Oct 17, 2019 · $$\operatorname {Ei} (x)=\operatorname {Ei} (-1)-\int_ {-x}^1\frac {e^ {-t}}t~\mathrm dt$$ which are both easily differentiated using the fundamental theorem of calculus, now that …

Apr 19, 2024 · This result can be obtained directly from a Maclaurin expansion of the function. By denoting \begin {equation} y=\mathrm {Ei}^ {-1} (x) \end {equation} the integral ...

$\operatorname {Ei} (x)$ is a special function and is generally agreed to be considered useful enough to have it's own place amongst the special functions.

Quiz: Spelling- 'ie' or 'ei'? This is a beginner/elementary-level quiz containing 10 multichoice quiz questions from our 'spelling and punctuation' category. Simply answer all questions and press …

Feb 18, 2013 · Intuition comes from knowledge and experience! Learning facts about complex exponentiation then making use of those facts to solve problems will build your experience.

Nov 1, 2025 · This isn't a complete answer as I'm not sure a closed form result exists, but the correct approach is outlined below whereas I believe there are some errors in the approach …

Oct 13, 2021 · Prove Euler's identity $e^ {i\theta} = \cos \theta + i \sin \theta$ using Taylor series. Then plug in $\theta = \pi$.

Feb 17, 2019 · Having an integral like $\int_ {2}^ {10} {\frac {x} {\ln x}}dx$ How does this function turns to an exponential integral of the form: $ \operatorname {Ei} (x)=-\int ...

Oct 19, 2021 · EDIT: I don't know why, but information on the web about the complex function $\operatorname {Ei} (s)$ is very scarce. But it's an important function used a lot in analytic …

Nov 10, 2025 · So I tried some u-sub like $\frac {\operatorname {Ei} (x)} {\ln x}$, $\frac {\operatorname {li} (x)} {\ln x}$ but I think it's some other u-substitute. (I tried to show effort but …