Advanced Calculus fi..

250 Advanced Calculus, Fifth EditionEXAMPLEThe integral// sin I dx dy,where D is the square domain 0 < x < 1,0 < y < 1, exists, even though the functionis badly discontinuous on the y axis, since I sin :I 5 1 in D.For any such bounded function f, one can approximate the integral over thewhole region arbitrarily closely by the integral over a smaller region, avoiding thediscontinuities. For if R is split into two regions RI , R2, overlapping only on boundarypoints, then4Further, if 1 f 1 p M in R, theni?// f b , y)dx dyR25 A4 . &. (4.90)where A2 is the area of R2, as follows from (4.47). If A2 is sufficiently small, theintegral over R1 will approximate the integral over R as closely as desired.True improper integrals arise, as for functions of one variable, when f (x, y) isunbounded in R. The most important case of this is when f (x, y) is defined andcontinuous but unbounded in a bounded domain D, with no information given aboutvalues off on the boundary of D. In this case the limit of the sum C f (x, y) AA willfail to exist because f is unbounded. The integral of f over D is termed improperand is assigned a value by the limit processprovided that the limit exists. Here R denotes a closed region contained, with itsboundary, in D, as suggested in Fig. 4.14. The limit process is understood as follows:The limit exists and has value K if, given t > 0, a particular region R1 can be foundsuch thatIIfor all regions R containing R1 and lying in D.What amounts to a special case of the preceding is the case of a function havingapoint discontinuity, that is, a function continuous in a domain D except at a singlepoint P of D. It would be natural in this case to isolate the trouble at P by integratingup to a small circle of radius h about P and then letting h approach 0 (Fig. 4.15).This definition of the improper integral is equivalent to the preceding, provided that'