Respuesta en el registro de la onda sinusoidal y de registro-cos integral con la diversión resultado
Huizeng Qin
Institute of Applied Mathematics,
Shandong University of Technology,
Zibo, Shandong, P. R. China(Email:qinhz_000@163.com)
El uso de
\begin{equation}
\int_{0}^{\frac{\pi }{2}}\sin ^{2x-1}u\cos ^{2y-1}u\ln ^{p}\sin u\ln
^{q}\cos udu=2^{-p-q-1}B_{p,q}(x,y) \tag*{(1)}
\end{equation}
donde $B(x,y)$ es la función Beta y $B_{p,q}(x,y)=\frac{\partial ^{p+q}}{
\partial x^{p}\partial y^{q}}B(x,y),$ tenemos
\begin{equation}
I(p,q)=\int_{0}^{\frac{\pi }{2}}\frac{\ln ^{p}\sin u\ln ^{q}\cos u}{\sin
u\cos u}du=2^{-p-q-1}B_{p,q}(0,0). \tag*{(2)}
\end{equation}
Por la derivada parcial de la identidad
\begin{equation}
xyB(x,y)=(x+y)(x+y+1)B(x+1,y+1) \tag*{(3)}
\end{equation}
en $x,y$ hemos
\begin{equation}
\begin{array}{c}
xyB_{p,q}(x,y)+pyB_{p-1,q}(x,y)+qxB_{p,q-1}(x,y)+qpB_{p-1,q-1}(x,y) \\
=(x+y)(x+y+1)B_{p,q}(x+1,y+1)+q(2x+2y+1)B_{p,q-1}(x+1,y+1) \\
+p(2x+2y+1)B_{p-1,q}(x+1,y+1)+q(q-1)B_{p,q-2}(x+1,y+1) \\
+2qpB_{p-1,q-1}(x+1,y+1)+p(p-1)B_{p-2,q}(x+1,y+1)
\end{array}
\etiqueta*{(4)}
\end{equation}
En (4) deje $x\rightarrow 0,y\rightarrow 0$, hay
\begin{equation}
\begin{array}{c}
B_{p-1,q-1}(0,0)=\frac{1}{p}B_{p,q-1}(1,1)+\frac{1}{q}B_{p-1,q}(1,1) \\
+\frac{q-1}{p}B_{p,q-2}(1,1)+2B_{p-1,q-1}(1,1)+\frac{p-1}{q}B_{p-2,q}(1,1)
\end{array}
\etiqueta*{(5)}
\end{equation}
y
\begin{equation}
\begin{array}{c}
B_{p,q}(0,0)=\frac{1}{p+1}B_{p+1,q}(1,1)+\frac{1}{q+1}B_{p,q+1}(1,1) \\
+\frac{q}{p+1}B_{p+1,q-1}(1,1)+2B_{p,q}(1,1)+\frac{p}{q+1}B_{p-1,q+1}(1,1)
\end{array}
\etiqueta*{(6)}
\end{equation}
Así tenemos
\begin{equation}
\begin{array}{c}
I(p,q)=2^{-p-q-1}(
\frac{B_{p,q+1}(1,1)}{q+1}+\frac{B_{p+1,q}(1,1)}{p+1} \\
+2B_{p,q}(1,1)+\frac{pB_{p-1,q+1}(1,1)}{q+1}+\frac{qB_{p+1,q-1}(1,1)}{p+1}
) \end{array}
\etiqueta*{(7)}
\end{equation}
Para $B_{p,q}(1,1)$ hay las siguientes relaciones de recurrencia
\begin{equation}
\begin{array}{c}
B_{p,q}(1,1)=(q-1)!p!(
\sum\limits_{k=0}^{p}\sum\limits_{j=0}^{q-1}C_{p+q-k-j-1}^{p-k}\frac{
(-1)^{p+q-k-j}B_{k,j}(1,1)}{k!j!} \\
-\sum\limits_{k=0}^{p-1}\sum\limits_{j=0}^{q-1}C_{p+q-k-j-1}^{p-k}\frac{(-1)^{p+q-k-j}\zeta (p+q-k-j)B_{k,j}(1,1)}{k!j!}
)\end{array}
. \etiqueta*{(8)}
\end{equation}
Por (8) podemos conseguir que
\begin{equation}
\begin{array}{c}
B_{0,0}(1,1)=1,B_{0,1}(1,1)=-1,B_{0,2}(1,1)=2,B_{0,3}(1,1)=-6, \\
B_{1,1}(1,1)=2-\frac{\pi ^{2}}{6},B_{1,2}(1,1)=-6+\frac{\pi ^{2}}{3}+2\zeta
(3), \\
B_{1,3}(1,1)=24-\pi ^{2}-\frac{\pi ^{4}}{15}-6\zeta (3),B_{2,2}(1,1)=24-
\frac{4\pi ^{2}}{3}-\frac{\pi ^{4}}{90}-8\zeta (3), \\
B_{2,3}(1,1)=-120+6\pi ^{2}+\frac{\pi ^{4}}{6}+36\zeta (3)-2\pi ^{2}\zeta
(3)+24\zeta (5) \\
B_{3,3}(1,1)=720-36\pi ^{2}-\pi ^{4}-\frac{23\pi ^{6}}{420}-216\zeta
(3)+12\pi ^{2}\zeta (3)+36\zeta ^{2}(3)-144\zeta (5)
\end{array}
\etiqueta*{(9)}
\end{equation}
Por (7),(9) y $B_{p,q}(1,1)=B_{q,p}(1,1)$ hemos
\begin{equation}
\begin{array}{c}
I(1,0)=-\frac{\pi ^{2}}{24},I(1,1)=\frac{\zeta (3)}{4},I(1,2)=-\frac{\pi ^{4}
}{576},I(1,3)=-\frac{\pi ^{2}\zeta (3)}{32}+\frac{9\zeta (5)}{16} \\
I(2,2)=-\frac{\pi ^{2}\zeta (3)}{24}+\frac{\zeta (4)}{2},I(2,3)=-\frac{\pi
^{6}}{2304}+\frac{9\zeta ^{2}(3)}{32}, \\
I(3,3)=-\frac{\pi ^{4}\zeta (3)}{128}-\frac{3\pi ^{2}\zeta (5)}{16}+\frac{
45\zeta (7)}{16}.
\end{array}
\etiqueta*{(10)}
\end{equation}
