De la integral definida, cociente del logaritmo y polinomio

Question

Yo estaba pensando en esta integral : $$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$$ Yo lo que hago es utilizar una Recíproca subsitution, fácil demostrar que: $$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{\lambda ^2x^2+\lambda x+1}\text{d}x =\frac{1}{\lambda}\int_0^{\infty}\frac{(\ln x-\ln \lambda)^2}{x^2+x+1}\text{d}x \\ =\frac{\ln ^2\lambda}{\lambda}\int_0^{\infty}\frac{1}{x^2+x+1}\text{d}x+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2x}{x^2+x+1}\text{d}x \\ =\frac{\ln ^2\lambda}{\lambda}\frac{2\pi}{3\sqrt{3}}+\frac{2}{\lambda}\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x$$ Pero hav ninguna idea en proceso el resto de la integral:( Alguien sabe cómo solucionarlo?

THX chicos! Me vino con otro método podría funcionar para esto: Recordar la mano GF $$\frac{1}{x^2+x+1}=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$ Entonces tenemos : $$\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\frac{2}{\left(k+1\right)^3}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$ Con $$\sum_{k=1}^{\infty}\frac{\sin (kx)}{k^3}=\frac{\pi ^2}{6}x-\frac{\pi}{4}x^2+\frac{x^3}{12}$$ ¿Qué podemos llegó?

Answer 1

Aquí es una solución de forma cerrada para la integral

$$\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{8\sqrt{3}}{243}\pi^3 \sim 1.768047624. $$

He introducido una técnica general, que se basa en la fracción parcial combinada con el uso de dilogarithm función, para resolver la integral

$$ \int _{a }^{b }\!{\frac {\ln \left( tx + u \right) }{m{x}^{2}+nx+p}}{dx}.$$

En su caso, en lugar de utilizar el dilogarithm función, vamos a utilizar el polylogarithm función de $\operatorname{Li}_{s}(z)$ y todo el problema se reduce a evaluar las integrales de la forma

$$ \int_{0}^{1} \frac{\ln(x)^2}{x-\alpha}= -2 Li_{3}\left(\frac{1}{a}\right), $$

donde

$$ \operatorname{Li}_{s}(z) = \sum_{k=1}^{\infty}\frac{z^k}{k^s}. $$

Answer 2

En primer lugar, permítanme decir que el cálculo de esta integral no es tan fácil, y según mis notas se evalúa a $\frac{16\pi^{3}}{81\sqrt{3}}.$

Sugerencia: Para evaluar $$\int_0^\infty \frac{(\log x)^2}{x^2+x+1}dx,$$ we take advantage of the fact that $$(a+1)^3-(a-1)^3 = 6a^2+2$$ and let $$f(x)=\frac{\left(\log x\right)^{3}}{x^{2}-x+1},$$ with a negative sign in the denominator. Consider the integral $\oint_{\mathcal{C}}f(x)dx$ where $\mathcal{C}$ is the counter clockwise oriented keyhole contour which is a circle of radius $R$, the circle of radius $\epsilon,$ and the two branches which extend to negative infinity on the upper and lower half plane. It is easy to see that as $R\rightarrow\infty,$ and as $\epsilon\rightarrow0,$ the two circles contribution goes to zero. In the limit, the integral on the branches becomes $$\int_{0}^{\infty}\frac{\left(\log x+\pi i\right)^{3}}{x^{2}+x+1}dx-\int_{0}^{\infty}\frac{\left(\log x-\pi i\right)^{3}}{x^{2}+x+1}dx.$$

Creo que se puede resolver desde aquí.

Answer 3

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

${\tt\large\mbox{Just another one:}}$

\begin{align} &\color{#66f}{\large\int_{0}^{1}{\ln^{2}\pars{x} \over 1 + x + x^{2}}\,\dd x} =\lim_{\mu \to 0}\,\partiald[2]{}{\mu} \int_{0}^{1}{\pars{1 - x}x^{\mu} \over 1 - x^{3}}\,\dd x \\[3mm]&=\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\int_{0}^{1} {x^{\mu/3} - x^{\pars{\mu + 1}/3} \over 1 - x}\,{1 \over 3}\,x^{-2/3}\dd x ={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\int_{0}^{1} {x^{\pars{\mu - 2}/3} - x^{\pars{\mu - 1}/3} \over 1 - x}\,\dd x \\[3mm]&={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\bracks{% \int_{0}^{1}{1 - x^{\pars{\mu - 1}/3} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{\pars{\mu - 2}/3} \over 1 - x}\,\dd x} \\[3mm]&={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu} \bracks{\Psi\pars{\mu + 2 \over 3} - \Psi\pars{\mu + 1 \over 3}} ={1 \over 27}\bracks{\Psi''\pars{2 \over 3} - \Psi''\pars{1 \over 3}} \\[3mm]&={1 \over 27}\,\bracks{\pi\,\totald[2]{\cot\pars{\pi z}}{z}}_{z\ =\ ^1/_3}\ =\ {2\pi^{3} \over 27}\,\color{#c00000}{\cot\pars{\pi \over 3}} \color{#f0f}{\csc^{2}\pars{\pi \over 3}} ={2\pi^{3} \over 27}\,\color{#c00000}{{1 \over \root{3}}}\,\color{#f0f}{\pars{2 \over \root{3}}^{2}} \\[3mm]&=\color{#66f}{\large{8\root{3} \over 243}\,\pi^{3}} \approx 1.7680 \end{align}

Answer 4

Reescribir \begin{align} \int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \int_0^1 \dfrac{1-x}{1-x^3} \ln^2x\ dx\\ & = \int_0^1\sum_{k=0}^{\infty} x^{3k}\ (1-x) \ln^2x\ dx\\ &=\sum_{k=0}^{\infty}\left(\int_0^1 x^{3k}\ln^2x\ dx-\int_0^1 x^{3k+1}\ln^2x\ dx\right).\tag1 \end{align} El uso de $$ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{para }\ n=0,1,2,\ldots\tag2 $$ a continuación, $(1)$ resulta ser \begin{align} \int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \sum_{k=0}^{\infty}\left[\frac{2}{(3k+1)^{3}}-\frac{2}{(3k+2)^{3}}\right]\\ &= \frac1{3^3}\sum_{k=0}^{\infty}\left[\frac{2}{\left(k+\frac13\right)^{3}}-\frac{2}{\left(k+\frac23\right)^{3}}\right].\tag3\\ \end{align} Considerar la serie representación de polygamma función $$ \psi_n(z)=(-1)^{n+1}\sum_{k=0}^\infty\frac{n!}{(k+z)^{n+1}}\quad;\quad\text{para}\ n>0\tag4 $$ y su reflejo fórmula $$ \psi_n(1-z)+(-1)^{n+1}\psi_n(z)=(-1)^{n}\pi\frac{d^n}{dz^n}\cuna(\pi z).\tag5 $$ El uso de $(4)$$(5)$, $(3)$ se convierte en \begin{align} \int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \frac1{3^3}\bigg[\psi_2\left(\frac23\right)-\psi_2\left(\frac13\right)\bigg]\\ &=\frac\pi{27}\left.\frac{d^2}{dz^2}\cot(\pi z)\right|_{z=\frac13}\\ &=\large\color{blue}{\frac{8\sqrt3}{243}\pi^3}. \end{align}