# Improper integral $\int_0^{1/2}\frac1{t^a|ln\ t|^b} dt$

I have the following improper integral: $$\int_0^{1/2}\frac1{t^a|\log\ t|^b} dt$$ for some $a,b\ge0$. There exists an $M>0$ such that: $$\int_0^{M}\frac1{t^a|\log\ t|^b} dt<\int_0^{M}\frac1{t^a} dt$$ and this last one converges for $a<1$, so we can say that if $a<1$ our integral is convergent. I'm not sure if the following is valid. For $a\ge1$: $$\int_0^{1/2}\frac1{t^a|\log\ t|^b} dt = \int_{-\infty}^{\log\frac1{2}}\frac1{e^{(a-1)x}|x|^b} dx>\int_{-\infty}^{\log\frac1{2}}\frac1{|x|^b} dx$$ This diverges for $b\le1$

Please tell me if there is something wrong. Also I would like to know what happens with the integral when both $a$ and $b$ are greater than 1. Is there a more direct way to solve this kind of improper integrals?

EDIT: If yo don't like or you don't understand what I´ve done, at least say how would you do it, please. I haven´t been taught how to solve them.

It is easy to check the convergence of this integral if we "transform" it into a different form.
Now, the original integral is written as follows : $$\int_{0}^{\frac{1}{2}}t^{-a}|\log t|^{-b}dt$$ First, we apply variable transformation $\log t = u$ to get $$\int_{-\infty}^{\log\frac{1}{2}}e^{u(1-a)}|u|^{-b}du$$ Now, we apply once more another variable transformation $u = -x$ to finally obtain $$\int_{\log 2}^{\infty}e^{-x(1-a)}|x|^{-b}dx$$ which is equivalent to $$\int_{\log 2}^{\infty}e^{-x(1-a)}x^{-b}dx$$ since our domain of integration is $[\log 2,\infty)$ in which the value of $|x|$ and $x$ are the same.
Now, this is the easy part. First, we consider the case of $a=1$ in which we have $$\int_{\log 2}^{\infty}x^{-b}dx$$ and we know that the integral converges $\forall b > 1$ in this case.

Next, we consider the case of $0\leq a < 1$ in which $-x(1-a) < 0$ and therefore we have $$0< \int_{\log 2}^{\infty}e^{-x(1-a)}x^{-b} \leq \int_{\log 2}^{\infty}e^{-x(1-a)}(\log2)^{-b}dx \leq (\log2)^{-b}\int_{0}^{\infty}e^{-x(1-a)}dx<\infty$$ Since the integral is bounded and nonnegative, we can assure its value is finite for any $b\geq 0$ (In fact, it still converges for $b<0$. You can compare it with Gamma Function).
Finally, when $a>1$, we have $-x(1-a) >0$ and thus the integral diverges since exponential function "grows faster" compared to "polynomial function of any degree".

To sum it up,
1. $a=1$, the integral converges for $b>1$.
2. $0\leq a <1$, the integral always converges for any $b \in \mathbb{R}^{+}\cup\{0\}$.
3. $a>1$, the integral always diverges for any $b \in \mathbb{R}^{+}\cup\{0\}$.

I skipped some details for rigorous arguments to give you a gap to fill as exercise. Good luck.