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I tried to evaluate the values of Fox H-function $H_{2,2}^{1,1}(t)$ at point $t>0$. We have $$ H_{2,2}^{1,1}\left[ t\left| \begin{array}{cc} (0,1) & (1,1)\newline (1,1) & (\alpha,1) \end{array} \right. \right] = \frac1{2\pi i} \int_L \frac{\Gamma(1-s)\Gamma(1+s)}{\Gamma(1+s)\Gamma(1-\alpha - s)} t^{-s}ds. $$ Contour $L$ must separate the poles of $\Gamma(1-s)$ and $\Gamma(1+s)$. Poles of $\Gamma(1-s)$ are natural numbers 1,2,3,..., whereas $\Gamma(1+s)$ has negative integers as poles, i.e., -1,-2,-3,.... Then we choose $L$ the imaginary axis and, after cancellation, obtain $$ H_{2,2}^{1,1}\left[ t\left| \begin{array}{cc} (0,1) & (1,1)\newline (1,1) & (\alpha,1) \end{array} \right. \right] = \frac1{2\pi i} \int_{0-i\infty}^{0+i\infty} \frac{\Gamma(1-s)}{\Gamma(1-\alpha - s)} t^{-s}ds. $$ Then, we need to find the residues of all poles in the left half-plane. However, function $\frac{\Gamma(1-s)}{\Gamma(1-\alpha - s)} t^{-s}$ has no poles in the left half-plane of the complex $s$-plane. Does this imply that Fox H-function with those parameters is equal to zero? I do not think so, but cannot figure out what I do wrong.

pabodu
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    The integral over the vertical line diverges if $\operatorname {Re} \alpha \geq 0$. Even if it converges, the integral over a large semicircle in the left half-plane is not negligible if $t > 1$. We have to choose a contour which makes the integral convergent. A possible choice is a left loop for $t < 1$ and a right loop for $t > 1$. – Maxim Aug 25 '20 at 23:05
  • I forgot to mention that $\alpha>0$. However, I do not understand why integral is divergent in this case. If it diverges, does it imply that Fox H-function is infinite? – pabodu Aug 30 '20 at 14:09

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