MHB Divergent or Convergent Integral

[tmt1]

I have:

$$\int_{1}^{2} \frac{1}{x lnx} \,dx$$

I can set $u = lnx$, therefore $du = \frac{1}{x} dx$ and $xdu = dx$. Plug that into the original equation:

$$\int_{1}^{2} \frac{x}{x u} \,du$$

Or

$$\int_{1}^{2} \frac{1}{ u} \,du$$

Therefore: $ln |u | + C$ and $ln |lnx | + C$

So I need to find the value of this equation where $x = 2$ and where $x = 1$, subtract the latter from the former.

Wolfram says $ln(ln2) = -0.3665$ and $ln(ln1) = - \infty$

so $ln(ln 2) - ln(ln1) = \infty $.

Is this correct?

 

[alyafey22]

Seems correct for me.

 

[MarkFL]

We are given:

$$I=\int_1^2 \frac{1}{x\ln(x)}\,dx$$

As we have division by zero on the left-hand limit, this is an improper integral, and I would write:

$$I=\lim_{t\to1}\left(\int_t^2 \frac{1}{x\ln(x)}\,dx\right)$$

Using the same $u$-substitution, we may write:

$$I=\lim_{t\to1}\left(\int_{\ln(t)}^{\ln(2)} \frac{1}{u}\,du\right)$$

Apply the FTOC:

$$I=\lim_{t\to1}\left(\ln\left(\frac{\ln(2)}{\ln(t)}\right)\right)=\infty$$

I agree, the integral diverges. :D