exponential integral

The antiderivative of the function

$$x\mapsto \frac{e^{-x}}{x}$$

is not expressible in closed form.  Thus such integrals (http://planetmath.org/ImproperIntegral) as

$${\int }_x^{\mathrm{\infty }}\frac{e^{-t}}{t}𝑑t\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{\int }_{\mathrm{\infty }}^{-x}\frac{e^{-t}}{t}𝑑t,$$

define certain non-elementary (http://planetmath.org/ElementaryFunction) transcendental functions.  They are called exponential integrals and denoted usually${\mathrm{E}}_1$ and $\mathrm{Ei}$, respectively.  Accordingly,

$${\mathrm{E}}_1(x):={\int }_x^{\mathrm{\infty }}\frac{e^{-t}}{t}𝑑t$$

$$\mathrm{Ei}x:={\int }_{\mathrm{\infty }}^{-x}\frac{e^{-t}}{t}𝑑t=-{\int }_{-x}^{\mathrm{\infty }}\frac{e^{-t}}{t}𝑑t:={\int }_{-\mathrm{\infty }}^x\frac{e^{-u}}{u}𝑑u.$$

Then one has the connection

$${\mathrm{E}}_1(x)=-\mathrm{Ei}(-x).$$

For positive values of $x$ the series expansion

$$\mathrm{Ei}x=\gamma +\ln x+\sum _{j=1}^{\mathrm{\infty }}\frac{x^j}{j!j},$$

where $\gamma$ is the http://planetmath.org/node/1883Euler–Mascheroni constant, is valid.

Note: Some authors use the convention  $\mathrm{Ei}x:={\int }_x^{\mathrm{\infty }}\frac{e^{-t}}{t}𝑑t$.

0.1 Laplace transform of $\frac{1}{t+a}$

By the definition of Laplace transform,

$$ℒ\{\frac{1}{t+a}\}={\int }_0^{\mathrm{\infty }}\frac{e^{-st}}{t+a}𝑑t.$$

The substitution (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)  $t+a=u$  gives

$$ℒ\{\frac{1}{t+a}\}={\int }_a^{\mathrm{\infty }}\frac{e^{as-su}}{u}𝑑u=e^{as}{\int }_a^{\mathrm{\infty }}\frac{e^{-su}}{u}𝑑u,$$

from which the substitution  $su=t$  yields

$$ℒ\{\frac{1}{t+a}\}=e^{as}{\int }_{as}^{\mathrm{\infty }}\frac{e^{-t}}{t}𝑑t,$$

i.e.

$ℒ\{{\displaystyle \frac{1}{t+a}}\}=e^{as}{\mathrm{E}}_1(as).$(1)

Using the rule (http://planetmath.org/LaplaceTransformOfDerivative)  $ℒ\{f^{\prime }(t)\}=sF(s)-f(0)$,  one easily derives from (1) the

$ℒ\{{\displaystyle \frac{1}{{(t+a)}^2}}\}={\displaystyle \frac{1}{a}}-s{e}^{as}{\mathrm{E}}_1(as).$(2)