Characterization of Exponential Integral Function

Theorem

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

where:

$\Ei$ denotes the exponential integral function
$\gamma$ denotes the Euler-Mascheroni constant
$x$ is a real number with $x > 0$.

Proof

We have, by Derivative of Exponential Function: Corollary 1:

$\dfrac \d {\d u} \paren {1 - e^{-u} } = e^{-u}$
$\displaystyle \int \frac {\d u} u = \ln u + C$

So:

 $\displaystyle \int_0^x \frac {1 - e^{-u} } u \rd u$ $=$ $\displaystyle \intlimits {\paren {1 - e^{-u} } \ln u} 0 x - \int_0^x e^{-u} \ln u \rd u$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \paren {1 - e^{-x} } \ln x - \lim_{x \mathop \to 0^+} \paren {1 - e^{-x} } \ln x - \paren {\int_0^\infty e^{-u} \ln u \rd u - \int_x^\infty e^{-u} \ln u \rd u}$ Sum of Integrals on Adjacent Intervals for Integrable Functions

We have:

 $\displaystyle \lim_{x \mathop \to 0^+} \paren {1 - e^{-x} } \ln x$ $=$ $\displaystyle \paren {\lim_{x \mathop \to 0^+} \frac {1 - e^{-x} - \paren {1 - e^0} } x} \paren {\lim_{x \mathop \to 0^+} x \ln x}$ Combination Theorem for Limits of Functions: Product Rule $\displaystyle$ $=$ $\displaystyle \frac \d {\d x} \paren {1 - e^{-x} }_{x = 0} \paren {\lim_{x \mathop \to 0^+} x \ln x}$ Definition of Derivative $\displaystyle$ $=$ $\displaystyle e^0 \times 0$ Derivative of Exponential Function, Limit of $x^m \paren {\ln x}^n$ $\displaystyle$ $=$ $\displaystyle 0$

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

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

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

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

We have by Derivative of Logarithm Function:

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

We therefore have:

 $\displaystyle \paren {1 - e^{-x} } \ln x + \gamma + \int_x^\infty e^{-u} \ln u \rd u$ $=$ $\displaystyle \paren {1 - e^{-x} } \ln x + \gamma + \paren {\intlimits {-e^{-u} \ln u} x \infty + \int_x^\infty \frac {e^{-u} } u \rd u}$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \paren {1 - e^{-x} } \ln x + \gamma - \lim_{u \mathop \to \infty} \paren {e^{-u} \ln u} + e^{-x} \ln x + \map \Ei x$ Definition of Exponential Integral Function $\displaystyle$ $=$ $\displaystyle \ln x + \gamma + \map \Ei x - \lim_{u \mathop \to \infty} \paren {e^{-u} \ln u}$

It remains to evaluate the limit on the right hand side.

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 $x^n e^{-a x}$:

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

So, by the Squeeze Theorem:

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

So:

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

Rearranging gives:

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

$\blacksquare$