>**Proposition 1:** The number of integer solutions of the equation 
$$
\sum_{i=1}^{k}x_i = N
$$
where $x_i\geq n_i$ for $i=1,\ldots,k$, is given by 
$$
\binom{N+k-1-n_1-n_2-...-n_k}{k-1}
$$

Now, to tackle the problem as stated, you need to apply Proposition 1 and invoke the [inclusion-exclusion principle][1], in the following sense:

For $i=1,...,k$, set as 
> $q_i$: the property of a solution of Proposition 1, to satisfy the condition 
$$
x_i> m_i
$$

Then,  if we denote:  
 

 - $N(q_i)$, the number of solutions (provided by Prop. 1) satisfying property $q_i$,  
 - $N(q_i q_j)$, the number of solutions (provided by Prop. 1) satisfying both properties $q_i$, $q_j$, .... etc,  

then we have (again applying Prop. 1) that: 
$$
N(q_1)=\binom{N-1-m_1-n_2-...-n_k}{k-1}, \\ N(q_2 q_3)=\binom{N-1-n_1-m_2-m_3-n_4-...-n_k}{k-1}
$$
etc... 

Now all you need to do to obtain a compact formula for the number of solutions satisfying your constraints, is to apply the inclusion-exclusion principle to determine the number of solutions produced by Proposition 1, **which have none** of the properties $q_i$ for $i=1,2,...,k$. 

**Example:** As an example of application of the previous method, consider the following special case of the OP:  
>Find the number of (positive) integer solutions of the equation
$$\sum_{i=1}^{k}x_i = N$$ for some positive integer $N$, given the constraints $1\leq x_i\leq \alpha$ for $i=1,\ldots,k$

The method described above gives: 
$$
{\small
\binom{N-1}{k-1}+\binom{k}{1}\binom{N-\alpha-1}{k-1}+\binom{k}{2}\binom{N-2\alpha-1}{k-1}+\binom{k}{3}\binom{N-3\alpha-1}{k-1}+\cdots
}
$$
where $\binom{..}{..}$ stands for the [generalized binomial coefficients][2] and the summation halts when zero terms appear. 


  [1]: https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
  [2]: https://en.wikipedia.org/wiki/Binomial_coefficient