Recently there were some discussions about divergent series which proved to be interesting. Here's another example which gives rise to an interesting representation of the harmonic number which at least I haven't seen before.
We start from the well-known asymptotic series
$$H_{n}\simeq \log (n)+\gamma +\frac{1}{2 n}-\sum _{k=1}^{\infty } \frac{B_{2 k}}{2 k n^{2 k}}\tag{1}$$
where $B_m$ are the Bernoulli numbers.
The sum is know to be (strongly) divergent but we shall nervertheless attempt to sum it.
The trick is to insert the integral representation of the Bernoulli numbers
$$B_{2 k}=-2 (-1)^k \int_0^{\infty } \frac{2 k t^{2 k-1}}{e^{2 \pi t}-1} \, dt\tag{2}$$
and do the sum under the integral:
$$-\sum _{k=1}^{\infty } \frac{2\ 2 (-1)^k k t^{2 k-1}}{2 k n^{2 k}}=\frac{2 t}{n^2+t^2}\tag{3}$$
Hence we get a tentative representation of the harmonic number in the form
$$\boxed{H_{n}=\log (n)+\gamma+\frac{1}{2 n}-\int_0^{\infty } \frac{2 t}{\left(e^{2 \pi t}-1\right) \left(n^2+t^2\right)} \, dt}\tag{4}$$
The integral is obviously convergent, and to my great surprise numerical tests and a plot showed that the equality sign turned out to be justified for real $n\gt0$.
Question
In the first place I had the question if this derivation was valid. But I believe now that the result shows that it was valid as a heuristic procedure. So what remains is the question how $(4)$ can be proved directly.
The formula $(4)$ is standard.
Browsing through Gradshteyn Ryshik for another purpose I found "my" formula listed as #6.3.21.
Still it was a nice exercise in giving a meaning to a divergent sum.