Difference between revisions of "Exponential integral E"

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$$\mathrm{Ei}(z) = \int_{-\infty}^x \dfrac{e^t}{t} dt; |\mathrm{arg}(-z)|<\pi$$
 
$$\mathrm{Ei}(z) = \int_{-\infty}^x \dfrac{e^t}{t} dt; |\mathrm{arg}(-z)|<\pi$$
 
and
 
and
The exponential integral is related to the [[logarithmic integral]] by the formula
+
$$E_1(z) = \displaystyle\int_z^{\infty} \dfrac{e^{-t}}{t}dt;|\mathrm{arg \hspace{2pt}}z<\pi|.$$
 +
=Properties=
 +
<div class="toccolours mw-collapsible mw-collapsed">
 +
<strong>Proposition:</strong> The exponential integral $\mathrm{Ei}$ is related to the [[logarithmic integral]] by the formula
 
$$\mathrm{li}(x)=\mathrm{Ei}( \log(x)).$$
 
$$\mathrm{li}(x)=\mathrm{Ei}( \log(x)).$$
 +
<div class="mw-collapsible-content">
 +
<strong>Proof:</strong> █
 +
</div>
 +
</div>
 +
 
=Videos=
 
=Videos=
 
[https://www.youtube.com/watch?v=TppV_yDY3EQ Laplace transform of exponential integral]<br />
 
[https://www.youtube.com/watch?v=TppV_yDY3EQ Laplace transform of exponential integral]<br />

Revision as of 01:32, 2 February 2015

The exponential integrals are $$\mathrm{Ei}(z) = \int_{-\infty}^x \dfrac{e^t}{t} dt; |\mathrm{arg}(-z)|<\pi$$ and $$E_1(z) = \displaystyle\int_z^{\infty} \dfrac{e^{-t}}{t}dt;|\mathrm{arg \hspace{2pt}}z<\pi|.$$

Properties

Proposition: The exponential integral $\mathrm{Ei}$ is related to the logarithmic integral by the formula $$\mathrm{li}(x)=\mathrm{Ei}( \log(x)).$$

Proof:

Videos

Laplace transform of exponential integral