I am trying to derive the value of the exponential integral $\int_{-\infty}^{x}\frac{e^{t}}{t}\mathrm{d}t$ when $x>0$. I understand that we are integrating through a singularity at $x=0$, so I tried to use the PV in Cauchy's sense, here is what I came up so far: \begin{align*} &\int_{-\infty}^{x}\frac{e^{t}}{t}\mathrm{d} t=\int_{-\infty}^{0}\frac{e^{t}}{t}\mathrm{d} t+\int_{0}^{x}\frac{e^{t}}{t}\mathrm{d} t=\lim_{\delta\to 0}\left(\int_{-\infty}^{-\delta}\frac{e^{t}}{t}\mathrm{d} t+\int_{\delta}^{x}\frac{e^{t}}{t}\mathrm{d} t\right)\\ =&\lim_{\delta\to 0}\left(\frac{e^{t}}{t}\bigg|_{-\infty}^{-\delta}-\int_{-\infty}^{-\delta}(-1)\frac{e^{t}}{t^{2}}\mathrm{d} t+\frac{e^{t}}{t}\bigg|_{\delta}^{x}-\int_{\delta}^{x}(-1)\frac{e^{t}}{t^{2}}\mathrm{d} t\right)\\ =&\lim_{\delta\to 0}\left(\frac{e^{-\delta}}{-\delta}-\lim_{t\to-\infty}\frac{e^{t}}{t}+\int_{-\infty}^{-\delta}\frac{e^{t}}{t^{2}}\mathrm{d} t+\frac{e^{x}}{x}-\frac{e^{\delta}}{\delta}+\int_{\delta}^{x}\frac{e^{t}}{t}\mathrm{d} x\right)\\ =&\frac{e^{x}}{x}+\lim_{\delta\to 0}\left(\frac{e^{-\delta}}{-\delta}-\frac{e^{\delta}}{\delta}\right)+\lim_{\delta\to 0}\left(\int_{-\infty}^{-\delta}\frac{e^{t}}{t^{2}}\mathrm{d} t+\int_{\delta}^{x}\frac{e^{t}}{t^{2}}\mathrm{d} x\right) \end{align*} The $\lim_{t\to-\infty}\frac{e^{t}}{t}$ is simply zero, that much I understand. I also expect the same method (integral by parts) can be applied infinitely to generate a series form of the value. However I am troubled by the $\lim_{\delta\to 0}\left(\frac{e^{-\delta}}{-\delta}-\frac{e^{\delta}}{\delta}\right)$ part. One would the two fractions to cancel each other so the term is zero, but obviously it diverges. Is there anything I did wrong?
TIA
