End of functor to product category is product of ends $\require{AMScd}$
This is exercise E11 in Fosco Loregian's This is the (co)end, my only (co)friend.

Suppose we have a functor $F:\mathcal{C}^{op}\times\mathcal{C}\to \prod_{i=1}^n \mathcal{A}_i$, and suppose that all ends $\int_cF_i(c,c)$ exist, with $F_i = \pi_i\circ F$.
Denote by $\mathcal{L}$ the product of these "partial" ends, i.e. $\mathcal{L} = \left(\int_cF_1,...,\int_cF_n\right)$ (the arguments are dropped for readability reasons).
Now let $\eta:X\xrightarrow{..}F$ be a dinatural transformation for some object $X$ in the target category. That is, we have the following commutative diagram for all $c,c'\in \mathcal{C}$  and all $f:c\to c'$:
\begin{CD}
X @>\eta_c>> F(c,c) \\
@V\eta_{c\prime}VV @VF(1,f)VV\\
F(c',c') @>F(f,1)>>F(c,c')
\end{CD}
Then we get for each $i$ a commutative diagram
\begin{CD}
X @>\eta^i_c\ =\ \pi_i\circ \eta_c>> F_i(c,c) \\
@V\eta^i_{c\prime}VV @VF_i(1,f)VV\\
F_i(c',c') @>F_i(f,1)\ =\ \pi_i\circ F(f,1)>>F_i(c,c')
\end{CD}
Since the end of $F_i$ exists, we obtain a unique morphism $m_i:X\to \int_cF_i$ s.t. $\eta^i$ factors through $\int_c F_i$ via $m_i$. But because we are looking at the product of the ends, we automatically get a unique morphism $m:X\to \mathcal{L}$ s.t. $m_i = \pi_i\circ m$.
The problem: In the exercise, part of the data we are given is that each $\mathcal{A}_i$ possess both an initial an a terminal object.
The question: Why the extra data? Why is what I did not sufficient to show that $\mathcal{L}$ is the end? 
 A: You don't need those conditions. Here's a proof for $n=2$ for ends: $$\begin{align}
&\mathcal A\times\mathcal B((A,B),\int_{C:\mathcal C}F(C,C)) \\
\cong\ & \int_{C:\mathcal C}\mathcal A\times\mathcal B((A,B),F(C,C)) \\
\cong\ & \int_{C:\mathcal C}\mathcal A(A,\pi_1F(C,C))\times\mathcal B(B,\pi_2F(C,C)) \\
\cong\ & \left(\int_{C:\mathcal C}\mathcal A(A,\pi_1F(C,C))\right)\times\left(\int_{C:\mathcal C}\mathcal B(B,\pi_2F(C,C))\right) \\
\cong\ & \mathcal A(A,\int_{C:\mathcal C}\pi_1F(C,C))\times\mathcal B(B,\int_{C:\mathcal C}\pi_2F(C,C)) \\
\cong\ & \mathcal A\times B((A,B),\left(\int_{C:\mathcal C}\pi_1F(C,C),\int_{C:\mathcal C}\pi_2F(C,C)\right)) \\
\end{align}$$
and for coends:
$$\begin{align}
&\mathcal A\times\mathcal B(\int^{C:\mathcal C}F(C,C),(A,B)) \\
\cong\ & \int_{C:\mathcal C}\mathcal A\times\mathcal B(F(C,C),(A,B)) \\
\cong\ & \int_{C:\mathcal C}\mathcal A(\pi_1F(C,C),A)\times\mathcal B(\pi_2F(C,C),B) \\
\cong\ & \left(\int_{C:\mathcal C}\mathcal A(\pi_1F(C,C),A)\right)\times\left(\int_{C:\mathcal C}\mathcal B(\pi_2F(C,C),B)\right) \\
\cong\ & \mathcal A(\int^{C:\mathcal C}\pi_1F(C,C),A)\times\mathcal B(\int^{C:\mathcal C}\pi_2F(C,C),B) \\
\cong\ & \mathcal A\times B(\left(\int^{C:\mathcal C}\pi_1F(C,C),\int^{C:\mathcal C}\pi_2F(C,C)\right),(A,B)) \\
\end{align}$$
Here's another proof. If $\pi_i$ is continuous and cocontinuous then by Theorem 1.16 $\pi_i\int_{C:\mathcal C}F(C,C)\cong\int_{C:\mathcal C}\pi_i F(C,C)$ and $\pi_i\int^{C:\mathcal C}F(C,C)\cong\int^{C:\mathcal C}\pi_i F(C,C)$. We have $\int_{C:\mathcal C}F(C,C) = (\pi_1\int_{C:\mathcal C}F(C,C),\pi_2\int_{C:\mathcal C}F(C,C))$ from which the result follows for ends, and similarly for coends.
Now where the existence of initial and terminal objects would come in handy is if you wanted to establish that $\pi_i$ was (co)continuous by showing that it had a (right/)left adjoint. Let's see what we'd need for $\pi_1:\mathcal A\times\mathcal B\to\mathcal A$ to have a left adjoint. $$\begin{align}
&\mathcal A(A,\pi_1 (X, Y)) \\
=\ & \mathcal A(A, X) \\
\cong\ & \mathcal A(A, X)\times 1 \\
\cong\ & \mathcal A(A, X)\times\mathcal B(?,Y) \\
\cong\ & \mathcal A\times B(LA, (X, Y))
\end{align}$$
If only there were an object of $\mathcal B$ to fill in that question mark so that $\mathcal B(?,Y)\cong 1$ for all $Y$. Oh wait, this is exactly the universal property of the initial object. $LA=(A,0)$ assuming $0$ is an initial object in $\mathcal B$. For the right adjoint, you need an object such that $\mathcal B(Y,?)\cong 1$ which is the universal property of the terminal object.
But we don't need $\pi_i$ to be (co)continuous, let alone these adjoints to exist, if we already know the (co)ends of the components exist. On the other hand, knowing the $\pi_i$ are (co)continuous allows us to show that the (co)ends of the components exist if the (co)end in the product category exists.
