convergence of $\int_0^{1/2}\frac{1}{r[log(r)]^{1+\varepsilon N}}dr$

Question

I have a problem involving belongships of a function to a space. After a while, I have arrived at an integral that I have to see if it is finite or not. The integral is as follows $$\int_0^{1/2}\frac{1}{r[log(r)]^{1+\varepsilon N}}dr$$ where $0<\varepsilon<<1$ and $N$ is a fixed integer.

I know that if $\varepsilon=0$, then the integral will be $log(log(r))$ which in zero will explode.

So, my question is: How do I compute the previous integral? (Which will be the same as asking For what value of $\varepsilon$ the integral is finite?)

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My guess:

I have tried doing a variable change as in the case of $\varepsilon=0$, because I think it will be, more or less, the same. Id est: $$\int_0^{1/2}\frac{1+[log(r)]^{1+\varepsilon N}-[log(r)]^{1+\varepsilon N}}{r[log(r)]^{1+\varepsilon N}}dr\ \xrightarrow{\text{variable change}}\ ?$$ taking $u=[log(r)]^{1+\varepsilon N}$, or something like this, but I do not manage to make progress.

Answer 1

The derivative of $-\frac1{\varepsilon N \log^{\varepsilon N}r}$ is $\frac1{r\log^{1+\varepsilon N}r}$

Answer 2

As @Winther suggest, let's try a substitution.

Take $u=\log(r)$. We have $du = \frac1r dr$ and we have $$ \int_0^{1/2}\frac1{r[\log(r)]^{1+\epsilon N}}dr = \int_{-\infty}^{\log(1/2)}\frac1{u^{1+\epsilon N}}du = \left[-\frac1{\epsilon N}u^{-\epsilon N}\right]^{\log(1/2)}_{-\infty} $$ I think you can finish the rest.