question 212462
prove that
∫_0 ^∞ (dx/( (√(x^4 +25x^2 +160))))=∫_0 ^∞ (dx/( (√(x^4 −95x^2 +2560))))
my attempt (but is it a proof?):
I_b ^a :=∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))=(2/( π(b)^(1/4) agm (1, ((√(a+2(√b)))/(2(b)^(1/4) )))))
∫_0 ^∞ (dx/( (√(x^4 +ax^2 +b))))=
[t=2arctan (x/( (b)^(1/4) ))]
=(1/( (b)^(1/4) ))∫_0 ^(π/2) (dt/( (√(1−((1/2)−(a/(4(√b))))sin^2 t))))=
[K (k^2 ):=∫_0 ^(π/2) (dt/( (√(1−k^2 sin^2 t))))]
=(1/( (b)^(1/4) ))K ((1/2)−(a/(4(√b))))
K (k^2 ) =(2/(πagm (1, (√(1−k^2 )))))
(1/( (b)^(1/4) ))K ((1/2)−(a/(4(√b)))) =(2/( π(b)^(1/4) agm (1, ((√(a+2(√b)))/(2(b)^(1/4) )))))
u_0 =p∧v_0 =q
u_(n+1) =((u_n +v_n )/2)∧v_(n+1) =(√(u_n v_n ))
agm (p, q) =lim_(n→∞) u_n =lim_(n→∞) v_n
I_(160) ^(25) =(1/( π((10))^(1/4) agm (1, ((√(25+8(√(10))))/(4((10))^(1/4) )))))
I_(2560) ^(−95) =(1/(2π((10))^(1/4) agm (1, ((√(−95+32(√(10))))/(8((10))^(1/4) )))))
I_(160) ^(25) =I_(2560) ^(−95)
⇔
agm (1, ((√(25+8(√(10))))/(4((10))^(1/4) )))=2agm (1, ((√(−95+32(√(10))))/(8((10))^(1/4) )))
which is true
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