convergence of the integral (with parameter) Find all values of the parameter $c$ for which the following integral is convergent:
$$ \int_{1}^{\infty}  \frac{dx}{(\ln x)^c(x^2-x)^{2c}}.$$
Please help me solve it or give a hint.
 A: Hints: $\int_0^1 \frac{dy}{y^\alpha}$ converges only for $\alpha<1$, $\int_1^\infty \frac{dy}{y^\alpha}$ converges only for $\alpha>1$,
$$\ln(1+y)\sim y, \quad y \to 0$$
and
$$\lim_{x \to +\infty} \frac{\ln x}{x^{\beta}}=0, \quad \forall \beta>0.$$
A: Near $x=1$, $\ln{x} \sim x-1$ so that the integrand in this neighborhood behaves as
$$\frac{1}{x^{2 c} (x-1)^{3 c}}$$
so for convergence of the integral, $c \lt 1/3$. 
As $x \to \infty$, however, the integrand behaves as
$$\frac{1}{(\ln{x})^c x^{4 c}}$$
In this case, we would need $c \gt 1/4$ for convergence of the integral.  (The log piece has no impact on the convergence criteria here.)  Therefore, the range of $c$ for which the integral converges is $1/4 \lt c \lt 1/3$.
A: You have to consider the convergence in 1 and in $+ \infty$. i will give you some hints for the letter problem. What you want to use is the ratio test.
First of all revrite the problem in this form: $$ \int_3^ \infty \frac{1}{x^{4c}ln^c(x)(1+o(1))}dx$$ Now if c>1/4 your integral converges by ratio test with 1/x^4c. To know what happens at c=1/4, you just need to solve $$\int_3^\infty \frac{1}{x} \frac{1}{ln^{1/4}(x)}dx,$$ (DONOTREAD!!   
what about a change of variable?)
P.S.: at 1 try using Taylor on the logarithm!
