Improper integral

1. Dec 10, 2007

frasifrasi

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?

2. Dec 10, 2007

quasar987

You have made the change of variables

$$\int_{-\infty}^{0}\frac{6}{5x-2}dx = \frac{6}{5}\int_{-\infty}^{-2}\frac{du}{u}$$

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. Dec 10, 2007

frasifrasi

wait, do you mean 5x - 2?

4. Dec 10, 2007

Hurkyl

Staff Emeritus
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. Dec 10, 2007

frasifrasi

So, what substitution should I use?

6. Dec 10, 2007

Dick

Hurkyl is trying to get you to recall ln(|u|) is also an antiderivative for 1/u.

7. Dec 10, 2007

frasifrasi

I am still unsure wth I should do -- u-subs or what?

Thank you.

8. Dec 10, 2007

Dick

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. Dec 10, 2007

quasar987

No, I mean 2-5x. Try it.

10. Dec 10, 2007

frasifrasi

where did that come from, though?

11. Dec 10, 2007

quasar987

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. Dec 10, 2007

frasifrasi

But how was that derived, can anyone explain?

13. Dec 10, 2007

quasar987

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. Dec 11, 2007

HallsofIvy

Staff Emeritus
If u is negative, |u|= -u. Since, on this range, 5x-2 is negative, |5x-2|= 2- 5x.

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