1

I am trying to calculate the following integral using Euler-Maclaurin formula. Found the end resault using an online intergrals calculator but I can't seem to get there on my own.

$$ \int_0^1 e^{-x^2} $$

I need an explanation on how to use the formula to calculate the integral and an explanation on what exactly is p in the formula.

Here is the Euler-Maclaurin's formula from wikipedia:

$$ \sum_{i=m}^n f(i) = \sum_{k=0}^{2p}\frac{1}{k!}\left(B^\ast_k f^{(k - 1)}(n) - B_k f^{(k - 1)}(m)\right) + R $$

Antonio Vargas
  • 25,541
user1640736
  • 333
  • Where is $p$ appearing ? – Claude Leibovici Jun 28 '14 at 11:36
  • There are a few ways of writing Euler-Maclaurin slightly differently as far as I can see (haven't tried to see how they are equivalent yet), but this might help http://math.stackexchange.com/questions/719401/euler-maclaurin-summation-for-e-x2 – bolbteppa Jun 28 '14 at 11:37

1 Answers1

1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

\begin{align} &\int_{0}^{1}\expo{-x^{2}}\,\dd x =\half\int_{0}^{2}\expo{-x^{2}/4}\,\dd x \\[3mm]&\approx\half\braces{\sum_{k = 1}^{1}\expo{-k^{2}/4} + \half\bracks{1 + \expo{-2^{2}/4}} - {1 \over 12}\bracks{-\,\half\,2\expo{-2^{2}/4}}} \\[3mm]&=\half\,\expo{-1/4} + {1 \over 4} + {7 \over 24}\,\expo{-1} =\color{#c00000}{0.746}6985619 \\[5mm]&\mbox{A more precise numerical integration yields}\quad 0.7468241328 \end{align}

The exact result is $\ds{\half\,\root{\pi}{\rm Erf}\pars{1}}$.

Information of the remainder can be seen in $\quad\large\tt\mbox{page 886}\quad$ of this table.

Felix Marin
  • 94,079
  • One thing I don't understand, if you describe erf(1) in the resault and do the simplifications then you get exactly $ \int_0^1 e^{-x^2} $. So isn't this basically just like saying 2 = 2*2/2 ? – user1640736 Jun 29 '14 at 00:52