}\int_{0}^{+\infty}e^{-u}\,du = \frac{1}{xe^x}.$ Additionally, $$ \frac{1}{xe^x}-E_1(x) = \frac{1}{x^2 e^x}\int_{0}^{+\infty}\frac{ue^{-u}}{1+\frac{u}{x}}\,du $$ and by the Cauchy-Schwarz inequality $$ \int_{0}^{+\infty}\frac{ue^{-u}}{1+\frac{u}{x}}\,du \leq \frac{\sqrt{x}}{2}.$$
As $k>n-1$ the integral in the decomposition you proposes diverges. You can write the original integral as $$\exp(-x)\int_0^\infty (1+u)^{-n}\exp(-ux)du$$ and integrate by parts several times. Firstly $$E_n(x)=e^{-x}\left[\frac{1}{x}-\frac{n}{x}\int_0^\infty (1+u)^{-n-1}e^{-ux}du\right]$$ which gives the first order term. Going on gives you the following terms. Up to the first order $$E_n(x)=\frac{e^{-x}}{x}+O(x^{-2}e^{-x})$$