The antiderivative of the function $$x \mapsto \frac{e^{-x}}{x}$$ is not expressible in closed form. Thus such integrals as $$\int_x^\infty\!\frac{e^{-t}}{t}\,dt \quad \mbox{and} \quad \int_\infty^{-x}\!\frac{e^{-t}}{t}\,dt,$$ define certain non-elementary transcendental functions. They are called exponential integrals and denoted usually ${\rm E}_1$ and ${\rm Ei}$ , respectively. Accordingly, $${\rm E}_1(x) \;:=\; \int_x^\infty\!\frac{e^{-t}}{t}\,dt$$ $${\rm Ei}\,x \;:=\; \int_\infty^{-x}\!\frac{e^{-t}}{t}\,dt \;=\; -\int_{-x}^\infty\!\frac{e^{-t}}{t}\,dt \;:=\; \int_{-\infty}^x\!\frac{e^{-u}}{u}\,du.$$ Then one has the connection $${\rm E}_1(x) \;=\; -{\rm Ei}\,(-x).$$ For positive values of $x$ the series expansion $${\rm Ei}\,x \;=\; \gamma+\ln{x}+\sum_{j=1}^\infty\frac{x^j}{j!j},$$ where $\gamma$ is the Euler-Mascheroni constant, is valid.

Note: Some authors use the convention ${\rm Ei}\,x \,:=\, \int_x^\infty\!\frac{e^{-t}}{t}\,dt$ .

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

By the definition of Laplace transform, $$\mathcal{L}\{\frac{1}{t\!+\!a}\} \;=\; \int_0^\infty\frac{e^{-st}}{t\!+\!a}\,dt.$$ The substitution $t\!+\!a = u$ gives $$\mathcal{L}\{\frac{1}{t\!+\!a}\} \;=\; \int_a^\infty\frac{e^{as-su}}{u}\,du \;=\; e^{as}\int_a^\infty\frac{e^{-su}}{u}\,du,$$ from which the substitution $su = t$ yields $$\mathcal{L}\{\frac{1}{t\!+\!a}\} \;=\; e^{as}\int_{as}^\infty\frac{e^{-t}}{t}\,dt,$$ i.e.

(1)

(2)