@Carl Love 

thanks very  much for your help>>>>>>>>>

but if i have this loop how i do try and catch on it

for na to 5 do for nb from 0 to 5 do printlevel := 4; timelimit(30, evalf(eval(v1[na, nb], {theta[1] = `λ1_B`[so], theta[2] = `λ2_B`[so], theta[3] = `α_B`[so], theta[4] = `η_B`[so], theta[5] = q_B[so], theta[6] = Px_[so], theta[7] = Py_[so]}))); V1[na, nb] := evalf(eval(v1[na, nb], {theta[1] = `λ1_B`[so], theta[2] = `λ2_B`[so], theta[3] = `α_B`[so], theta[4] = `η_B`[so], theta[5] = q_B[so], theta[6] = Px_[so], theta[7] = Py_[so]})) end do end do

@Kitonum 

but  If the time limit is reached before the expression is evaluated, timelimit generates the error `time expired`. I don't want to get the error. I just want to break a certain loop only

@ 

the summation includes this integration so I named it II and calculated it first. But after evaluating II how can i evaluate the summations

here is my file. 

Download ask.mw

@dharr 

i tried your suggestion and it really helps me a lot thanks a lot for your help

@Adam Ledger 

for your first question why do I assign the values of N_x , n_x,R_x & R_y after you have defined the functions f[1],f[2],f[3] and f[4]?  because this code I uploaded here is just a small part of my whole code . and only this part I have a problem with.

and this is the reason for so many packages that I applied in my worksheet.

@vv 

thanks for your replying sir. but is not alpha =0 is the trivial solution by putting one of the variables equal zero and solve the system of equation on one variable.  is there any replacement to fsolve that help to solve more complex system of equations

@Carl Love Thanks a lot for your helping sir you always  help me in different situations 

thanks a lot for your help it's really working for me@acer 

@acer  what about this sir the comand expand doesnt work here

Download ask.mw
 

@tomleslie 

thanks, a lot, for your help 

but I totally understand what do you mean sir

but mathematically when Z tends to infinity  how integ tends to zero

@vv thanks, a lot, for your help 

@acer 

thanks, a lot, sir  your solution is the most simple one and really works for me 

@Christopher2222 

suppose that 
b:=[1,2,3] i want to put a for loop by saying for first time randomly eliminate 1 item of members a  then for second iteration  randomly eliminate 2 and so on, your comond sir is eliminate the comon members in both sets

@Carl Love 

sir i cant understand this line you give then how i apply it .is there amuch simple way to randomlly remove anumber of n element of alist