Improper integrals with u-substitution $\int\limits_1^\infty \frac{\log x}{x}\,dx$ I am going over improper integrals and I have a question about approaching a certain problem.
The question is the the integral $$\int_1^\infty \frac{\log x}{x}\,dx$$
This integral does not converge, and I am wondering how I go about this.
The most obvious (to me) solution would be to take the limit as $t \to \infty$ of the integral $\int_1^t \frac{\log x}{x}\,dx$
To integrate this expression, the easiest way would be to do $u$-substitution, where $u = \log x$ and $du = (1/x)\,dx$.
The question I have is in regards to switching the bounds to be in terms of $u$. Would the bounds now be $\log(t)$ and $\log(1) = 0$?
 A: You approach is perfectly sound!  Indeed, we can write the improper integral of interest as
$$\int_1^{\infty}\frac{\log x}{x}\,dx=\lim_{L\to \infty}\int_1^L\frac{\log x}{x}\,dx$$
And then, we can make a "$u$" substitution with $u=\log x$ and $du=\frac{1}{x}\,dx$.  The new limits of integration are $0$ and $\log L$.  Therefore, we can write
$$\begin{align}
\int_1^{\infty}\frac{\log x}{x}\,dx&=\lim_{L\to \infty}\int_0^{\log L}u\,du\\\\
&=\lim_{L\to \infty}\left.\left(\frac12 u^2\right)\right|_0^{\log L}\\\\
&=\lim_{L\to \infty}\frac 12\log^2 L\\\\
&=\infty
\end{align}$$
And we are done!
A: $$\int_1^\infty \frac{\log x}{x}\,\text{d}x=$$
$$\lim_{a\to \infty} \int_1^a \frac{\log x}{x}\,\text{d}x=$$

Substitute $u=\ln(x)$ and $\text{d}u=\frac{1}{x}\text{d}x$:

$$\lim_{a\to \infty} \int_1^a x\,\text{d}x=$$
$$\lim_{a\to \infty} \frac{1}{2}\left[u^2\right]_{1}^{a}=$$
$$\lim_{a\to \infty} \left[\frac{\ln^2(x)}{2}\right]_{1}^{a}=$$
$$\lim_{a\to \infty} \left(\frac{\ln^2(a)}{2}-\frac{\ln^2(1)}{2}\right)=$$
$$\lim_{a\to \infty} \left(\frac{\ln^2(a)}{2}-\frac{0}{2}\right)=$$
$$\lim_{a\to \infty} \left(\frac{\ln^2(a)}{2}-0\right)=$$
$$\lim_{a\to \infty} \frac{\ln^2(a)}{2}=$$
$$\frac{1}{2}\lim_{a\to \infty} \ln^2(a)=$$
$$\frac{1}{2}\left(\lim_{a\to \infty} \ln(a)\right)^2=$$
$$\frac{1}{2}\left(\infty\right)^2=\infty$$
A: Yes,
$$\int_{x=1}^t \frac{\log x}{x}\,dx=\int_{u=0}^{\log(t)} u\,du.$$
This is a standard change of variable and there is no singularity.
