Why can't Mathematica evaluate the limit of this hypergeometric function? Is there any way to find the answer?

Question

I have the following function

    f = 1/Gamma[(1 + d)/2] \[Pi]^(
  1/2 (-1 + d)) ((
    Sqrt[2] (-1 + d) r^5 (2 + r^(-4 + 2 d)) Hypergeometric2F1[-(1/2), 
      1/(-4 + 2 d), 1 + 1/(-4 + 2 d), -(1/2) r^(-4 + 2 d)])/(
    2 r^4 + r^(2 d)) + (
    r^(-1 + d) (6 - 4 d + (
       Sqrt[2] (2 - 3 d + d^2) r^(
        2 + d) (2 + r^(-4 + 2 d)) Hypergeometric2F1[1/2, (3 - 2 d)/(
         4 - 2 d), (7 - 4 d)/(
         4 - 2 d), -(1/2) r^(-4 + 2 d)])/((-3 + 2 d) (2 r^4 + r^(
          2 d))) + 
       2 (-1 + d) Log[(2 r^d)/(r^d + r^2 Sqrt[2 + r^(-4 + 2 d)])]))/(
    2 (-1 + d)))

Mathematica can not evaluate the limit of $f$ when $r$ goes to infinity.

Limit[f, r -> Infinity, Assumptions -> d >= 4]

Is there anyway that this limit can be evaluated by Mathematica? Note that if I first put $d=4$ or $5$ or ... and then take the limit, Mathematica find the answer.

Limit[f /. d -> 4, r -> Infinity]
-(2/9) 2^(3/4) Sqrt[\[Pi]] (9 Gamma[-(3/4)] + 4 Gamma[1/4]) Gamma[5/4]

But I need the general relation as a function of $d$.

Answer 1

f[d_, r_] = 1/Gamma[(1 + d)/2] π^(1/2 (-1 + d))((Sqrt[2] (-1 + d) r^5 (2 + r^(-4 + 2 d)) Hypergeometric2F1[-(1/2), 1/(-4 + 2 d), 1 + 1/(-4 + 2 d), -(1/2) r^(-4 + 2 d)])/(2 r^4 + r^(2 d)) + (r^(-1 + d) (6 - 4 d + (Sqrt[2] (2 - 3 d + d^2) r^(2 + d) (2 + r^(-4 + 2 d)) Hypergeometric2F1[1/2, (3 - 2 d)/(4 - 2 d), (7 - 4 d)/(4 - 2 d), -(1/2) r^(-4 + 2 d)])/((-3 + 2 d) (2 r^4 + r^(2 d))) + 2 (-1 + d) Log[(2 r^d)/(r^d + r^2 Sqrt[2 + r^(-4 + 2 d)])]))/(2 (-1 + d)));

F[d_] := F[d] = Limit[f[d, r], r -> ∞]

Table[F[d] == -((2^(1/2 (-3 + 1/(-2 + d))) π^(-1 + d/2) *
                Gamma[-((-1 + d)/(2 (-2 + d)))] *
                Gamma[1/(-4 + 2 d)])/Gamma[(1 + d)/2]), {d, 4, 20}] // FullSimplify
(*    {True, True, True, True, True, True, True, True, True, True, True, True, True, True, True, True, True}    *)

So it looks like the limit is

$$ -\frac{2^{\frac{1}{2} \left(\frac{1}{d-2}-3\right)} \pi ^{\frac{d}{2}-1} \Gamma \left(-\frac{d-1}{2 (d-2)}\right) \Gamma \left(\frac{1}{2 d-4}\right)}{\Gamma \left(\frac{d+1}{2}\right)} $$

Answer 2

Asymptotic evaluates to

asy=Asymptotic[f, r -> Infinity]
(*ConditionalExpression[ ..., 0 < d < 1]*)

and restricts parameter to 0<d<1.

Simplify[asy, 0 < d < 1]
(*(Sqrt[2] (-1 + d) \[Pi]^(1/2 (-1 + d))r)/Gamma[(1 + d)/2]*)

$$ \frac{\sqrt{2} (d-1) \pi ^{\frac{d-1}{2}} r}{\Gamma \left(\frac{d+1}{2}\right)} $$

shows asy~r! The limit doesn't exist for r -> Infinity!

Answer 3

Using Asymptotic with an assumption about d ought to do the job (as in Ulrich's answer):

Asymptotic[f, r->Infinity, Assumptions->d>4]

But it takes too long. On the other hand, it looks like you can help Mathematica by shifting the d variable:

asym = Asymptotic[
    FullSimplify[f /. d -> z + 4], 
    r -> Infinity, 
    Assumptions -> z > 0
];

Unfortunately, the output still depends on r:

FreeQ[asym, r]

False

However, if you repeat using Asymptotic on the above output:

asym2 = Asymptotic[asym, r->Infinity, Assumptions -> z > 0];

The output no longer depends on r:

FreeQ[asym2, r]

True

Shifting back and simplifying yields:

r = FullSimplify[asym2 /. z -> d - 4]

-((2^(1/2 (-3 + 1/(-2 + d))) π^(-1 + d/2) Gamma[1/(2 (-2 + d))] Gamma[-((-1 + d)/(2 (-2 + d)))])/Gamma[(1 + d)/2])

This is equivalent to Roman's answer:

Simplify @ Equal[
    r,
    -((2^(1/2 (-3+1/(-2+d))) π^(-1+d/2)*Gamma[-((-1+d)/(2 (-2+d)))]*Gamma[1/(-4+2 d)])/Gamma[(1+d)/2])
]

True