关于H-Fox 函数

 

........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,

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

 

posted @ 2016-05-16 11:02  张文彪  阅读(1126)  评论(0编辑  收藏  举报