Analytical approximation of an integral I think there is no analytical solution for
$$
\int_{K}^{\infty} \frac{exp(-x)}{x} dx
$$
where $K > 0$. Instead, is there an analytical approximation?
 A: There is an asymptotic approximation for large $K$ which can be obtained by repeated integration by parts.
This can also be obtained from Watson's lemma by making the change of variables $x=K(1+y)$ to get
$$
\begin{align}
\int_K^\infty \frac{e^{-x}}{x}\,dx &= e^{-K} \int_0^\infty \frac{e^{-K y}}{1+y}\,dy \\
&\sim e^{-K} \sum_{n=0}^{\infty} (-1)^n \int_0^\infty e^{-Ky} y^n\,dy \\
&= e^{-K} \sum_{n=0}^{\infty} \frac{(-1)^n n!}{K^{n+1}}
\end{align}
$$
as $K \to \infty$.
A: Consider
$$
f(k)=e^k\int_k^\infty\frac{e^{-t}}{t}\mathrm{d}t\tag{1}
$$
Differentiating $(1)$, we get
$$
\begin{align}
f(k)
&=\frac1k+f'(k)\\
&=\frac1k-\frac1{k^2}+f''(k)\\
&=\frac1k-\frac1{k^2}+\frac2{k^3}+f'''(k)\\
&=\frac1k-\frac1{k^2}+\frac2{k^3}-\frac6{k^4}+f''''(k)\\
&\vdots\\
&=\frac1k-\frac1{k^2}+\frac2{k^3}-\frac6{k^4}+\dots+(-1)^n\frac{n!}{k^{n+1}}+f^{(n+1)}(k)\tag{2}
\end{align}
$$
Note that since
$$
f(k)=\int_0^\infty\frac{e^{-t}}{t+k}\mathrm{d}t\tag{3}
$$
we have
$$
\begin{align}
|f^{(n+1)}(k)|
&=\int_0^\infty\frac{(n+1)!\,e^{-t}}{(t+k)^{n+2}}\mathrm{d}t\\
&\le\frac{(n+1)!}{k^{n+2}}\tag{4}
\end{align}
$$
Combining $(1)$, $(2)$, and $(4)$ yields
$$
\int_k^\infty\frac{e^{-t}}{t}\mathrm{d}t=e^{-k}\left(\frac1k-\frac1{k^2}+\frac2{k^3}-\frac6{k^4}+\dots+(-1)^n\frac{n!}{k^{n+1}}+O\left(\frac1{k^{n+2}}\right)\right)\tag{5}
$$
$(4)$ says that the error is smaller than the first term omitted.
A: In particular, for $E_1(x) = \int_{x}^{\infty}t^{-1}e^{-t}dt$, my favorite bound is
\begin{align}
\frac{1}{2}\log\left(1+\frac{2}{x}\right) < e^x E_1(x) < \log\left(1+\frac{1}{x}\right),\,x>0
\end{align}
which will be quite tight for large $x$.
A: Another possibility, due to E.J. Weniger, is to use a so-called factorial series to approximate your exponential integral.
We start from the asymptotic expansion in robjohn's and Antonio's answers:
$$f(k)\sim\exp(-k)\sum_{n=0}^\infty \frac{(-1)^n n!}{k^{n+1}}$$
Weniger gives the remarkable formula
$$\sum_{j=0}^\infty \frac{c_j}{z^{j+1}}=\sum_{\ell=0}^\infty \frac1{(z)_{\ell+1}}\sum_{r=0}^\ell\left[{{\ell}\atop{r}}\right]c_r$$
where $(z)_k$ is a Pochhammer symbol, and $\left[{{n}\atop{k}}\right]$ is a Stirling cycle number. Applied to your exponential integral, we have the formula
$$f(k)=\exp(-k)\sum_{\ell=0}^\infty \frac1{(k)_{\ell+1}}\sum_{r=0}^\ell(-1)^r\left[{{\ell}\atop{r}}\right]r!$$
For instance, taking $k=3$, the tenth partial sum gives a value $\approx0.013049262$, and the twentieth partial sum gives a value $\approx0.013048481$; compare this with $E_1(3)\approx0.0130483811$. As with asymptotic series, this approximation performs much better for larger arguments.
