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 …

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 …

Apr 19, 2024 · The Exponential integral is defined by $$ \\mathrm{Ei}(x) = \\int_{-\\infty}^x \\frac{e^t}{t} \\mathrm dt, $$ and has the following expansion $$ \\mathrm{Ei}(x ...

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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.

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

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 ...

Mar 12, 2005 · First, it's not "e.i" it's "i.e." Both "i.e." and "e.g." are from Latin and have different meanings and uses: i.e. = "id est" which means approximately "that is [to say]" Use it to …

May 28, 2015 · In this answer Cleo posted the following result without a proof: $$\begin {align}\int_0^\infty\operatorname {Ei}^4 (-x)\,dx&=24\operatorname {Li}_3\!\left (\tfrac14\right) …

I was reading R. Wong's book on Asymptotic Approximations of Integrals, and I'm having problems with the derivation of the asymptotic expansion of the exponential integral which he …