While reviewing (as an instructor/test editor) a second semester calculus exam, I came across the following problem:
Find the volume of the solid created by revolving around the $x$-axis the area under the curve $$y = \frac{1}{x\ln x}$$ to the right of the vertical line $x=e$.
This involves calculating the integral $$\pi \int_{e}^{\infty}\frac{1}{x^2(\ln x)^2}dx. $$
As far as I can see, this integral is well beyond an undergraduate calculus course. The best even I can do with it is to say the integral converges (comparison test with the relevant integral of $\frac{1}{x^2}$.)
Am I missing a way to solve this integral?
We have the integral $I$ given by
$$I=\pi\int_e^{\infty} \frac{1}{x^2(\log x)^2}\,dx$$
Enforce the substitution $x\to e^x$. Then we have
$$I=\pi\int_1^{\infty} \frac{e^{-x}}{x^2}\,dx$$
Now, integration by parts reveals
$$\begin{align} I&=\pi\left(-\left.\frac{e^{-x}}{x}\right|_{1}^{\infty}-\int_1^{\infty} \frac{e^{-x}}{x}\,dx\right)\\\\ &=\pi\left(e^{-1}-\int_1^{\infty}\frac{e^{-x}}{x}\,dx\right)\\\\ &=\pi/e+\pi \,\text{Ei}(-1) \end{align}$$
where $\text{Ei}(x)$ is the Exponential Integral
$$\text{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}{t}\,dt$$