4
$\begingroup$

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 $\log(1) = 0$?

$\endgroup$
3
$\begingroup$

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!

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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


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


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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.