Power Series Expansion for Exponential Integral Function/Formulation 2

Theorem

Let $\Ei: \R_{>0} \to \R$ denote the exponential integral function:

$\map \Ei x = \ds \int_{t \mathop \to -\infty}^{t \mathop = x} \frac {e^t} t \rd t$

Then:

$\ds \map \Ei x$

$=$

$\ds \gamma + \ln x + \sum_{n \mathop = 1}^\infty \frac {x^n} {n \times n!}$

$\ds $

$=$

$\ds \gamma + \ln x + \frac x {1 \times 1!} + \frac {x^2} {2 \times 2!} + \frac {x^3} {3 \times 3!} + \dots$

where $\gamma$ denotes the Euler-Mascheroni constant.

$\ds \map \Ei x$

$=$

$\ds \gamma + \ln x + \int_0^x \frac {e^u - 1} u \rd u$

$\ds $

$=$

$\ds \gamma + \ln x + \int_0^x \frac 1 u \paren {\sum_{n \mathop = 0}^\infty \frac {u^n} {n!} - 1} \rd u$

$\ds $

$=$

$\ds \gamma + \ln x + \int_0^x \frac 1 u \paren {\sum_{n \mathop = 1}^\infty \frac {u^n} {n!} } \rd u$

$\ds $

$=$

$\ds \gamma + \ln x + \sum_{n \mathop = 1}^\infty \frac 1 {n!} \paren {\int_0^x u^{n - 1} \rd u}$

$\ds $

$=$

$\ds \gamma + \ln x + \sum_{n \mathop = 1}^\infty \frac {x^n} {n \times n!}$

$\blacksquare$