Derivative of Exponential Integral Function

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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:

$\dfrac \d {\d x} \paren {\map \Ei x} = -\dfrac {e^{-x} } x$


Proof

\(\ds \map {\frac \d {\d x} } {\map \Ei x}\) \(=\) \(\ds \map {\frac \d {\d x} } {-\gamma - \ln x + \int_0^x \frac {1 - e^{-t} } t \rd t}\) Characterization of Exponential Integral Function
\(\ds \) \(=\) \(\ds -\frac 1 x + \frac 1 x - \frac {e^{-x} } x\) Derivative of Constant, Derivative of Natural Logarithm, Fundamental Theorem of Calculus: First Part (Corollary)
\(\ds \) \(=\) \(\ds -\frac {e^{-x} } x\)

$\blacksquare$