#OK to post homework
# Edna Jones, Math 640 (Spring 2018), Homework Assignment 6

# 1. get procedure CheckZ5(N) to work by first creating a list of floating-point numbers for (phi(x)-x)/x^2 from 2 to N and then adding them, i.e. replacing integration by summation

#th(x): sum(log(p), primes p <= x)
th := proc(x) local i, su:
    su := 0;
    for i from 1 while ithprime(i) <= x do
        su :=  su + log(ithprime(i));
    od:
    su;
end:

# CheckZ5(N): Checks V in Zagier's paper using summation instead of integration
CheckZ5 := proc(N) local x:
    evalf(add((th(x)-x)/x^2, x=2..N));
end:

# CheckZ5(1000) returns -1.561336322
# CheckZ5(10000) returns -1.615454926
# It appears to grow slowly enough so that the sum might converge.


# 2. It has been read and understood.

# 3. Write a procedure gT(f,t,z,T) that inputs an expression f in a variable t, a complex number z, and a real number T, and outputs the integral of f(t)*exp(-z*t) for t from 0 to T. Experiment with several f and see or which ones limit of gT(0) converges, and whether it goes to g(0).

gT := proc(f,t,z,T)
    evalf(Int(f*exp(-z*t), t=0..T));
end:

# gT(sin(t),t,0,1000) returns 0.4376209237.
# gT(sin(t),t,0,10000) returns 1.952155368.
# int(sin(t),t=0..infinity) returns undefined.
# If f(t) = sin(t), {gT(0)} does not converge.

# gT(exp(-t),t,0,1000) returns 1.000000000.
# gT(exp(-t),t,0,10000) returns 1.000000000.
# evalf(limit(gT(exp(-t),t,0,T),T=infinity)) returns 1.000000000.
# int(exp(-t),t=0..infinity) returns 1, which agrees with evalf(limit(gT(exp(-t),t,0,T),T=infinity)).
# That is, if f(t) = exp(-t), then the limit of gT(0) converges to g(0).

# gT(1/(t+1),t,0,1000) returns 6.908754779.
# gT(1/(t+1),t,0,10000) returns 9.210440367.
# gT(1/(t+1),t,0,100000) returns 11.51293546.
# int(1/(t+1),t=0..infinity) returns infinity.
# If f(t) = 1/(t+1), {gT(0)} does not converge.

# gT(1/(t+1)^2,t,0,1000000) returns 0.9999990000.
# gT(1/(t+1)^2,t,0,1000000000) returns 0.9999999990.
# gT(1/(t+1)^2,t,0,1000000000000) returns 1.000000000.
# evalf(limit(gT(1/(t+1)^2,t,0,T),T=infinity)) returns 1.000000000.
# int(1/(t+1)^2,t=0..infinity) returns 1.
# Thus, if f(t) = 1/(t+1)^2, then the limit of gT(0) converges to g(0).


# 4. It's optional.