I was attempting a problem, when I was met with a confusion regarding the limits of integration. The problem was the following

$$ \int_{1}^{2} x^{2} \mathrm{d}(\ln x) $$

To clarify any confusion, $\ln x$ is inside the differential. My first intuition was to write $\mathrm{d}(\ln x)$ as $\frac{1}{x}\mathrm{d}x$, and proceed with the integral as usual to yield $\frac{3}{2}$ as the answer. However the answer given is $\frac{e^4 - e^2}{2}$.

I thought the limits of the function then would represent values of $\ln x$ from $1$ to $2$ and not the values of $x$ and that apparently yields the correct answer. But upon researching to confirm this, I found on Wikipedia Rienmann-Stieltjes Integral, where they imply

$$ \int_{a}^{b} \mathrm{f}(x) \mathrm{d}\mathrm{g}(x) = \int_{a}^{b} \mathrm{f}(x) \mathrm{g}^{\prime}(x)\mathrm{d}(x) $$

When $f(x)$ is bounded and $g(x)$ is increasing, continuous & differentiable in $[a, b]$. The same can be found out in Apostol's Mathematical Analysis Ch7 Sec1.

From this it appears my answer is correct. So is the book wrong and the limits do represent values of $x$ from $1$ to $2$ or is there something wrong? Thanks.

  • $\begingroup$ $g$ increasing and continuous does not gives possibility write $g'$. $\endgroup$
    – zkutch
    Jun 17, 2023 at 13:05
  • $\begingroup$ @zkutch I have edited the question $\endgroup$ Jun 17, 2023 at 13:10
  • $\begingroup$ @zkutch A monotonic function is differentiable almost everywhere (for the Lebesgue measure), so $g'$ would be defined in regards to the integral, but you are right to worry about it in the sense that even Wikipedia says this: "It may be the case that $g$ has jump discontinuities, or may have derivative zero almost everywhere while still being continuous and increasing (for example, $g$ could be the Cantor function or “Devil's staircase”), in either of which cases the Riemann–Stieltjes integral is not captured by any expression involving derivatives of $g$." $\endgroup$
    – Bruno B
    Jun 17, 2023 at 13:14
  • 1
    $\begingroup$ Regarding your question, this post math.stackexchange.com/questions/3847107/… answers it really well (with different limits of integration though). $\endgroup$
    – stange
    Jun 17, 2023 at 13:16
  • 2
    $\begingroup$ You are right: $\int_1^2 x^2 \mathrm{d} \ln x = \int_1^2 x^2 \frac{1}{x} \mathrm{d} x = \frac{3}{2}$. $\endgroup$
    – jjagmath
    Jun 17, 2023 at 13:16

1 Answer 1


First of all, this notation is awful and is evidently prone to misunderstanding. Here, they mean that $$\int_{\ln(t) = 1}^{\ln(t)=2} t^2 d(\ln(t)) = \int_{e}^{e^2} t^2 \frac{1}{t} dt = \frac{e^4-e^2}{2}.$$ But this probably originated from the variable change $x = \ln(t)$ in the integral $$\int_1^2 (e^x)^2 dx,$$ which is obviously a stupid way to calculate this integral.

In summary, never do what they did here! Always be clear, so try to avoid notations like $d(g(x)),$ which is for physicists.

  • $\begingroup$ So it means my answer is correct given the information given in the question right? $\endgroup$ Jun 17, 2023 at 13:17
  • 2
    $\begingroup$ Yes, the problem originates from the notation chosen in the book, not from you. $\endgroup$ Jun 17, 2023 at 13:19
  • 2
    $\begingroup$ Btw a slight mistake from your side, not physicians! They are physicists! $\endgroup$ Jun 17, 2023 at 13:22
  • $\begingroup$ Yes sorry physicians are paramedics right? As you can tell english is not my mother tongue $\endgroup$ Jun 17, 2023 at 13:23
  • $\begingroup$ You are right, mine neither. $\endgroup$ Jun 17, 2023 at 13:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .