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To begin with I have a series of interest, $\sum_n f_n$ that diverges everywhere $f_n: \mathbb{R}^2 \rightarrow \mathbb{R}$. However, $\int_z \sum_n \partial_z f_n$ has a nice closed form, lets call it $f$

(note to obtain this $f$ a change to complex coordinates was made and then the derivative and integration were with respect to $z$, I'm just saying this explicitly because below I state $\sum \partial_x f_n$ was not something I could find in a closed form and this could seem contradictory to the above)

Of course, the exchange of partial derivatives, the integral and the sum are only "legit" as I understand it if the original sum converged for at least one point and the sum of partial derivatives was uniformly convergent. These conditions are not met in this case.

However, ultimately I am only interested in the expressions $\sum \partial_{x} f_n ,\sum \partial_{y} f_n, \sum \partial_{xy} f_n $ and $\sum \partial_{yy} f_n$. Moreover, I believe these expressions are at least locally uniformly convergent (and definitely converge for any fixed point unlike the original sum above).

My main problem is that I can't find a closed form for these sums in a straight forward manner. However, numerically I can check that indeed it seems that $f_x = \sum \partial_{x} f_n$ etc. That is, summing the first few million terms seems to check out.

As of now my approach was just to hand wave that yes $f$ doesn't really make sense, but it turns out that it is sufficient to then describe the necessary derivatives. This is rather unsatisfying and not rigorous. Is this situation common? Are there things I could cite giving weight to this approach or someway to make it more rigorous?

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  • $\begingroup$ Do you have an exact form for $f_n$? $\endgroup$ – Alex R. Apr 2 '15 at 18:19
  • $\begingroup$ Yes, it's $\log( (x-nL)^2 + y^2 + \delta^2)$ where L and $\delta$ are constants $\endgroup$ – Fractal20 Apr 2 '15 at 18:47
  • $\begingroup$ Ok so is your question to find a closed form for the sum of the derivative of your $f$? $\endgroup$ – Alex R. Apr 2 '15 at 19:01
  • $\begingroup$ It is but the details of my actual problem are rather long. For example the $f_x$ terms are $\sim 1/n$ so they diverge as well. But in the actual context $f_x$ shows up as a difference that results in $\sim 1/n^2$. But in each case there are details that make it more complicated in the actual situation I have. That is why I'm hoping to avoid wading into all those nasty little details and in someway just use the strange expression for f $\endgroup$ – Fractal20 Apr 2 '15 at 19:11

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