Integrations involving hypergeometrics

[CAF123]

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!

 

[geoffrey159]

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 !

 

[CAF123]

Hi geoffrey159, thanks for your reply.

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

Isn't the integrand $$F(u.z) = \frac{u^{1+\epsilon} (1-z)^{\epsilon} z}{(s/t + z)(1-uz)}?$$

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

Ah, a<0 in my case but I don't think you used this fact anywhere below?

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 $|uz| < 1$.

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 :)

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 :-(

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?