Alternative expressions for the integral.
En primer lugar, I=π/2∫0ln(1−xcotx)dx=π/2∫0ln(sinx−xcosx)dx−π/2∫0ln(sinx)dx=π2ln2+I1,
donde I1 permite la integración por partes: I1=π/2∫0ln(sinx−xcosx)dx=xln(sinx−xcosx)|π/2 0−π/2∫0x2sinxsinx−xcosxdx, I1=−π/2∫0x2sinxsinx−xcosxdx=−π/2∫0x21−xcotxdx=−J21, donde
Jmn=π/2∫0xm(1−xcotx)ndx.
Por otro lado, J21=π/2∫0x2(1−xcotx+xcotx)1−xcotxdx=π324+I2, donde I2=π/2∫0x3cotx1−xcotxdx=π/2∫0x3tanx−xdx.
Fórmulas (3) no son adecuados para los cálculos numéricos.
Pero la integración por partes es posible, I2=14π/2∫01tanx−xdx4=14x4tanx−x|π/20+14π/2∫0x4(1+tan2x−1)(tanx−x)2dx, I2=14π/2∫0x4(1−xcotx)2dx=14J42,
I=π2ln2−π324−14J42.
Fórmula (4) proporciona cálculos numéricos adecuados vía Wolfram Alpha mediante la expresión
![Expression]()
con el resultado
![Result]()
y la posterior construcción de la serie en las funciones elementales a través de las transformaciones en forma de J42=π/2∫0x4((1−xcotx)2+2xcotx(1−xcotx)+x2cot2x)(1−xcotx)2dx=π/2∫0(x4+2x5cotx1−xcotx+x6cot2x(1−xcotx)2)dx=π/2∫0x4dx+π/2∫0(2x5cotx1−xcotx+x6cot2x(1−xcotx)2)dx, J42=π5160+I3+I4, donde I3=2π/2∫0x5cotx1−xcotxdx=2π/2∫0x5tanx−xdx=13π/2∫01tanx−xdx6=13x6tanx−x|π/20+13π/2∫0x6(1+tan2x−1)(tanx−x)2dx=13J62, I4=π/2∫0x6cot2x(1−xcotx)2dx=π/2∫0x6(tanx−x)2dx=17π/2∫01(tanx−x)2dx7=27x7(tanx−x)3|π/20+27π/2∫0x7(1+tan2x−1)(tanx−x)3dx=27π/2∫0x7cotx(1−xcotx)3dx=27π/2∫0x6(1−(1−xcotx))(1−xcotx)3dx=27(J63−J62),
Por lo tanto,
I=π2ln2−π324−π5640−184J62−114J63.
Cálculos numéricos mediante Mathcad Alpha mediante la fórmula (6)
![Formula (6)]()
conduce al mismo resultado, lo que confirma la corrección del planteamiento.
Recurrence relations.
Para la arbitrariedad m,n J_{mn} = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\,(x\cot x + (1-x\cot x))^n \dfrac{x^m}{(1 - x\cot x)^n}\,\mathrm dx = \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\sum\limits_{k=0}^n\binom nk\dfrac{x^{m+k}\cot^k x}{(1 - x\cot x)^k}\,\mathrm dx = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{k=1}^n\dfrac{\dbinom nk}{m+k+1} \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{\mathrm dx^{m+k+1}}{(\tan x - x)^k} = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}}\\ + \sum\limits_{k=1}^n\dfrac{\dbinom nk}{m+k+1} \left(\dfrac{x^{m+k+1}}{(\tan x - x)^{k}}\bigg|_{\,0}^{\Large^\pi\hspace{-1pt}/_2} + k\int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^{m+k+1}(1+\tan^2x-1)}{(\tan x-x)^{k+1}}\,\mathrm dx\right)\\ = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{k=1}^n\dfrac{k}{m+k+1} \dbinom nk \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^{m+2}(x\cot x)^{k-1}}{(1 -x\cot x)^{k+1}}\,\mathrm dx\\ = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{k=1}^n\dfrac{k}{m+k+1} \dbinom nk \int\limits_0^{\Large^\pi\hspace{-1pt}/_2}\dfrac{x^{m+2}(1-(1-x\cot x))^{k-1}}{(1 -x\cot x)^{k+1}}\,\mathrm dx\\ = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{k=1}^n\dfrac{k}{m+k+1} \dbinom nk \sum\limits_{j=0}^{k-1}(-1)^{k-1-j}\dbinom{k-1}j J_{m+2,\,j+2},
J_{mn} = \dfrac{\pi^{m+1}}{(m+1)2^{m+1}} + \sum\limits_{j=0}^{n-1} F_{j} J_{m+2,\,j+2},\tag7
donde
F_{j} = \sum\limits_{k=j+1}^n (-1)^{k-1-j} \dfrac{k}{m+k+1}\dbinom nk \dbinom{k-1}j.\tag8
Si (m,n)=(2,1),\ entonces F_{0} = \sum\limits_{k=1}^1 (-1)^{k-1} \dfrac{k}{2+k+1}\dbinom1k \dbinom{k-1}0 =\dfrac14, J_{21} = \dfrac{\pi^{3}}{3\cdot2^3} + \sum\limits_{j=0}^0 F_{j} J_{4,\,j+2} = \dfrac{\pi^{3}}{24} + J_{42}.
Si (m,n)=(4,2),\ entonces F_{0} = \sum\limits_{k=1}^2 (-1)^{k-1} \dfrac{k}{4+k+1}\dbinom2k \dbinom{k-1}0 =\dfrac13 - \dfrac27 = \dfrac{1}{21}, F_{1} = \sum\limits_{k=2}^2 (-1)^{k} \dfrac{k}{4+k+1}\dbinom2k \dbinom{k-1}1 =\dfrac27, J_{42} = \dfrac{\pi^{5}}{5\cdot2^5} + \sum\limits_{j=0}^1 F_{j} J_{2,\,j+2} = \dfrac{\pi^5}{160} + \dfrac1{21}J_{62} + \dfrac27J_{63}.
Asimismo, , J_{62} = \dfrac{\pi^7}{896}+\dfrac1{36}J_{82}+\dfrac29J_{83}\tag9 (véase también Prueba de Wolfram Alpha ).
![J62 Test]()
Además de , J_{63} = \dfrac{\pi^7}{896}+\dfrac1{120}J_{82} + \dfrac1{15}J_{83} + \dfrac3{20}J_{84}.\tag{10}
\color{brown}{\textbf{Simple series.}}
Los resultados obtenidos no son la mejor manera de obtener las series requeridas de la longitud arbitraria.
\boxed{ \begin{matrix} I & = & -3.35333726288947201778500718670823032009876022464933939598 \\ \frac\pi2\ln2 & = &1.088793045151801065250344449118806973669291850184643147162 \\ J_{21} & = & 4.442130308041273083035351635930890531086461245854584994170 \\ \frac{\pi^3}{24} & = & 1.291928195012492507311513127795891466759387023578546153922 \\ J_{42} & = & 12.60080845211512230289535403253999625730829688910415536099 \\ \frac{\pi^5}{160} & = & 1.912623029908009082892133187771472540501879416425468690959 \\ J_{62} & = & 9.357325953756236734147158157553707227832359838953032605558 \\ J_{63} & = & 35.84909465209885681432007993043088180418373451454989791084 \\ \frac{\pi^7}{896} & = & 3.370862977429455432493534032446475258836420173320761453966 \\ J_{82} & = & 13.21743446830609099759197972403428192140938899336281280188 \\ J_{83} & = & 25.28690408493225448274231109747825862030555487117486858192 \\ J_{84} & = & 102.2743092725712233044348622015074565154951081384648503713 \\ \end{matrix}}
Por otro lado, el uso de la simple serie Laurent para la función g(y) = \dfrac{35}{1-y\sqrt{15}\cot y\sqrt{15}} = \dfrac7{y^2}-\sum\limits_{i=0}^\infty c_iy^{2i}\tag{11}
![g(y), Laurent series]()
da series evidentemente convergentes J_{21} = \dfrac1{35}\int\limits_0^{\Large^\pi\hspace{-1pt}/_2} \left(7 - \sum\limits_{i=0}^\infty c_i\left(\dfrac{x^2}{15}\right)^{i+1}\right)\,\mathrm dx,
J_{21} = \dfrac32\pi - \dfrac3{14}\pi\sum\limits_{i=0}^\infty \dfrac{c_i}{2i+3}\left(\dfrac{\pi^2}{60}\right)^{i+1}\,\mathrm dx,\tag{12}
en el que el primer 8 términos proporcionan la precisión de 8 dígitos decimales.