# About the extensions from Confluent Hypergeometric Function of the Second Kind

I know that $\int_0^\infty t^{a-1}(1+t)^{c-a-1}e^{-yt}~dt=\Gamma(a)U(a,c,y)$ , where $\text{Re}(a),\text{Re}(y)>0$ .

How about $\int_0^\infty t^{a-1}(1+t)^{c-a-1}(1+xt)^{-b}e^{-yt}~dt$ and $\int_0^\infty t^{a-1}(1+t)^{c-a-1}(1+x_1t)^{-b_1}\cdots(1+x_nt)^{-b_n}e^{-yt}~dt$ , where $\text{Re}(a),\text{Re}(y)>0$ ? Do some definitions in http://en.wikipedia.org/wiki/Humbert_series can help us?

• What are you trying for? In any case the Laplace transform of t^c * Phi2(;;) (Humbert's function) seems to resemble what you want: look at apps.nrbook.com/bateman/Vol1.pdf section 4.24 (5). Or mathworld.wolfram.com/AppellHypergeometricFunction.html (without the exp(-yt). Just as an aside any of the real line integrals of functions with a Taylor expansion and a factor of exp(-y*t) can be reduced too a Taylor series in t sum(p_n(y)*t^n) where p_n is a polynomial of order n. – rrogers Feb 24 '15 at 13:58
• @rrogers I asked the $\int_0^1$ version before (math.stackexchange.com/questions/812748) and it had solved perfectly. Now I ask about the $\int_0^\infty$ version. – Harry Peter Feb 27 '15 at 16:05
• :I will give it a try; but it might be some time. You might try adopting the argument that leads to Bateman Integral Transforms Volume 1, section 5.4 equation (9). authors.library.caltech.edu/43489/1/Volume%201.pdf Its the complex domain form though; and the exp(yt) would have to be transformed on the coefficients (y) and such. – rrogers Mar 1 '15 at 14:25
• I did manage to do your 2 term (1+xx) equation with Reduce-algebra and it reflects the fact that we end up with summing sequences of HyperGeometric functions which can be calculated and could be encoded recursively, but I am not sure anybody is interested. The fundamental idea is to apply: dlmf.nist.gov/16.5.E3 to a fabricated hypergeometric representation of the product (1+x*t).... . For instance, see apps.nrbook.com/bateman/Vol3.pdf Page 268 19.11 (3) . Reduce-algebra can be beat up to handle many problems of this type. – rrogers May 26 '17 at 17:58
• Instead of 16.5.E3 one can realize that $int(e^{-t}*t^{a-1},0,inf)=\Gamma(a)$ Which actually proves 16.5.E3 . – rrogers May 26 '17 at 18:01