I am wishing to test the convergence of the following integral and find the value of (where convergent):
$$\int_{0}^{1/2} \frac{1}{x(\log(1/x))^c} dx$$
For all $ c \in \mathbb{R} $.
My first step is factoring out the $\frac{1}{(-1)^c}$ of the integral.
Like so:$$\frac{1}{(-1)^c} \int_{0}^{1/2} \frac{1}{x(\log(x))^c} dx$$
However, for some non-integers of c, this will give me a complex number. So when I find the limit of a variable tending to 0 (for the integral limit), like this
$$\lim_{t \to 0^+}\frac{1}{(-1)^c} \int_{t}^{1/2} \frac{1}{x(log(x))^c} dx$$
Won't I have that, because it's complex, there exists no such limit? Or can I factor this out of the limit, so the existence of the limit is irrespective of this constant - Can you do this with improper integrals?
Thank you.
$\int_0^\frac{1}{2}\dfrac{1}{x\left(\ln\dfrac{1}{x}\right)^c}~dx$
$=\int_\infty^{\ln2}\dfrac{1}{e^{-x}\left(\ln\dfrac{1}{e^{-x}}\right)^c}~d(e^{-x})$
$=\int_{\ln2}^\infty\dfrac{1}{x^c}~dx$
The integral converges when $c<-1$ .