Since the infinity sum of sin(x) from x=0 to infinity (sin(0) + sin(1)+ ...) don't converge, we can use the Abel-Plana formula to calculate a renormalized sum of the former divergent sum:

sum(f(t),t=0 to infinity)  =  f(0)/2  + integrate(f(t),t,0,infinity) - i * integrate((f(i*t) - f(-i*t))/(exp(2*pi*t)-1) , t, 0 ,infinity)

 But today I only show the problem,  waiting to someone to try solve this ridle. 

Edit: A minor fix to the result. 

Welcome to Omnimaga indeed

And I am a little confused by your notation.  When you say Sum, do you mean integral?

I'm gonna think about that, it looks inteesting


And btw, you sir deserve 1 internet for making WolframAlpha crazy (first time I actually see something like that on it) :

Wow, this looks absolutely intriguing. I cannot yet arrive to this same conclusion, but I have some math that may be of interest. Please note that I figured this all out on my own (including the Euler-Maclaurin formula. I only just recently learned that it was already discovered). Anyways, this does not at all answer your question, but I think you will appreciate this math, especially if you understand the Riemann Zeta function.

Use the Euler-Maclaurin formula to show that:
sum(sin(n),n,0,x)=sum(sin(n),n,1,x)=integral(sin(x))+sin(x)/2+cos(x)/12--cos(x)/(30*4!)+cos(x)/(42*6!)--cos(x)(30*8!)+...
=-|B₀|cos(x)+|B₁|sin(x)+|B₂|cos(x)/2!+|B₄|cos(x)/4!+|B₆|cos(x)/6!+|B₈|cos(x)/8!+|B₁₀|cos(x)/10!...
(where B_n is the Bernoulli numbers)
Then we have that sum(sin(n),n,1,x)=B₁sin(x)+cos(x)(-B₀+|B₂|/2!+|B₄|/4!+|B₆|/6!+|B₈|/8!+...)=B₁sin(x)-cos(x)+cos(x)sum(|B_2n|/(2n)!,n,1,x)

To be mathematically tricky, I will make that -cos(x) go into the sum, even though all the other terms are positive. How? Note that 2^0=1, but 2^n, where n is an integer greater than 0 is even. Then, (-1)²ⁿ=-1 for n=0 and 1 for all the rest. So:
sum(sin(n),n,1,x)=sin(x)/2+cos(x)sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)

Note that this is a constant: sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)

Now we step completely away from this. How do we write sum(sin(n),n,1,x) in an alternative way? First, look at sum(aⁿ,n,0,x)=a⁰+a¹+a²+a³+...+a^x. If we add a^x+1, we will have the following:
a^x+1+sum(aⁿ,n,0,x)=sum(aⁿ,n,0,x+1)
Reindexing the right side a few times:
a^x+1+sum(aⁿ,n,0,x)=a⁰+sum(aⁿ,n,1,x+1)
a^x+1+sum(aⁿ,n,0,x)=1+sum(aⁿ⁺¹,n,0,x)
Now the sums have the same index, so we can combine them:
a^x+1-1=sum(aⁿ⁺¹,n,0,x)-sum(aⁿ,n,0,x)
a^x+1-1=sum(aⁿ⁺¹-aⁿ,n,0,x)
a^x+1-1=sum(aⁿ(a-1),n,0,x)
a^x+1-1=(a-1)sum(aⁿ,n,0,x)
(a^x+1-1)/(a-1)=sum(aⁿ,n,0,x)


Now how is this useful? Recall that e^bi=cos(b)+isin(b) (if you want, I can give a clever little proof of that using the Maclaurin series of e^x). Then, if we take only the imaginary part of that, we will have sin(b). This means, if we want sin(0)+sin(1)+sin(2)+...+sin(x), we can simply do this:
sum(sin(n),n,0,x)=Im(sum(eⁿⁱ,n,0,x))=Im((e^i(x+1)-1)/(eⁱ-1))

So reindexing, we get:
sum(sin(n),n,1,x)=Im(sum(eⁿⁱ,n,1,x))=Im((e^i(x+1)-1)/(eⁱ-1)-1)
sum(sin(n),n,1,x)=Im(sum(eⁿⁱ,n,1,x))=Im((e^i(x+1)-1)/(eⁱ-1))          (this is because "-1" is not imaginary, so it doesn't matter)

Now we look back
Im((e^i(x+1)-1)/(eⁱ-1))=sin(x)/2+cos(x)sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)
Im((e^i(x+1)-1)/(cos(x)(eⁱ-1)))=tan(x)/2+sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)
Im((e^i(x+1)-1)/(cos(x)(eⁱ-1)))=tan(x)/2+sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)


And all I can get from this is that the constant "sum((-1)²ⁿ|B_2n|/(2n)!,n,0,x)" changes values This is why I love the Bernoulli numbers when manipulating them with alternating functions.

xeda strikes again O.o

This sum doesn't converge, but the average value of a finite sum sum(sin(x),x,0,i) is approximately 0.91524, and 2+(1/2)*cotan(1/2) = 2.91524386... I think that 2+ doesn't really belong.

Also, solving the infinite series the easy way (multiply and cancel) gives cot(1/2)/2 without the 2+:

sin(x) = (e^ix - e^-ix)/2i
sum(sin(x),x,0,inf) = (... - e^-3i - e^-2i - e^-i + 0 + e^i + e^2i + e^3i + ...)/2i
sum(sin(x),x,0,inf)*e^(-i/2) = (... - e^(-5i/2) - e^(-3i/2) + 0 + e^(i/2) + e^(3i/2) + e^(5i/2) + ...)/2i
sum(sin(x),x,0,inf)*e^(i/2)  = (... - e^(-5i/2) - e^(-3i/2) - e^(-i/2) + 0 + e^(3i/2) + e^(5i/2) + ...)/2i
sum(sin(x),x,0,inf)*(e^(-i/2) - e^(i/2)) = (e^(-i/2) + e^(i/2))/2i
sum(sin(x),x,0,inf)*-2i*sin(1/2) = cos(1/2)/i
sum(sin(x),x,0,inf) = cot(1/2)/2

That is pretty cool o.o
@Goplat: That is brilliant o.o

After some review, the sum of sin(x) from x=0 to infinity is really (1/2)*cotan(1/2).