Let $ f(x)=\int_x^\infty \frac{e^{-t}}{t}\mathrm dt.$ Show that $f(x)\sim e^{-x}(\frac{1}{x}-\frac{1}{x^2}+\frac{1\cdot2}{x^3}-\cdots)$

Question

Let $\displaystyle f(x)=\int_x^\infty \frac{e^{-t}}{t}\mathrm dt.$ Use integration by parts to show that $f(x)\sim e^{-x}(\frac{1}{x}-\frac{1}{x^2}+\frac{1\cdot2}{x^3}-\cdots).$

We need to show that $f(x)\sim e^{-x}(\frac{1}{x}-\frac{1}{x^2}+\frac{1\cdot2}{x^3}-\dots)$ i.e. $\vert f(x)-e^{-x}(\frac{1}{x}-\frac{1}{x^2}+\frac{1\cdot2}{x^3}-\dots-\frac{(-1)^n1\cdot2\dots\cdot(n-1)}{x^n}\vert<\frac{\epsilon}{|x|^n}$

By integration by parts, $f(x)=e^{-x}(\frac{1}{x}-\frac{1}{x^2}+\frac{1\cdot2}{x^3}-\dots-\frac{(-1)^n1\cdot2\dots\cdot(n-1)}{x^n})-n\int_x^\infty e^{-t}t^{-n-1}\mathrm dt$

From here one can see that a bound is needed for $|-n\int_x^\infty e^{-t}t^{-n-1}\mathrm dt|$

so I could conclude $\vert f(x)-e^{-x}(\frac{1}{x}-\frac{1}{x^2}+\frac{1\cdot2}{x^3}-\dots-\frac{(-1)^n1\cdot2\dots\cdot(n-1)}{x^n}\vert<\frac{\epsilon}{|x|^n}$

Could give me a detailed explanation of how to find a bound for $|-n\int_x^\infty e^{-t}t^{-n-1}\mathrm dt|$?

Answer 1

First off, notice that if you want to prove the asymptotic expansion $$f(x)\sim e^{-x}\left(\frac{1}{x}-\frac{1}{x^2}+\frac{1\cdot2}{x^3}-\dots\right)$$ then you actually need to show that $$\left\vert f(x)-e^{-x}\left(\frac{1}{x}-\frac{1}{x^2}+\frac{1\cdot2}{x^3}-\dots-\frac{(-1)^n1\cdot2\dots\cdot(n-1)}{x^n}\right)\right\vert\leq \epsilon\frac{e^{-x}}{|x|^{n+1}},$$ that is, that the remainder is $O(e^{-x}x^{-(n+1)})$, the next term in the expansion. The remainder you have after integration by parts is (notice the correction) $$\left|(-1)^n n! \int_x^\infty e^{-t}t^{-n-1}\;dt\right| \leq n! \frac{1}{x^{n+1}}\int_x^\infty e^{-t}\;dt \le n! \frac{e^{-x}}{x^{n+1}}$$ which satisfies the required bound (I'm assuming $x > 0$, since otherwise the initial integral in not defined).

Answer 2

Just take the simplest, naive bound: $$\begin{align*} n\int_x^\infty e^{-t}t^{-n-1}\,\mathrm{d}t &= n e^{-x}\int_0^\infty \frac{e^{-t}}{(t+x)^{n+1}}\,\mathrm{d}t\\ &\leq n e^{-x}\int_0^\infty e^{-t}\cdot\frac{1}{x^{n+1}}\,\mathrm{d}t\\ &=\frac{ne^{-x}}{x^{n+1}} \end{align*}$$ would do if $|x|\gg 0$.

By the way, you are actually asked to show $$ \left| e^x f(x)-\left(\frac{1}{x}-\frac{1}{x^2}+\dots+\frac{(-1)^{n-1}(n-1)!}{x^n}\right) \right|\leq\frac{\varepsilon}{x^n} $$ i.e. you can't use the $e^{-x}\to 0$ in the numerator and so have to rely on the extra $x$ in the denominator.