True or False

Calculus Level 3

True or False?

∫ 0 1 ∫ 0 1 y 1 − x 2 y 2 d x d y = ∑ n = 0 ∞ 1 ( 2 n + 1 ) ( 2 n + 2 ) \int_0^1 \int_0^1 \dfrac y{1-x^2 y^2} \, dx \; dy = \sum_{n=0}^\infty \dfrac1{(2n+1)(2n+2) }

True False

Jose Sacramento by Jose Sacramento , 66, Portugal

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2 solutions

Jose Sacramento
Apr 4, 2017

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Kushal Bose
Apr 5, 2017

∫ 0 1 ∫ 0 1 y 1 − x 2 y 2 d x d y = ∫ 0 1 ∫ 0 1 ( y + y . ( x 2 y 2 ) + y . ( x 2 y 2 ) 2 + y . ( x 2 y 2 ) 3 + . . . . . ∞ ) d y d x = ∫ 0 1 ( 1 / 2 + x 2 / 4 + x 4 / 6 + . . . ∞ ) d x = 1 / 2 + 1 / 3.4 + . . . + ∞ \int_0^1 \int_0^1 \dfrac y{1-x^2 y^2} \, dx \; dy \\=\int_0^1 \int_0^1 (y+y.(x^2y^2)+y.(x^2y^2)^2+y.(x^2y^2)^3+.....\infty) \, dy \, dx \\ =\int_{0}^{1} (1/2+x^2/4 + x^4/6 +... \infty) \, dx \\ =1/2 +1/3.4 +...+ \infty

Hence the equality holds

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