Convergence of definite integrals given unknowns.

Question

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 $a>1$ or $a<1$

Answer 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 $a > 1$? When $a < 1$? What does the integral become in the case that $a = 1$?

Answer 2

Take $x=e^t$, then $$I=\int_{1}^{2} \frac{dx}{x \ln^a x}=\int_{0}^{\ln 2} \frac{dt}{t^a}=\frac{1}{1-a}|_{0}^{\ln 2}= \frac{1}{1-a}[(\ln 2)^{1-a}-0^{1-a}]= \text{finite if}~ a<1.$$

Answer 3

We have that

$$\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx=\int_{0}^{1} \frac{1}{(1+x)(\ln (1+x))^{a}} dx$$

with $(\ln (1+x))^{a} \sim x^a $ as $x \to 0^+$ and therefore

$$\frac{1}{(1+x)(\ln (1+x))^{a}} \sim \frac1{x^a}$$

and we can conclude by limit comparison test.