关于H-Fox 函数

末蓝、 2021-12-22 03:27 229阅读 0赞

........**We arrive at the following results which provide the sine and cosine transforms of the H-function**

$$\\int\_\{0\}^\{\\infty\}x^\{\\rho\}\\sin (ax)H\_\{p,q\}^\{m,n\}\\bigg\[bx^\{\\sigma\}\\Bigg|\{\}\_\{(b\_\{q\},B\_\{q\})\}^\{(a\_\{p\},A\_\{p\})\}\\Bigg\]dx=\\frac\{2^\{\\rho-1\}\\sqrt\{\\pi\}\}\{a^\{\\rho\}\}H\_\{p+2,q\}^\{m,n+1\}\\Bigg\[b\\left(\\frac\{2\}\{a\}\\right)^\{\\sigma\}\\Bigg|\{\}\_\{(b\_\{q\},B\_\{q\})\}^\{(\\frac\{1-\\rho\}\{2\},\\frac\{\\rho\}\{2\}),(a\_\{p\},A\_\{p\}),(\\frac\{2-\\rho\}\{2\},\\frac\{\\rho\}\{2\}\}\\Bigg\]$$

where $a,\\alpha,\\sigma>0,\\rho,b\\in C,|\\arg b|<\\frac\{\\pi \\alpha\}\{2\}$,

$$Re(\\rho)+\\sigma \\min\_\{1\\leq j\\leq m\}Re\\left(\\frac\{b\_\{j\}\}\{B\_\{j\}\}\\right)>-1;Re(\\rho)+\\sigma\\max\_\{1\\leq j\\leq n\}\\frac\{a\_\{j\}-1\}\{A\_\{j\}\}<1$$

**And**

$$\\int\_\{0\}^\{\\infty\}x^\{\\rho\}\\cos (ax)H\_\{p,q\}^\{m,n\}\\bigg\[bx^\{\\sigma\}\\Bigg|\{\}\_\{(b\_\{q\},B\_\{q\})\}^\{(a\_\{p\},A\_\{p\})\}\\Bigg\]dx=\\frac\{2^\{\\rho-1\}\\sqrt\{\\pi\}\}\{a^\{\\rho\}\}H\_\{p+2,q\}^\{m,n+1\}\\Bigg\[b\\left(\\frac\{2\}\{a\}\\right)^\{\\sigma\}\\Bigg|\{\}\_\{(b\_\{q\},B\_\{q\})\}^\{(\\frac\{2-\\rho\}\{2\},\\frac\{\\rho\}\{2\}),(a\_\{p\},A\_\{p\}),(\\frac\{1-\\rho\}\{2\},\\frac\{\\rho\}\{2\})\}\\Bigg\]$$

where $a,\\alpha,\\sigma>0,\\rho,b\\in C,|\\arg b|<\\frac\{\\pi \\alpha\}\{2\}$,

$$Re(\\rho)+\\sigma \\min\_\{1\\leq j\\leq m\}Re\\left(\\frac\{b\_\{j\}\}\{B\_\{j\}\}\\right)>0;Re(\\rho)+\\sigma\\max\_\{1\\leq j\\leq n\}\\frac\{a\_\{j\}-1\}\{A\_\{j\}\}<1$$

**Specially,**

$$E\_\{\\alpha,\\beta\}(z)=\\sum\_\{k=0\}^\{\\infty\}\\frac\{z^\{k\}\}\{\\Gamma(\\alpha z+\\beta)\}=H\_\{1,2\}^\{1,1\}\\Bigg\[-z\\Bigg|\_\{(0,1),(1-\\beta,\\alpha)\}^\{(0,1)\}\\Bigg\]$$

**Set** $z=-a x^\{2\}$, we have

$$E\_\{\\alpha,\\beta\}(-a x^\{2\})=H\_\{1,2\}^\{1,1\}\\Bigg\[a x^\{2\}\\Bigg|\_\{(0,1),(1-\\beta,\\alpha)\}^\{(0,1)\}\\Bigg\]$$

Thus,

$$\\int\_\{0\}^\{\\infty\}\\cos (kx)E\_\{\\alpha,\\beta\}(-ax^\{2\})dx=\\frac\{\\pi\}\{k\}H\_\{1,1\}^\{1,0\}\\Bigg\[\\frac\{k^\{2\}\}\{a\}\\Bigg|\_\{(1,2)\}^\{(\\beta,\\alpha)\}\\Bigg\]$$

转载于:https://www.cnblogs.com/zhangwenbiao/p/5497351.html