I am trying to calculate the Abel summation formula for a function of the form
$$
\sum _{j=1}^x \sum _{i=1}^x \phi \left( x-i j \right)
$$
where the function $\phi$ meets the requirements for Abel's formulation. The summation formula states that
$$
\begin{align}
\sum _{i=1}^x \phi (i)
&= \lfloor x\rfloor \phi (x) -\int_1^x \lfloor u\rfloor \phi' (u) \mathrm{d}u \\
&= \lfloor x\rfloor \phi (x) -\int_1^x \lfloor u\rfloor \frac {\delta}{\delta u} \phi (u) \mathrm{d}u
\end{align}
$$
(I have included that last line to clarify the basis of differentiation.)
The first summation, indexed on $i$, seems fine. Set $\psi (i) = \phi (x-i j)$. Then
$$
\begin{align}
\sum _{i=1}^x \phi (x-i j)
&= \sum _{i=1}^x \psi (i) \\
&= \lfloor x\rfloor \psi (x) -\int_1^x \lfloor u\rfloor \frac {\delta}{\delta u} \psi (u) \mathrm{d}u \\
&= \lfloor x\rfloor \phi \left( x (1-j) \right) -\int_1^x \lfloor u\rfloor \frac {\delta}{\delta u} \phi \left( x-u j \right) \mathrm{d}u
\end{align}
$$
As a simple test, define $\phi (x) := x^3$. Tabulating the results in Mathematica suggests the formula is correct (the Mathematica code is there in case the later error is in my coding – in which case, I will re-ask my question on the appropriate forum.):
Obviously, the fact that this particular example works is not proof that the formula is correct, but at least it's reassuring. However, I run into trouble on the second summation, indexed over $j$.
Set $\mu (j) = \sum _{i=1}^x \phi (x-i j) = \sum _{i=1}^x \psi (i)$ which we have just formulated above. Then
$$
\begin{align}
\sum _{j=1}^x \sum _{i=1}^x \phi (x-i j)
&= \sum _{j=1}^x \mu (j) \\
&= \lfloor x\rfloor \mu (x) -\int_1^x \lfloor u\rfloor \frac {\delta}{\delta u} \mu (u) \mathrm{d}u \\
&= \lfloor x\rfloor \left( \lfloor x\rfloor \phi \left( x (1-x) \right) -\int_1^x \lfloor u\rfloor \frac {\delta}{\delta u} \phi \left( x (1-u) \right) \mathrm{d}u \right) -\int_1^x \lfloor u\rfloor \frac {\delta}{\delta u} \left( \lfloor x\rfloor \phi \left( x (1-u) \right) -\int_1^x \lfloor u\rfloor \frac {\delta}{\delta u} \phi \left( x-u^2 \right) \mathrm{d}u \right) \mathrm{d}u
\end{align}
$$
However, a simple test proves this wrong. Using $\phi (x) := x^3$ again, Mathematica produces the following:
So. Something is clearly wrong… But what?
Best Answer
I believe the error is in your last line. The last term should have the two $u$'s labeled differently, so it should read $$ -\int_1^x \lfloor u\rfloor \frac\delta{\delta u}\big(\lfloor x\rfloor \phi(x(1-u)) - \int_1^x \lfloor w\rfloor \frac\delta{\delta w} \phi(x-uw)\,dw\big)\,du. $$ Please let me know if this works.