About the definition of Day's convolution I'm struggling with the definition of Day's convolution.
Given a monoidal category $(\mathcal C,\otimes, I)$, there is a way to define a monoidal product on the category of presheaves $\widehat{\mathcal C}$. This monoidal product, called the Day's convolution, has nice properties (e.g. it makes $\widehat{\mathcal C}$ a closed monoidal category). The definition of Day's convolution given in nlab (and any other source I can get) is the following: for all presheaves $F,G$ on $\mathcal C$,
$$ F\otimes G := \int^{c,d \in \mathcal C} F(c)\times G(d) \times \hom_{\mathcal C}(-,c\otimes d). $$
My problem is that I can't infer from the (abusive notation of the) coend what is the source category of the diagram. Usually, a coend is the universal dinatural cocone over a diagram $\Delta \colon \mathcal E^{\rm op}\times\mathcal E \to \mathcal F$. Here, obviously $\mathcal F$ should be $\widehat{\mathcal C}$. Bu what should be $\mathcal E$?
I'm tempted to take $\mathcal C \times \mathcal C$ because $(c,d)\mapsto F(c)\times G(d)$ and $(c,d) \mapsto \hom_{\mathcal C}(-,c\otimes d)$ have opposite variances, but then, taking the coend of the pointwise product of those functors, I should get
$$ \int^{c,d,c',d'\in\mathcal C} F(c)\times G(d)\times \hom_{\mathcal C}(-,c'\otimes d'). $$
What am I missing here?
Edit. $\mathcal E$ should be $\mathcal C$. My misunderstanding came from notations. Following the notation of a colimit of a diagram $D \colon \mathcal E \to \mathcal F$ as $ \operatorname{colim}_{e \in \mathcal E} D(e)$, I thought that the exponent of the integral was designating the source category of the diagram, when actually the coend of $D \colon {\mathcal E}^{\rm op} \times \mathcal E \to \mathcal F$ is denoted $\int^{e \in \mathcal E} D(e,e)$.
(A special thanks to Mariano Suárez-Alvarez who has been really patient with me.)
 A: $\def\E{\mathbf E}\def\D{\mathbf D}$It's not counterintuitive at all when you understand how coends work! 
It takes a while but once you understand it, you understand it forever. According to the definition of coend as a universal cowedge, writing $\int^{e, e'}F(e,e')$ means that you are considering a functor $F\colon \E\times \E^{op}\times \E\times \E^{op} \to \D$ which is mute in the third and fourth $\E$-component; taking its coend, then, would give you $\varinjlim_{e,e'}F$, which is generally different from $\int^e F$ (see Exercise 1.17 here).
In the end (!), you're right that it is something similar to a matter of taste in choosing the notation. But several notations are to be preferred to others because they grasp the "right" meaning of the structure we want to describe.
Now, there are several practical reasons why one should prefer denoting a coend as something like $\int^c F(c,c)$. Are you familiar with the Einstein convention in tensor calculus? Whenever ou are summing over repeated indices the sum symbol can be suppressed. Something similar (explain to yourself why! Or read Def 5.3 ibi.) happens with co/ends.
Now for your question: once you understand the basic machinery behind coends, you can effortlessly prove something more general, working out the exercises in Appendix A ibi.
