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4 1) (1 0 2 2 π n n x n M N dx x x P + = ∫ + = In a previous note above we examined the properties of Legendre polynomials and in the process found that integrals of the form-) 1 arctan() 1 () (1 0 2 2 2 a a M N dx x a x P n n x n + = ∫ + + = For the special case of a=1, one obtains the following identity for π-]) 1 ([ 4 1 0 2 2 dx x P N M n n n ∫ + + − = π Running through the first few integers n one finds-1 ] 3 [ 3 1) (2 3 [) 2 (4 1 0 2 2 = − + = ∫ + + − − = n for dx x x P π π 2 ] 51 160 [ 51 160 ] 1 3 20 [) 2 / 17 (4 1 0 2 4 = + − + = ∫ + + = n for dx x P π π and-3 ] 205 644 [ 205 644 1) (5 161 [ 41 4 1 0 2 6 = + − + = ∫ + + − − = n for dx x x P π π What is clear from these results is that the first term on the right of the equality gets ever closer to the true value of π while the second term gets progressively smaller approaching zero as n goes to infinity. This means that one has a way to approximate π by simply looking at

Evaluation of the integral ∫₀^{𝑝}𝑢ⁿ𝑒^{-𝑢²}(𝑢+𝑥)⁻¹𝑑𝑢