Malmsten's Proof of the Integral Theorem

x = X, y = Y ;and letsx 1 , x 2 , . . . , x n−1be values betweenThen Malmsten constructs the sumy 1 , y 2 , . . . , y n−1x 0 and X resp y 0 and Y.S n = (x 1 − x 0 ) · p(x 0 , y 0 ) + (y 1 − y 0 ) · q(x 0 , y 0 )+(x 2 − x 1 ) · p(x 1 , y 1 ) + (y 2 − y 1 ) · q(x 1 , y 1 )+ . . . +With the condition+(X − x n−1 ) · p(x n−1 , y n−1 ) + (Y − y n−1 ) · q(x n−1 , y n−1 ).∂p(x, y)dyMalmsten claims that as n → ∞ and the differencesdecrease indefinitely,=x 1 − x 0 , x 2 − x 1 , . . . , X − x n−1 ,∂q(x, y), (7)dxy 1 − y 0 , y 2 − y 1 , . . . , Y − y n−1(8)lim S n =∫ Xx 0p(x, Y )dx +∫ Yy 0q(x 0 , y)dy,=∫ Xx 0p(x, y 0 )dx +∫ Yy 0q(X, y)dy,i.e lim S n is independent of n as well as of the choice of the values in (8).In short, the last expression says that the limit of any sequence, followingthe conditions described above, is equal to the integrations along two specificpaths, i.e. from one corner of a rectangle to the opposite via the contours(clockwise respectively counter clockwise).Malmstens approach is to treat the terms of the sum S n by replacingthem with approximate integrals and error terms. This new sum has the15