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Integrations involving hypergeometrics

  1. Nov 27, 2015 #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!
     
  2. jcsd
  3. Nov 27, 2015 #2
    Wow, this looks awful !

    Let's write ## F(u,z) = \frac{u^{1+\epsilon} (1-z)^\epsilon }{1-uz} \frac{1}{1+au} ##, where ##a = s/t##.

    I make the assumption that ##a,\epsilon >0##, I hope its OK for what you need.

    The function ##F## is continuous on ##V = [0,1]^2 - \{1,1\} = [0,1]\times [0,1[ \cup [0,1[ \times [0,1] ##, so you can always assume ##|uz| < 1##.

    Therefore you can express ##F## in terms of a convergent serie on ##V## : ## F(u,z) = S_N(u,z) + R_N(u,z) ##,
    where ##S_N (u,z) = \sum_{n = 0}^N \frac{u^{1+n +\epsilon}z^n (1-z)^{\epsilon}}{1+au}##, and ##R_N(u,z) = \sum_{n = N+1}^{+\infty} \frac{u^{1+n +\epsilon} z^n (1-z)^{\epsilon}}{1+au}##.

    In particular, I'm 99.5% sure that on any compact ##K \subset V##, the serie is uniformly convergent, so ##\int_K R_N ## will tend to 0 as ##N\to \infty##, so that on ##K##, you can exchange the sum and the double integral :

    ##\int_K F(u,z) \ dudz = \sum_{n = 0}^{\infty} \int_K \frac{u^{1+n +\epsilon}z^n (1-z)^{\epsilon}}{1+au}##.

    You can consider the compacts ##K_{\alpha,\beta} = [0,\alpha]\times [0,\beta] ##, for ##\alpha,\beta < 1##

    Then the double integration is a product of two single integrations.

    Now you need someone more knowledgeable than me to tell you whether it is true that ##\int_V F(u,z) = \lim_{(\alpha,\beta)\to(1,1)} \int_{K_{\alpha,\beta}} F(u,z) ##. If yes you are done, otherwise my post is useless :-(


    Hope there aren't too many mistakes here !
     
  4. Nov 28, 2015 #3

    CAF123

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    Hi geoffrey159, thanks for your reply.
    Isn't the integrand $$F(u.z) = \frac{u^{1+\epsilon} (1-z)^{\epsilon} z}{(s/t + z)(1-uz)}?$$
    Ah, a<0 in my case but I don't think you used this fact anywhere below?
    So here you are saying that the integrand is continuous on the domain ##[0,1]^2## except when ##u## and ##z## are both 1? So if we keep the sets open then we can always have |uz|<1? I come from a physics background so I probably said that very loosely :)
    Ok I will speak to my prof about this. If we assume that indeed the limit is true, then I have $$\sum_{n=0}^{\infty} \int_0^1 dz \int_0^1 du \frac{u^{1+\epsilon+n} z^{n+1}(1-z)^{\epsilon}}{(s/t+z)}.$$ This then gives two single integrations which I can write like $$ \sum_{n=0}^{\infty} \frac{t}{s} \left(\int_0^1 du\,u^{1+n+\epsilon}\right) \frac{\Gamma(n+2) \Gamma(1+\epsilon)}{\Gamma(n+3+\epsilon)} {}_2F_1\left(1, n+2, n+3+\epsilon, -\frac{t}{s}\right) = \frac{t}{s}\sum_{n=0}^{\infty} \frac{1}{n+2+\epsilon} \frac{\Gamma(n+2) \Gamma(1+\epsilon)}{\Gamma(n+3+\epsilon)} {}_2F_1\left(1, n+2, n+3+\epsilon, -\frac{t}{s}\right)$$ I don't really see how to calculate this sum, any thoughts here?
     
    Last edited: Nov 28, 2015
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