Difference between revisions of "Exponential integral E"
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<strong>Proposition:</strong> The exponential integral $\mathrm{Ei}$ is related to the [[logarithmic integral]] by the formula | <strong>Proposition:</strong> The exponential integral $\mathrm{Ei}$ is related to the [[logarithmic integral]] by the formula | ||
$$\mathrm{li}(x)=\mathrm{Ei}( \log(x)); n=0,1,2,\ldots, \mathrm{Re}(z)>0$$ | $$\mathrm{li}(x)=\mathrm{Ei}( \log(x)); n=0,1,2,\ldots, \mathrm{Re}(z)>0$$ | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> █ | ||
| + | </div> | ||
| + | </div> | ||
| + | |||
| + | <div class="toccolours mw-collapsible mw-collapsed"> | ||
| + | <strong>Proposition:</strong> The following series representation holds for $\mathrm{Ei}$: | ||
| + | $$\mathrm{Ei}(x) = \gamma + \log x + \displaystyle\sum_{k=1}^{\infty} \dfrac{x^k}{kk!}; x>0.$$ | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
Revision as of 01:36, 2 February 2015
The exponential integrals are $$\mathrm{Ei}(z) = \int_{-\infty}^x \dfrac{e^t}{t} dt; |\mathrm{arg}(-z)|<\pi,$$ $$E_1(z) = \displaystyle\int_z^{\infty} \dfrac{e^{-t}}{t}dt;|\mathrm{arg \hspace{2pt}}z<\pi|,$$ and $$E_n(z)=\displaystyle\int_1^{\infty} \dfrac{e^{-zt}}{t^n} dt.$$
Properties
Proposition: The exponential integral $\mathrm{Ei}$ is related to the logarithmic integral by the formula $$\mathrm{li}(x)=\mathrm{Ei}( \log(x)); n=0,1,2,\ldots, \mathrm{Re}(z)>0$$
Proof: █
Proposition: The following series representation holds for $\mathrm{Ei}$: $$\mathrm{Ei}(x) = \gamma + \log x + \displaystyle\sum_{k=1}^{\infty} \dfrac{x^k}{kk!}; x>0.$$
Proof: █
Videos
Laplace transform of exponential integral
