# Improper integral

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?

quasar987
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Gold Member
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.

wait, do you mean 5x - 2?

Hurkyl
Staff Emeritus
Gold Member
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.

So, what substitution should I use?

Dick
Homework Helper
So, what substitution should I use?

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

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

Thank you.

Dick
Homework Helper
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.

quasar987
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Gold Member
wait, do you mean 5x - 2?

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

where did that come from, though?

quasar987
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Gold Member
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.

But how was that derived, can anyone explain?

quasar987
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Gold Member
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

HallsofIvy
Homework Helper
If u is negative, |u|= -u. Since, on this range, 5x-2 is negative, |5x-2|= 2- 5x.