Q:

A simple yet intriguing integral

Consider indefinite integral of :
$$\begin{align} \int \frac{1}{n \log n} \, dn & = \int \frac{1}{t} \,dt  \, \, \text{;   where $t = \log n$} \\
& = \log t + C \\
& = \log( \log n) + C \\
\end{align}$$
Now if done by parts:
$$\begin{align} \int \frac{1}{n \log n} \, dn & = \frac{\log n}{\log n} - \int \log n \, \frac{d}{dn} \left(\frac{1}{\log n}\right) \, dn \\
& = 1 - \int \frac{\log n}{{\log n}^2} \, (-1) \cdot \left(\frac{1}{n}\right) \, dn \\
& = 1 + \int \frac{1}{n \log n} \, dn \\
\end{align}$$
which is absurb because
$$I = 1 + I$$
$$0 = 1$$
Which is absurd (note that I have not added or ignored any arbitrary constant, just equated I=I.
The same holds true even if the integral be definite from some c1 to some c2
So, where is the fallacy? Is it possibly in the first step of integration by parts? Should an arbitrary constant be introduced in that step? Or is it connected to rinemann summation?

A:

The arbitrary constant is there all the time, it is just a convention not to write it as long as there is still an indefinite integral term present (because an indefinite integral is always determined only up to functions with 0 derivative).
Thus the equality $I=1+I$ is correct, because in both cases $I$ is only determined up to a constant. Of course, in such arithmetic, cancellation is only valid up to a constant which is why the equality $1=0$ (modulo constants) should not appear too shocking.

