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) | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {e^{-x} } x\) |
$\blacksquare$