Mathematica question regarding relation between Pi and its convergents

I am researching my conjecture

(-1)^n*(Pi?A002485(n)/A002486(n))=(Abs(m)2^j)^(-1)Integrate[(x^l(1-x)^(2(j+2))*(k+(m+k)*x^2))/(1+x^2),{x,0,1}] (1)

where integer n = 3,4,5,… serves as the index for terms in OEIS A002485(n) and A002486(n) sequences, and { j, k, l, m} are some signed integers (to be found experimentally or otherwise).

The following cases were found by me and were confirmed using Maple

  1. 22/7 - Pi = Int(x^4*(1-x)^4*/(1+x^2),x = 0 .. 1) with n=3, j=0, k=1, l=4,  m=-1
    
  2. Pi - 333/106 = 1/530*Int(x^5*(1-x)^6*(197+462*x^2)/(1+x^2),x = 0 .. 1) with n=4, j=1, k=197, l=5, m=265
    
  3. 355/113 - Pi = 1/3164Int(x^8(1-x)^8*(25+816*x^2)/(1+x^2),x = 0 .. 1) with n=5, j=2, k=25, l=8, m=791

  4. Pi - 103993/33102 = 1/755216Int(x^14(1-x)^12*(124360+77159x^2)/(1+x^2),x = 0 .. 1) with n=6, j=4, k=124360, l=14, m= -47201

  5. 104348/33215 - Pi = 1/38544Int(x^12(1-x)^12*(1349-1060*x^2)/(1+x^2),x = 0 .. 1) with n=7, j=4, k=1349, l=12, m=-2409

While manipulating expressions in Wolfram Cloud Development Platform and WolframAlpha I came across the following parametric identity

Sqrt[Pi] = (1/(2^j)*((k Gamma[5 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2,7/2 + j + l/2}, -1])/Gamma[6 + 2 j + l] + ((k + m) Gamma[7 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 7/2 + j, 4 + j}, {4 + j + l/2,9/2 + j + l/2}, -1])/Gamma[8 + 2 j + l]))/(2^(-5 - 3 j -l) Gamma[5 + 2 j] Gamma[1 + l] (k HypergeometricPFQRegularized[{1, 5/2 + j,3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1] +1/2 (3 + j) (5 + 2 j) (k + m) HypergeometricPFQRegularized[{1,7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1])) (2)

which indeed gave Sqrt[Pi] for each set of {j,k,l,m} given in above-listed cases 1), 2), 3), 4), 5)

I presume that above identity (2) will yield Sqrt[Pi] for other (infinite) number of sets of {j,k,l,m}.

Is it an interesting identity?

Thanks,

Best Regards,

Alexander R. Povolotsky

I was notified that Maple simplifies the expression on the right hand side of (2) to sqrt(Pi).
It seems to be true for arbitrary j,k,l,m.