No sé cómo tratar con esta integral:
$$ \int_ {-1} ^ {1} {\ln\left(\,\left\vert\,z-x\,\right\vert\,\right)\over \,\sqrt{\vphantom{\large A} \,1 - x ^ {2} \,} \,} \, x\ {\rm d},, $$ donde $z$ es un número complejo.
Respuestas
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happylions
Puntos
21
Que $f(z)=\int_{-1}^1\dfrac{1}{\sqrt{1-x^2}}\ln|z-x|~dx$,
Entonces $\dfrac{df(z)}{dz}=\int_{-1}^1\dfrac{1}{(z-x)\sqrt{1-x^2}}dx$
Según http://www.wolframalpha.com/input/?i=int1%2F%28%28z-x%29%281-x%5E2%29%5E%281%2F2%29%29%2Cx%2C-1%2C1,
$\dfrac{df(z)}{dz}=\dfrac{\pi}{\sqrt{z^2-1}}$
$f(z)=\int\dfrac{\pi}{\sqrt{z^2-1}}dz=\ln(z+\sqrt{z^2-1})+C$
Pero no sé cómo encontrar $C$.
Felix Marin
Puntos
32763
$\newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$ $\ds{\int_{-1}^{1}{\ln\pars{\verts{z - x}} \\raíz{1 - x^{2}}}\,\dd x: {\large ?}}$
De ${\tt \mbox{@paul2357paul answer}}$ es claro que $\ds{C =\fermi\pars{1} =\int_{-1}^{1}{\ln\pars{\vértices{1 - x}} \\raíz{1 - x^{2}}}\,\dd x}$. Así, vamos a evaluar:
\begin{align} \color{#66f}{\Large C}& =\int_{-1}^{1}{\ln\pars{\verts{1 - x}} \over \root{1 - x^{2}}}\,\dd x =\int_{0}^{1}{\ln\pars{1 - x^{2}} \over \root{1 - x^{2}}}\,\dd x =\lim_{\mu\ \to\ -1/2}\partiald{}{\mu} \int_{0}^{1}\pars{1 - x^{2}}^{\mu}\,\dd x \\[3mm]&=\lim_{\mu\ \to\ -1/2}\partiald{}{\mu} \int_{0}^{1}\pars{1 - x}^{\mu}\,\half\,x^{-1/2}\,\dd x =\half\,\lim_{\mu\ \to\ -1/2}\partiald{}{\mu} \int_{0}^{1}x^{-1/2}\pars{1 - x}^{\mu}\,\dd x \\[3mm]&=\half\,\lim_{\mu\ \to\ -1/2}\partiald{}{\mu} \bracks{% \Gamma\pars{1/2}\Gamma\pars{\mu + 1} \over \Gamma\pars{\mu + 3/2}} =\color{#66f}{\large -\ln\pars{2}\pi} \approx {\tt -2.1774} \end{align}
A continuación, $$\color{#66f}{\large% \int_{-1}^{1}{\ln\pars{\verts{z - x}} \\raíz{1 - x^{2}}}\,\dd x =\pi\ln\pars{z + \raíz{z^{2} - 1}} - \ln\pars{2}\pi} $$
