如图，如何先令alpha取数值，后运算

h1[\[Alpha]_] := ((((left 25 \[Pi] mp^2)/
8 ((-(3/(2 n)))^(1/(
1 - 2 \[Alpha])) ((-(3^(1 - \[Alpha])/2^(1 - \[Alpha])))^(
1/(2 \[Alpha] - 1)) n - (2 \[Alpha] (n + 1) na)/(
2 \[Alpha] - 1)))^((2 \[Alpha] - 1)/(
2 \[Alpha] n + 2 \[Alpha])*2 n))^((n + 1)/
2) /. {left -> -nt Sqrt[(1 - \[Phi]1^(2 \[Alpha]))/(
2 \[Alpha] - 1)] (E^3.044*10^-10)^2 /. \[Phi]1 -> (-((2 n)/(
3 h1 ((-((3 h1^(2 \[Alpha]))/(2 n)))^(1/(
1 - 2 \[Alpha])) ((-(3^(1 - \[Alpha])/2^(
1 - \[Alpha])))^(1/(2 \[Alpha] - 1)) n - (
2 \[Alpha] (n + 1) na)/(2 \[Alpha] - 1)))^((
2 \[Alpha] - 1)/(2 \[Alpha] n + 2 \[Alpha])*(n + 1)))))^(
1/(2 \[Alpha] - 1))}) /. {nt -> (-4 n^2)/(
3 (((2 \[Alpha] - 1)/(
2 \[Alpha]*n +
2 \[Alpha])*((-3^(1 - \[Alpha]) 2^(\[Alpha] - 1))/n^(
1 - 2 \[Alpha]))^(1/(2 \[Alpha] - 1)) - na)*(
2 \[Alpha]*n + 2 \[Alpha])/(2 \[Alpha] - 1)*(-3/(2 n))^(1/(
1 - 2 \[Alpha])))^((2 \[Alpha] - 1)/\[Alpha])) /. {First[
FindRoot[
ns - 1 == (-8 n^2 - 4 n)/(
3 (((2 \[Alpha] - 1)/(
2 \[Alpha]*n +
2 \[Alpha])*((-3^(1 - \[Alpha]) 2^(\[Alpha] - 1))/
n^(1 - 2 \[Alpha]))^(1/(2 \[Alpha] - 1)) - na)*(
2 \[Alpha]*n + 2 \[Alpha])/(2 \[Alpha] - 1)*(-3/(2 n))^(
1/(1 - 2 \[Alpha])))^((2 \[Alpha] -
1)/\[Alpha])) /. {ns -> 0.96, na -> 55}, {n, -3}]],
na -> 55}}) /. {mp -> 1, na -> 55,
First[FindRoot[
ns - 1 == (-8 n^2 - 4 n)/(
3 (((2 \[Alpha] - 1)/(
2 \[Alpha]*n +
2 \[Alpha])*((-3^(1 - \[Alpha]) 2^(\[Alpha] - 1))/n^(
1 - 2 \[Alpha]))^(1/(2 \[Alpha] - 1)) - na)*(
2 \[Alpha]*n + 2 \[Alpha])/(2 \[Alpha] - 1)*(-3/(2 n))^(1/(
1 - 2 \[Alpha])))^((2 \[Alpha] - 1)/\[Alpha])) /. {ns ->
0.96, na -> 55}, {n, -3}]]}

FunctionDomain[h1[\[Alpha]], \[Alpha]]

和这帖实质上是一个问题：
https://tieba.baidu.com/p/8320899790
FunctionDomain（至少现在）也算不了黑箱。简例：
func = FunctionInterpolation[x, {x, -1, 1}];
FunctionDomain[{func@x, x^2 < 1}, x]