I'm having some difficulties solving this improper integral:
$$\int_{2}^{+\infty} \frac{1}{x \log^2{x}} dx.$$
Taking the limit as $b$ approaches infinity we have
$$\lim_{b\to\infty}\int_{2}^{b} \frac{1}{x \log^2{x}} dx.$$
What is the appropriate way to compute this improper integral? Is there a method to simplify the expression of the integrand?
On
As you figured out, we have the following:
$$\lim_{b\to\infty}\int_{2}^{b} \frac{1}{x \log^2{x}} dx$$
We can use $u$-substitution to solve this integral: If $u=\log(x)$, then $\frac{du}{dx}=\frac{1}{x}$ and thus $du=\frac{dx}{x}$. Therefore, the integral becomes the following:
$$\lim_{b\to\infty}\int_{2}^{b} \frac{1}{u^2} du$$
The indefinite integral of $u^{-2}$ is $-u^{-1}$. Since $u=\log(x)$, this means the indefinite integral is $-\frac{1}{\log(x)}+C$. Thus, the definite integral is:
$$\lim_{b\to\infty}(-\frac{1}{\log(b)}-(-\frac{1}{\log(2)}))$$
Now, as $b\to\infty$, $\log(b)\to\infty$, meaning $\frac{1}{\log(b)}\to0$. Thus, we can ignore the first part of our difference since it goes to $0$. Therefore, we have eliminated $b$ from the limit and by cancelling out the double negative, we are simply left with $\frac{1}{\log(2)}$, which is the answer.
On
$$\log ^2 x= (\log x)^2=(\frac{\ln x}{\ln 10})^2=\frac{\ln ^2x}{\ln ^2 10}$$ Substituting this to the integral we have:
\begin{align} \int \frac{1}{x\log ^{2}x} dx = \ln ^2 10 \int \frac{1}{x} \cdot \frac{1} { \ln ^2x} \space dx &= \ln ^2 10 \int \frac{1} { \ln ^2x} \space d(\ln x)=\frac {-\ln ^2 10}{\ln x} +c \\ F(x) &= \frac {-\ln ^2 10}{\ln x} +c \\ \lim_{x \to +\infty} F(x) &= c \\ F(2) &= -\frac {\ln ^2 10}{\ln 2} +c \end{align} So the definite integral is $\frac {\ln ^2 10}{\ln 2} \simeq 7.6 $.
I checked with calculator also.
Hint
I suppose that $\log$ is in fact $\ln$. (If it's not, you can easily adapt my answer by using the fact that $\log x=\frac{\ln x}{\ln 10}$). $$\frac{1}{x\ln^2x}=\frac{u'}{u^2}$$ where $u=\ln(x)$.