# Integral $\int_0^b \frac{1-\exp\{-x\}}{x}\text{d}x\qquad 0<b<\infty$

Is there any closed form expression for the definite integral $$\int_0^b \frac{1-\exp\{-x\}}{x}\text{d}x\qquad 0<b<\infty$$ as I could not find one in Gradshteyn and Ryzhik Table of Integrals?

Concerning the antiderivative $$\int\frac{e^{x}}{x}dx=\text{Ei}(x)$$ where appears the exponential integral function which would present problems around $x=0$. But, provided $b>0$ $$\int_0^b \frac{1-e^ {-x}}{x}dx=\gamma+\log (b)+\Gamma (0,b)$$
• Thank you. I had also found this formula on a website, as well as the series expansion $$- \sum_{k=1}^\infty \frac{(-z)^k}{k\,(k!)}$$ by a simple integration. I was wondering if there was something more "explicit" but apparently not. – Xi'an Jan 15 '15 at 9:51
• Well, we could start here a long philosophical discussion ! Is $\sin(x)$ explicit ? Là est la question. Cheers :-) – Claude Leibovici Jan 15 '15 at 9:57
This integral is known as the the complementary exponential integral $\mathrm{Ein}(b)$. It is entire, see http://dlmf.nist.gov/6.2.E3 for the definition and http://dlmf.nist.gov/6.6.E4 for the power series.