Improper integrals with u-substitution $\int\limits_1^\infty \frac{\log x}{x}\,dx$

Question

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 $\log(1) = 0$?

Answer 1

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!

Answer 2

$$\int_1^\infty \frac{\log x}{x}\,\text{d}x=$$ $$\lim_{a\to \infty} \int_1^a \frac{\log x}{x}\,\text{d}x=$$

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

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$$\lim_{a\to \infty} \int_1^a x\,\text{d}x=$$ $$\lim_{a\to \infty} \frac{1}{2}\left[u^2\right]_{1}^{a}=$$ $$\lim_{a\to \infty} \left[\frac{\ln^2(x)}{2}\right]_{1}^{a}=$$ $$\lim_{a\to \infty} \left(\frac{\ln^2(a)}{2}-\frac{\ln^2(1)}{2}\right)=$$ $$\lim_{a\to \infty} \left(\frac{\ln^2(a)}{2}-\frac{0}{2}\right)=$$ $$\lim_{a\to \infty} \left(\frac{\ln^2(a)}{2}-0\right)=$$ $$\lim_{a\to \infty} \frac{\ln^2(a)}{2}=$$ $$\frac{1}{2}\lim_{a\to \infty} \ln^2(a)=$$ $$\frac{1}{2}\left(\lim_{a\to \infty} \ln(a)\right)^2=$$ $$\frac{1}{2}\left(\infty\right)^2=\infty$$

Answer 3

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.