Difference between revisions of "Exponential integral E"
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<strong>Theorem (Derivative):</strong> $$\dfrac{d}{dz} E_n(z) = -E_{n-1}(z); n=1,2,3,\ldots$$ | <strong>Theorem (Derivative):</strong> $$\dfrac{d}{dz} E_n(z) = -E_{n-1}(z); n=1,2,3,\ldots$$ | ||
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| + | <strong>Proof:</strong> █ | ||
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| + | <strong>Theorem (Relation to incomplete gamma):</strong> The following formula holds: | ||
| + | $$E_n(z)=z^{n-1}\Gamma(1-n,z),$$ | ||
| + | where $\Gamma$ denotes the [[incomplete gamma function]]. | ||
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<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
Revision as of 03:42, 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: █
Theorem: The exponential integral $\mathrm{Ei}$ has series representation $$\mathrm{Ei}(x) = \gamma + \log x + \displaystyle\sum_{k=1}^{\infty} \dfrac{x^k}{kk!}; x>0,$$ where $\gamma$ denotes the Euler-Mascheroni constant.
Proof: █
Theorem: The exponential integral $E_1$ has series representation $$E_1(z)=-\gamma-\log z - \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^kz^k}{kk!}; |\mathrm{arg}(z)|<\pi,$$ where $\gamma$ denotes the Euler-Mascheroni constant.
Proof: █
Theorem (Symmetry): The following symmetry relation holds: $$E_n(\overline{z})=\overline{E_n(z)}.$$
Proof: █
Theorem (Recurrence): The following recurrence holds: $$E_{n+1}(z) = \dfrac{1}{n}[e^{-z}-zE_n(z)];(n=1,2,3,\ldots).$$
Proof: █
Theorem (Continued fraction): The following formula holds: $$E_n(z)=e^{-z} \left( \dfrac{1}{z+} \dfrac{n}{1+} \dfrac{1}{z+} \dfrac{n+1}{1+} \dfrac{2}{z+} \ldots \right); |\mathrm{arg} z|<\pi.$$
Proof: █
Theorem: The following value is known: $$E_n(0)=\dfrac{1}{n-1}; n>1.$$
Proof: █
Theorem: The following closed form expression is known: $$E_0(z)=\dfrac{e^{-z}}{z}.$$
Proof: █
Theorem (Derivative): $$\dfrac{d}{dz} E_n(z) = -E_{n-1}(z); n=1,2,3,\ldots$$
Proof: █
Theorem (Relation to incomplete gamma): The following formula holds: $$E_n(z)=z^{n-1}\Gamma(1-n,z),$$ where $\Gamma$ denotes the incomplete gamma function.
Proof: █
Videos
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