Learn the Truth about Integrals of 1/x: Debunking Common Misconceptions

[Eclair_de_XII]

TL;DR Summary

I've read somewhere that if a function is integrable over some domain, then a scalar multiple of that function is also integrable over the same domain. Moreover, the integral of a scalar multiple of a function is the scalar multiple of the integral of that function. But when I think about the integral of the function $f(x)=\frac{1}{x}$, I start to get confused.

Let's take an integral$−\int_1^e\frac{dx}{x}$. On one hand, this is equal to $-\ln(x)|_1^e$. But on the other, $−\int_1^e\frac{dx}{x}=\int_1^e\frac{dx}{-x}$. If I assume that the integral of this is $\ln⁡(-x)|_1^e$, then I'd be really stupid since $\ln$ is not even defined over the negative real numbers. I am very sure I am misunderstanding something very important in these statements I have cited. Or maybe I forgot some key statements that go along with these. Either way, I'm confused, and it's been bothering me all day. Can someone help me point out my mistakes? Thanks.

 

[Antarres]

Indeed, when you integrate $\tfrac{1}{x}$, you need to take into account that logarithm is defined on the positive subset of the real line, so it is correct to write:
$$\int \frac{dx}{x} = \ln{\vert x \vert} + C$$
That's the correct formula, so the confusion shouldn't arise in that case. Namely, we have:
$$-\int_1^e \frac{dx}{x} = -(\ln{\vert x\vert})\vert_1^e = -1$$
$$\int_1^e \frac{dx}{-x} = -\int_1^e \frac{d(-x)}{-x} = -(\ln{\vert -x\vert})\vert_1^e = -1$$

Is that clearer now? You made a mistake also, when you put a scalar in the integral, you forgot to change the variable(x into -x), so you didn't get the extra minus sign, besides missing the absolute value.

 

[Eclair_de_XII]

Okay, I got it. Thank you for explaining it to me.

 

[Antarres]

You're welcome!

 

[PeroK]

Okay, I got it. Thank you for explaining it to me.

If you've got your degree, you can stop tormenting yourself with mathematics now!

 

[Saracen Rue]

Indeed, when you integrate $\tfrac{1}{x}$, you need to take into account that logarithm is defined on the positive subset of the real line, so it is correct to write:
$$\int \frac{dx}{x} = \ln{\vert x \vert} + C$$
That's the correct formula, so the confusion shouldn't arise in that case. Namely, we have:
$$-\int_1^e \frac{dx}{x} = -(\ln{\vert x\vert})\vert_1^e = -1$$
$$\int_1^e \frac{dx}{-x} = -\int_1^e \frac{d(-x)}{-x} = -(\ln{\vert -x\vert})\vert_1^e = -1$$

Is that clearer now? You made a mistake also, when you put a scalar in the integral, you forgot to change the variable(x into -x), so you didn't get the extra minus sign, besides missing the absolute value.

Hi, I was just reading through this thread and became quite confused when I say you state that $\int_1^e \frac{dx}{-x} = -\int_1^e \frac{d(-x)}{-x}$, specifically where $dx$ becomes $d(-x)$. I'm almost certainly a complete novice when it comes to calculus but I am still quite confused as to how/why you are integrating with respect to negative x. Any chance you could explain it to a beginner? Thanks for your time :)

 

[Antarres]

That is just a shorter notation for substitution. We have either:
$$dx = -d(-x)$$
due to linearity of differential. In this case we have integration in variable $-x$, without performing the change of variables explicitly(since we integrate $d(-x)$).
Or we can perform a substitution: $u = -x, du = -dx$ in which case we get:
$$\int_1^e \frac{dx}{-x} = \int_{-1}^{-e} \frac{-du}{u}$$
When we perform a change of variables, we must change the boundaries of integration, that's why 1 goes to -1, and e goes to -e. And now when we integrate in $u$, we get the same result.

So basically, you can look at $dx$ in an integral, as a mark of what variable you're integrating. But also, it is a differential, so you can for example do something like $2xdx = d(x^2)$, where you effectively change the integration variable from $x$ to $x^2$ without explicitly doing it. When you write this way, you don't change the boundaries, since you haven't changed variables explicitly, and in the end result you substitute $x$ with the boundaries, not $x^2$.

Hope I didn't make that sound too confusing.

 

[WWGD]

This is my take:
In a general sense, $d/dx(ln(f(x)) =\frac {f'(x)dx}{f(x)}$

In your case , to have an integral of a composite function f(x) , you put it in above form: $f(x)=-x$ given

$\frac {dx}{-x} =\frac {d(-x)(-1)}{-x}= \frac {-1}{-x} d(-x)$