Problem: Prove that this notion of convergence of f(x) dx is independent of the choice of the point a, at which you "cut" f(x) dx into f f(x) dx and f f(x) dr. ∞

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Problem: Prove that this notion of convergence of f(x) dx is independent of the choice of the point a, at
which you "cut" f(x) dx into fa f(x) dx and f f(x) dx.
Guide: There are two parts you need to show, the convergent case and the divergent case. In other words, you
need to show that: Fix a real number a.
(1) [Convergent case] If f f(x) dx and f f(x) dx both converge, then for all real numbers b, the improper
integrals f f(x) dr and f f(x) dx must also converge and
[ f(x) dx + + √°*° f(x) dx = [*®*® f(x) dx + √°*® f(x) dx.
(2) [Divergent case] If either fª f(x) dx or f f(x) dx diverge, then for all real numbers b, either f f(x) dx
or f f(x) dx must also diverge.
Thus after showing the two cases above, we will have proved that in the definition (the two bullet points right after
(1)) of f(x) dx above, the number a can be replaced with any other real numbers.
Transcribed Image Text:Problem: Prove that this notion of convergence of f(x) dx is independent of the choice of the point a, at which you "cut" f(x) dx into fa f(x) dx and f f(x) dx. Guide: There are two parts you need to show, the convergent case and the divergent case. In other words, you need to show that: Fix a real number a. (1) [Convergent case] If f f(x) dx and f f(x) dx both converge, then for all real numbers b, the improper integrals f f(x) dr and f f(x) dx must also converge and [ f(x) dx + + √°*° f(x) dx = [*®*® f(x) dx + √°*® f(x) dx. (2) [Divergent case] If either fª f(x) dx or f f(x) dx diverge, then for all real numbers b, either f f(x) dx or f f(x) dx must also diverge. Thus after showing the two cases above, we will have proved that in the definition (the two bullet points right after (1)) of f(x) dx above, the number a can be replaced with any other real numbers.
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