Integral help (substitution and boundaries)

[Kqwert]

Homework Statement

Hello,
I´ve added an image where the task is shown. I am wondering about the integral bounds here. When doing the substitution, shouldn't the bounds be inf and 0? When using that u = ln x, i.e. u(0) =inf, u(1) = 0.

Homework Equations

The Attempt at a Solution

 

[Master1022]

Homework Statement

Hello,
I´ve added an image where the task is shown. I am wondering about the integral bounds here. When doing the substitution, shouldn't the bounds be inf and 0? When using that u = ln x, i.e. u(0) =inf, u(1) = 0.

If they are simply going step-by-step and changing the limits, then you are correct. Otherwise, there may be some intermediary steps to get to that point where they have further manipulated the integral.

Just wondering why use substitution when you can use reverse chain rule? Then you don't have to worry about the limits changing...

\int_{0}^{\infty} \frac{ln(x)}{x} = \frac{1}{2} ( ln(\infty)^2 - ln(0)^2)

 

[Ray Vickson]

Homework Statement

Hello,
I´ve added an image where the task is shown. I am wondering about the integral bounds here. When doing the substitution, shouldn't the bounds be inf and 0? When using that u = ln x, i.e. u(0) =inf, u(1) = 0.

Homework Equations

The Attempt at a Solution

The image is guilty of sloppy reasoning. The actual definition of the improper integral involved is
$$\int_0^1 \frac{\ln x}{x} \, dx \equiv \lim_{h \to 0+} \int_h^1 \frac{\ln x}{x} \, dx.$$
You can evaluate the integral $\int_h^1 \ln(x)/x \, dx$ and then examine the limit.