Convergence of definite integrals given unknowns. Given the following equation,
$$
\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx
$$
I am supposed to find the condition on the constant $a$ such that the above integral is convergent.
Substituting $u = \ln(x)$, we get
$$
\frac{du}{dx} = \frac{1}{x} 
$$
Where the integral
$$
\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx = \int_{0}^{\ln 2} \frac{du}{dx}\frac{1}{u^{a}}dx = \int_{0}^{ln2} \frac{1}{u^{a}}du
$$
simplifies to
$$
\frac{(\ln 2)^{1-a}}{1-a}
$$
When $a = 1$, the above diverges. However, I am unsure of what values the above equation converges to, i.e when $a&gt;1$ or $a&lt;1$
Let's be a bit more formal here. First of all, it indeed suffices to consider the transformed integral
$$
\int_0^{\ln(2)} \frac 1{u^a}\,du = \int_0^{\ln(2)} u^{-a}\,du.
$$
Note that because of the possible division by zero, this is an improper integral. In other words, the quantity that we're after is the limit
$$
\lim_{b \to 0^+} \int_b^{\ln(2)} u^{-a}\,du.
$$
In the case that $a \neq 1$, this can be written as
$$
\lim_{b \to 0^+} \int_b^{\ln(2)} u^{-a}\,du = \lim_{b \to 0^+} \left(\frac{\ln(2)^{1-a}}{1-a} - \frac{b^{1-a}}{1-a}\right).
$$
What can we say about this limit when $a &gt; 1$? When $a &lt; 1$? What does the integral become in the case that $a = 1$?
Take $x=e^t$, then
$$I=\int_{1}^{2} \frac{dx}{x \ln^a x}=\int_{0}^{\ln 2} \frac{dt}{t^a}=\frac{1}{1-a}|_{0}^{\ln 2}= \frac{1}{1-a}[(\ln 2)^{1-a}-0^{1-a}]= \text{finite if}~ a&lt;1.$$