Characterization of Exponential Integral Function/Formulation 1

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


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

Then:
 * $\ds \map \Ei x = -\gamma - \ln x + \int_0^x \frac {1 - e^{-u} } u \rd u$

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

Proof
We have, by Derivative of $e^{a x}$:


 * $\map {\dfrac \d {\d u} } {1 - e^{-u} } = e^{-u}$

By Primitive of Reciprocal:


 * $\ds \int \frac {\d u} u = \ln u + C$

So:

We have:

Also applying Definite Integral to Infinity of $e^{-x} \ln x$ gives:


 * $\ds \int_0^x \frac {1 - e^{-u} } u \rd u = \paren {1 - e^{-x} } \ln x + \gamma + \int_x^\infty e^{-u} \ln u \rd u$

We have, by Primitive of $e^{a x}$:


 * $\ds \int e^{-u} \rd u = -e^{-u} + C$

We have by Derivative of Logarithm Function:


 * $\dfrac \d {\d u} \paren {\ln u} = \dfrac 1 u$

We therefore have:

It remains to evaluate the limit on the.

We have, for $u > 1$:


 * $e^{-u} \ln u > 0$

and by Bounds of Natural Logarithm:


 * $e^{-u} \ln u < u e^{-u} - e^{-u}$

We have, from Limit to Infinity of $x^n e^{-a x}$:


 * $\ds \lim_{u \mathop \to \infty} \paren {u e^{-u} - e^{-u} } = 0$

So, by the Squeeze Theorem:


 * $\ds \lim_{u \mathop \to \infty} \paren {e^{-u} \ln u} = 0$

So:


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

Rearranging gives:


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