Primitive of Exponential Integral Function

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:

$\ds \int \map \Ei x \rd x = x \map \Ei x - e^{-x} + C$

Proof

By Derivative of Exponential Integral Function, we have:

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

So:

\(\ds \int \map \Ei x \rd x\)

\(=\)

\(\ds \int 1 \times \map \Ei x \rd x\)

\(\ds \)

\(=\)

\(\ds x \map \Ei x - \int \paren {-x \frac {e^{-x} } x} \rd x\)

Integration by Parts

\(\ds \)

\(=\)

\(\ds x \map \Ei x + \int e^{-x} \rd x\)

\(\ds \)

\(=\)

\(\ds x \map \Ei x - e^{-x} + C\)

Primitive of $e^{a x}$

$\blacksquare$