## 4 Practical Examples  
### Example I  
During studies about wall shear stress occurring in pipe walls when an unsteady pipe flow takes place, an interesting convolution integral appears during the transformations needed to obtain the final time domain solution:  
\[\frac{8\
u\upsilon_{0}\mu}{R^{3}}\int_{0}^{t}\sum_{i=1}^{\infty}e^{-\frac{\
u }{R^{2}}\eta_{i}^{2}(t-u)}{\sum_{j=1}^{\infty}e^{-\frac{\
u}{R^{2}}\xi_{j}^{2 }u}}du \tag{43}\]  
where \(\xi_{j}\) is the root of the Bessel function of zero order \(J_{0}\big{(}\xi_{j}\big{)}=0\) and \(\eta_{j}\) is the root of the Bessel function of the second order \(J_{2}\big{(}\eta_{j}\big{)}=0\). The above convolution defined by Equation (43) shows up as a result of the shear stress (in time domain) at the valve when a second possible analytical solution (Equation (28) from [PERSON]'s thesis [23]) is used for modelling the wall shear stress (nowadays, the commonly used solution is the one defined by Equation 35 in [PERSON]'s thesis [23]).  
This Cauchy product of two infinite exponential generalized Dirichlet's series Equation (43) has an analytical solution in the following form:  
\[8\frac{\upsilon_{0}\mu}{R}{\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{e^{- \frac{\
u}{R^{2}}\xi_{j}^{2}}-e^{-\frac{\
u}{R^{2}}\eta_{j}^{2}t}}} \tag{44}\]  
The analytical solution discussed in this paper facilitates the simplification of the aforementioned double sum into more manageable single sums. The transformation is shown step by step below. First, let us write our solution in the form of two double series:  
\[8\frac{\upsilon_{0}\mu}{R}\bigg{(}\sum_{j=1}^{\infty}e^{-\frac{\
u}{R^{2}} \xi_{j}^{2}}\sum_{i=1}^{\infty}\frac{1}{\eta_{i}^{2}-\xi_{j}^{2}}-\sum_{i=1}^ {\infty}e^{-\frac{\
u}{R^{2}}\eta_{j}^{2}t}{\sum_{j=1}^{\infty}\frac{1}{\eta _{i}^{2}-\xi_{j}^{2}}}\bigg{)}=8\frac{\upsilon_{0}\mu}{R}\bigg{(}\sum_{j=1}  
Compared with the solutions presented in Appendix A of this paper (A10 and A13), it can be seen that  
\[\left\{\begin{array}{c}\sum_{i=1}^{\infty}\frac{1}{\eta_{i}^{2}-\xi_{j}^{2}} \equiv\sum_{k}^{\infty}1\frac{1}{\beta_{2,k}^{2}-\beta_{0,n}^{2}}=\frac{2}{\beta _{0,n}^{2}}-\frac{1}{4(2-1)}\ where:\ p=2\\ \sum_{j=1}^{\infty}\frac{1}{\xi_{j}^{2}-\eta_{i}^{2}}\equiv\sum_{k}^{\infty}1 \frac{1}{\beta_{0,k}^{2}-\beta_{2,n}^{2}}=\frac{1}{4(1+0)}\ where:\ p=0\end{array}\right. \tag{46}\]  
So,  
\[\left\{\begin{array}{c}\sum_{i=1}^{\infty}\frac{1}{\eta_{i}^{2}-\xi_{j}^{2}} =\frac{2}{\xi_{j}^{2}}-\frac{1}{4}\ where:\ p=2\\ \sum_{j=1}^{\infty}\frac{1}{\xi_{j}^{2}-\eta_{i}^{2}}=\frac{1}{4}\ where:\ p=0 \end{array}\right. \tag{47}\]  
Thus, Equation (45) can be written with the help of Equation (47)  
\[8\frac{v_{0}\mu}{R}\left(\sum_{j=1}^{\infty}e^{-\frac{x}{R^{2}}t\xi_{j}^{2}} \left(\frac{2}{\xi_{j}^{2}}-\frac{1}{4}\right)+\frac{1}{4}\sum_{i=1}^{\infty} e^{-\frac{x}{R^{2}}\eta_{i}^{2}t}\right)=8\frac{v_{0}\mu}{R}\left(2\sum_{j=1}^{ \infty}\frac{e^{-\frac{x}{R^{2}}t\xi_{j}^{2}}}{\xi_{j}^{2}}-\frac{1}{4}\sum_{ j=1}^{\infty}e^{-\frac{x}{R^{2}}t\xi_{j}^{2}}+\frac{1}{4}\sum_{i=1}^{\infty}e^{- \frac{x}{R^{2}}\eta_{i}^{2}t}\right) \tag{48}\]  
The convolutional integral solution Equation (48) played a crucial role in demonstrating that the analytical solution for the unsteady wall shear stress at the downstream end valve section of a single pipe, induced by a rapid (instantaneous) valve closure, ultimately simplifies to  
\[\tau_{valve}=\frac{2v_{0}\mu}{R}\sum_{i=1}^{\infty}e^{-\frac{x}{R^{2}}\eta_{i} ^{2}t} \tag{49}\]  
A similar solution was recently derived [24] in a slightly simpler way.