This is my first question in this forum; I hope it is an appropriate question. The Wolframalpha website tells me that $ \int\nolimits_0^z\frac{1-e^x}{x} dx = \log (-z)+\Gamma(0, -z)+\gamma\quad \text{for}\quad \Re(z)<0. $ I tried to prove this for myself, but I would appreciate it if you could give me some help. This is not a homework question, and I would appreciate it if someone could point me to a reference or tell me what you expect to happen when $ℜ(z)>0$. Thanks.
How this integral $ \int_0^z\frac{1-e^x}{x} dx$ is connected to the Gamma function and Euler constant?
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0@Henning: The incomplete gamma and exponential integral functions are essentially equivalent, anyway. I presume the use incomplete gamma since the internal algorithms for integration start out with hypergeometrics and specialize accordingly, incomplete gamma being one of the more convenient special cases. – 2011-10-25
3 Answers
I believe that this works for both positive and negative $z$.
By definition and change of variables $ \Gamma(0,-z)=\int_{-z}^\infty e^{-x}\frac{\mathrm{d}x}{x}=-\int_{-\infty}^z e^x\frac{\mathrm{d}x}{x}\tag{1} $ where the principal value is taken where needed. Applying $(1)$ to the integral from $w$ to $z$: $ \int_w^z\frac{1-e^x}{x}\mathrm{d}x=\log|z|-\log|w|+\Gamma(0,-z)-\Gamma(0,-w)\tag{2} $ According to $(2)$, $ \int_0^z\frac{1-e^x}{x}\mathrm{d}x=\log|z|+\Gamma(0,-z)-C\tag{3} $ where $ C=\lim_{w\to0}(\log|w|+\Gamma(0,|w|))\tag{4} $ We can use $\Gamma(0,|w|)$ in $(4)$ since either $\Gamma(0,w)$ or $\Gamma(0,-w)$ is defined by a principal value integral, so we have $ \lim_{w\to0}(\Gamma(0,-w)-\Gamma(0,w))=0 $ Therefore, $ \begin{align} C &=\lim_{w\to0}(\log|w|+\Gamma(0,|w|))\\ &=\lim_{w\to0}\left(\log|w|+\int_{|w|}^\infty e^{-x}\frac{\mathrm{d}x}{x}\right)\\ &=\lim_{w\to0}\left(\log|w|-\log|w|\;e^{-|w|}+\int_{|w|}^\infty\log(x)e^{-x}\mathrm{d}x\right)\\ &=\int_0^\infty\log(x)e^{-x}\mathrm{d}x\\ &=-\gamma\tag{6} \end{align} $ where $\gamma$ is the Euler-Mascheroni Constant. Combining $(3)$ and $(6)$, we get $ \int_0^z\frac{1-e^x}{x}\mathrm{d}x=\log|z|+\Gamma(0,-z)+\gamma\tag{7} $ If there is interest, I can append a proof that $\int_0^\infty\log(x)e^{-x}\mathrm{d}x=-\gamma$.
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1Thank Robjonh, beautiful explanation! – 2011-10-26
First of all, you misquote WolframAlpha. It give $\int_0^z \left( 1- \mathrm{e}^x \right) \frac{\mathrm{d} x}{x} = \gamma + \log(-z) + \Gamma(0, -z)$. Notice $\Gamma(0, -z)$ instead of $\Gamma(0, z)$.
This is done using the fundamental theorem of calculus:
Let $ F(x) = \int \left( 1- \mathrm{e}^x \right) \frac{\mathrm{d} x}{x} = \log(x) - \operatorname{Ei}(x) $ Then, $\int_0^z \left( 1- \mathrm{e}^x \right) \frac{\mathrm{d} x}{x} = F(z) - \lim_{x\to 0^+} F(x) = F(z) + \gamma$. The latter limit follows from the Taylor series for the exponential integral.
The connection between $\Gamma(0,-z) = \int_{-z}^\infty \mathrm{e}^{-x} \frac{\mathrm{d} x}{x}$ and $\operatorname{Ei}(z) = \int_{-\infty}^z \mathrm{e}^{x} \frac{\mathrm{d} x}{x}$ is well known.
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0you are right to say that I quoted the wrong WolframAlpha. Thanks for the reply. One question, their answer is valid for all z? Thank you. – 2011-10-25
$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}...$
$\frac{1-e^{x}}{x}=-1-\frac{x}{2!}-\frac{x^{2}}{3!}...$
$\int_{0}^{z}\frac{1-e^{x}}{x} dx=-z-\frac{z^{2}}{2.2!}-\frac{z^{3}}{3.3!}...$
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0How does this help the OP? – 2012-03-21