Improper integral

  • Thread starter frasifrasi
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  • #1
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Regarding the integral from -infinity to 0

of 6/(5x-2)


--> I arrive at 6/5*ln(u), is this the right thing?

How would I evaluate something like ln(-2), or do I just assume it is divergent since it the other limit will come out to neg infinity?
 

Answers and Replies

  • #2
quasar987
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You have made the change of variables

[tex]\int_{-\infty}^{0}\frac{6}{5x-2}dx = \frac{6}{5}\int_{-\infty}^{-2}\frac{du}{u}[/tex]

But obviously, it's not true that ln(u) is an antiderivative of 1/u on the domain (-\infinity,-2) simply because ln(u) is not defined there. So your change of variable was not so useful. Try u=2-5x.
 
  • #3
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wait, do you mean 5x - 2?
 
  • #4
Hurkyl
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ln(u) is not the antiderivative of 1/u over the nonzero reals. ln(u) is a simplification that is only valid for positive u.
 
  • #5
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So, what substitution should I use?
 
  • #6
Dick
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So, what substitution should I use?

Hurkyl is trying to get you to recall ln(|u|) is also an antiderivative for 1/u.
 
  • #7
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I am still unsure wth I should do -- u-subs or what?

Thank you.
 
  • #8
Dick
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I am still unsure wth I should do -- u-subs or what?

Thank you.

You already did the u-subs correctly. The antiderivative is (6/5)*ln(|5x-2|). Now you just have to think what happens if you put the limits in.
 
  • #9
quasar987
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  • #10
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where did that come from, though?
 
  • #11
quasar987
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it comes from choosing a change of coordinate such that the bounds of the integral become positive, so that now ln(u) is an antiderivative to the integrand.
 
  • #12
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But how was that derived, can anyone explain?
 
  • #13
quasar987
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Because with u(x)=5x-2, the upper bound of the intregral became -2. So using u(x) = -(5x-2), instead, it becomes -(-2)=2
 
  • #14
HallsofIvy
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If u is negative, |u|= -u. Since, on this range, 5x-2 is negative, |5x-2|= 2- 5x.
 

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