- #1
CAF123
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I am considering the following integral $$\int_{-t}^{\infty} \frac{dy}{y-s} \frac{1}{y^2} \frac{1}{y^{\epsilon}} {}_2F_1(1,1,2+\epsilon, -t/y)$$ Rewriting the hypergeometric using its integral representation and making a change of variables ##y=-t/u## I obtain the integral, up to some numerical factors, $$\int_0^1 \int_0^1 dz\, du (1-uz)^{-1}u^{1+\epsilon} (1-z)^{\epsilon} (1+\frac{us}{t})^{-1}$$ Then a partial fraction decomposition gives $$\frac{1}{(1-uz)} \frac{1}{(1+\frac{us}{t})} = \frac{z}{\frac{s}{t}+z} \frac{1}{(1-uz)} + \frac{\frac{s}{t}}{\frac{s}{t}+z}\frac{1}{(1+\frac{us}{t})}$$ which gives $$\int_0^1 \int_0^1 dz\, du\, u^{1+\epsilon} (1-z)^{\epsilon} \left(\frac{z}{s/t + z} \frac{1}{1-uz} + \frac{s/t}{s/t+z}\frac{1}{1+us/t}\right)\,\,(1)$$ The second term there can be written as $$\int_0^1 \int_0^1 dz\, du\, u^{1+\epsilon} (1-z)^{\epsilon} \frac{s/t}{s/t+z}\frac{1}{1+us/t}=
{}_2F_1\left(1,1,2+\epsilon,-t/s\right)\int_0^1 du\, \frac{u^{1+\epsilon}}{1+us/t},$$ so I am left with what looks like a non trivial integral over u. (*)
The first term in (1) still looks less tractable because of the coupled dependence (1-uz). Maybe I can use a geometric series here but I am unsure of how this would help. (**)
Any ideas how to tackle (*) and/or (**), would be great!
Thanks!
{}_2F_1\left(1,1,2+\epsilon,-t/s\right)\int_0^1 du\, \frac{u^{1+\epsilon}}{1+us/t},$$ so I am left with what looks like a non trivial integral over u. (*)
The first term in (1) still looks less tractable because of the coupled dependence (1-uz). Maybe I can use a geometric series here but I am unsure of how this would help. (**)
Any ideas how to tackle (*) and/or (**), would be great!
Thanks!