Integrating odd functions with infinite discontinuity:

  • Thread starter atqamar
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  • #1
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If an odd function has an infinite discontinuity in its domain, can it be integrated (such that a convergent finite emerges) with that domain included?

For example: [tex]\int_{-1}^2 \frac{1}{x^{-3}} dx[/tex]. Intuitively, it can be simplified to [tex]\int_1^2 \frac{1}{x^{-3}} dx[/tex] and thus the infinite discontinuity at 0 is removed.

If that is not doable, can an integral converge if the end points of the domain are infinite discontinuities?

For example: Does [tex]\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} tan(x) dx = 0[/tex]?

If these kinds of functions are split in two, and limits to [tex]\infty[/tex] are taken, then algebraic manipulation of infinities are required.

Any insight would be appreciated.
 

Answers and Replies

  • #2
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You should look into improper integrals?
 
  • #3
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Certainly, these are all improper integrals. But my question is whether the above integrals would converge or diverge. Intuitively/Geometrically, they should converge; but like I said, once the improper integral is carried out, one has to deal with infinities.
 
  • #4
Gib Z
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Look up "Cauchy Principal Value". Strictly, the integrations over the discontinuities are divergent, but they can be assigned a Cauchy Principal Value that appeals to your geometric intuition.
 
  • #5
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Thanks a lot Gib Z! That helped immensely.
 

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