## 6 Examples  
In this section, we obtain some results about the generalized distance Estrada index of some typical graphs. This would be instrumental in interpreting this measure further in the case of more complex graphs. We also give an expression for \(D_{\alpha}E(G)\) of a (transmission) regular graph \(G\), in terms of the distance eigenvalues as well as adjacency eigenvalues of \(G\), and describe the generalized distance Estrada index of some graphs obtained by operations. As mentioned in introduction of the paper, for \(\alpha=0\) the generalized distance matrix \(D_{\alpha}(G)\) is equivalent to the distance matrix \(D(G)\) and for \(\alpha=\frac{1}{2}\), twice the generalized distance matrix \(D_{\alpha}(G)\) is the same as the distance signless Laplacian matrix \(D^{\mathcal{Q}}(G)\). Therefore, if in particular we put \(\alpha=0\) and \(\alpha=\frac{1}{2}\) in all the results obtained in this paper, we obtain the corresponding bounds for the distance Estrada index \(DEE(G)\) and the distance signless Laplacian Estrada index \(D^{\mathcal{Q}}_{E}E(G)\), respectively. **Theorem 11**. : _Suppose that \(G\) is a \(k\)-transmission regular graph of order \(n\) having distance eigenvalues \(\mu_{1},\mu_{2},\ldots,\mu_{n}\). Then,_  
\[D_{\alpha}E(G)=\sum_{i=1}^{n}e^{(1-\alpha)\mu_{i}}.\]  
**Proof. ** Note that the generalized distance spectrum of the graph \(G\) consists of \(\alpha k+(1-\alpha)\mu_{1}\geq\alpha k+(1-\alpha)\mu_{2}\geq\cdots\geq\alpha k +(1-\alpha)\mu_{n}\), where \(\mu_{1}\geq\cdots\geq\mu_{n}\) is the distance spectrum of \(G\). Also, it is easy to see that \(W(G)=\frac{nk}{2}\). Then, \(D_{\alpha}E(G)=\sum_{i=1}^{n}e^{\alpha k+(1-\alpha)\mu_{i}-ak}=DEE(G)=\sum_{i=1 }^{n}e^{(1-\alpha)\mu_{i}}\). \(\square\)  
**Theorem 12**. : _Let \(G\) be an \(r\)-regular graph of order \(n\), size \(m\) and diameter at most 2. If \(\{r,\lambda_{2},\ldots,\lambda_{n}\}\) are the eigenvalues of the adjacency matrix \(A(G)\) of \(G\), then_  
\[D_{\alpha}E(G)=e^{n(2n-r-2)-2\alpha(n^{2}-n-m)}+\sum_{i=2}^{n}e^{\alpha(n \lambda_{i}+rn+2n+2m)-n(\lambda_{i}+2)}.\]  
**Proof. ** The transmission of each vertex \(v\in V(G)\) can be computed as \(\mbox{Tr}(v)=d(v)+2(n-d(v)-1)=2n-d(v)-2\) and the Wiener index \(W(G)\) of \(G\) becomes \(W(G)=n^{2}-n-m\). Also,  
\[D_{\alpha}(G)=\alpha\mbox{Tr}(G)+(1-\alpha)D(G) = \alpha(2n-r-2)I+(1-\alpha)(2J-2I-A(G))\] \[= \alpha((2n-r-2)I-2J+2I+A(G))+2J-2I-A(G),\]  
where \(J\) is the all ones matrix. Then,  
\[D_{\alpha}E(G) = \sum_{i=1}^{n}e^{\partial_{i}-\frac{2\alpha W(G)}{n}}=e^{(2n-r-2 )-\frac{2\alpha(n^{2}-n-m)}{n}}+\sum_{i=2}^{n}e^{(\alpha(2n+\lambda_{i}-r)- \lambda_{i}-2)-\frac{2\alpha(n^{2}-n-m)}{n}}\] \[= e^{n(2n-r-2)-2\alpha(n^{2}-n-m)}+\sum_{i=2}^{n}e^{\alpha(n \lambda_{i}+rn+2n+2m)-n(\lambda_{i}+2)}.\]  
\(\Box\)  
We denote by \(G\times H\) the _cartesian product_ of two graphs \(G\) and \(H\). It is the graph with vertex set \(V(G)\times V(H)\) and two vertices \((u_{1},u_{2})\) and \((v_{1},v_{2})\) are adjacent if and only if \(u_{1}=v_{1}\) and \(u_{2}v_{2}\in E(H)\) or \(u_{2}=v_{2}\) and \(u_{1}v_{1}\in E(G)\). **Corollary 4**. : _Let \(G\) be an \(r\)-regular graph of diameter at most 2 with an adjacency matrix \(A\) and \(\mbox{spec}(G)=\{r,\lambda_{2},\ldots,\lambda_{n}\}\). Then, the generalized distance Estrada index of \(H=G\times K_{2}\) is_  
\[D_{\alpha}E(H)=n-1+e^{(1-\alpha)(5n-2r-4)}+e^{(\alpha-1)n}+\sum_{i=2}^{n}e^{( \alpha-1)(2\lambda_{i}+4)}.\]  
**Proof. ** Let \(V(G)=\{v_{1},v_{2},\ldots,v_{n}\}\), \(V(K_{2})=\{w_{1},w_{2}\}\). As  
\(d_{H}((v_{i},w_{j}),(v_{s},w_{t}))=d_{G}(v_{i},v_{s})+d_{K_{2}}(w_{j},w_{t})=d _{G}(v_{i},v_{s})+1\), we see that all vertices of \(H\) have the same transmission and \(\mbox{Tr}_{H}(v_{i},w_{j})=5n-2r-4\). So \(\mbox{Tr}(H)=(5n-2r-4)I\). Then \(W(H)=\frac{n(5n-2r-4)}{2}\). Note that \(H=G\times K_{2}\) has distance spectrum (see in [37]) \(\mbox{spec}(H)=\left\{5n-2(r+2),-2(\lambda_{i}+2),-n,0^{[n-1]}\right\}\), for \(i=2,\ldots,n\).