Where does the Euler-Mascheroni constant come from here?

Question

As you probably know $\int \frac{e^{x}}{x}dx=\operatorname{Ei}(x)+C$. We can solve this using Taylor series: $$\begin{align}\operatorname{Ei}(x)+C=\int \frac{e^{x}}{x}dx&=\int \frac{\sum_{n=0}^{\infty}\frac{x^{n}}{n!}}{x}dx\\&=\int \sum_{n=0}^{\infty}\frac{x^{n-1}}{n!}dx\\&=\int\frac{1}{x}+\sum_{n=1}^{\infty}\frac{x^{n-1}}{n!}dx\\&=\ln(x)+\sum_{n=1}^{\infty}\int\frac{x^{n-1}}{n!}dx\\&=\ln(x)+\sum_{n=1}^{\infty}\frac{x^{n}}{n!n}\end{align}$$ Now when I plug these values into wolfram alpha I'm off by the Euler-Mascheroni constant. It has a $+C$ in i so I'm not surprised its off by a constant but my question is why the Euler-Mascheroni constant. The true formula is: $$\operatorname{Ei}(x)=\gamma+\ln(x)+\sum_{n=1}^{\infty}\frac{x^{n}}{n!n}$$

Answer 1

Alright we know that $\operatorname{Ei(x)}=-E_{1}(-x)$. We also know that $E_{1}(x)=\int_{x}^{\infty}\frac{e^{-t}}{t}dt$ So if we find the series representation of $E_1(x)$ We can fint the series representation of $\operatorname{Ei(x)}$ we're golden: $$E_{1}(x)=\int_{x}^{\infty}\frac{e^{-t}}{t}dt=\int_{x}^{1}\frac{e^{-t}}{t}dt+\int_{1}^{\infty}\frac{e^{-t}}{t}dt=\int_{1}^{\infty}\frac{e^{-t}}{t}dt-\int_{1}^{x}\frac{e^{-t}}{t}+\frac{1}{t}-\frac{1}{t}dt=\int_{1}^{\infty}\frac{e^{-t}}{t}dt-\int_{1}^{x}\frac{e^{-t}-1}{t}dt-\int_{1}^{x}\frac{1}{t}dt=\int_{0}^{1}\frac{e^{-t}-1}{t}dt-\ln(x)-\int_{0}^{1}\frac{e^{-t}-1}{t}dt-\int_{0}^{x}\frac{e^{-t}-1}{t}dt=-\gamma-\ln(x)-\int_{0}^{x}\frac{e^{-t}-1}{t}dt=-\gamma-\ln(x)-\sum_{n=1}^{\infty}\frac{(-1)^nx^{n}}{n!n}$$ Now just substitute and you get our formula for $\operatorname{Ei}(x)$: $$\operatorname{Ei(x)}=\gamma+\ln(x)+\sum_{n=1}^{\infty}\frac{x^{n}}{n!n}$$