Evaluation of $\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)$? I would like to evaluate the sum
$$
\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)
$$
Where $\operatorname{Si}$ is the sine integral, defined as:
$$\operatorname{Si}(x) := \int_0^x \frac{\sin t}{t}\, dt$$
I found that the sum could be also written as
$$
-\sum\limits_{n=0}^\infty \int_n^\infty \frac{\sin t}{t}\, dt
$$
Anyone have any ideas?

We want (changing the sign and starting with $n=1$) :
$$\tag{1}S(0)= -\sum_{n=1}^\infty \left(\mathrm{Si}(n)-\frac{\pi}{2}\right)$$
Let's insert a 'regularization parameter' $\epsilon$ (small positive real $\epsilon$ taken at the limit $\to 0^+$ when needed) :
$$\tag{2} S(\epsilon) = \sum_{n=1}^\infty \int_n^\infty \frac {\sin(x)e^{-\epsilon x}}x\,dx$$
$$= \sum_{n=1}^\infty \int_1^\infty \frac {\sin(nt)e^{-\epsilon nt}}t\,dt$$
$$= \int_1^\infty \sum_{n=1}^\infty \Im\left( \frac {e^{int-\epsilon nt}}t\right)\,dt$$
$$= \int_1^\infty \frac {\Im\left( \sum_{n=1}^\infty e^{int(1+i\epsilon )}\right)}t\,dt$$
(these transformations should be justified...)
$$S(\epsilon)= \int_1^\infty \frac {\Im\left(\dfrac {-e^{it(1+i\epsilon)}}{e^{it(1+i\epsilon)}-1}\right)}t\,dt$$
But 
$$\Im\left(\dfrac {-e^{it(1+i\epsilon)}}{e^{it(1+i\epsilon)}-1}\right)=\Im\left(\dfrac {i\,e^{it(1+i\epsilon)/2}}2\frac{2i}{e^{it(1+i\epsilon)/2}-e^{-it(1+i\epsilon)/2}}\right)$$
Taking the limit $\epsilon \to 0^+$ we get GEdgar's expression :
$$\frac {\cos(t/2)}{2\sin(t/2)}=\frac {\cot\left(\frac t2\right)}2$$
To make sense of the (multiple poles) integral obtained :
$$\tag{3}S(0)=\int_1^\infty \frac{\cot\left(\frac t2\right)}{2t}\,dt$$
let's use the cot expansion applied to $z=\frac t{2\pi}$  :
$$\frac 1{2t}\cot\left(\frac t2\right)=\frac 1{2\pi t}\left[\frac {2\pi}t-\sum_{k=1}^\infty\frac t{\pi\left(k^2-\left(\frac t{2\pi}\right)^2\right)}\right]$$
$$\frac 1{2t}\cot\left(\frac t2\right)=\frac 1{t^2}-\sum_{k=1}^\infty\frac 2{(2\pi k)^2-t^2}$$
Integrating from $1$ to $\infty$ the term $\frac 1{t^2}$ and all the terms of the series considered as Cauchy Principal values $\ \displaystyle P.V. \int_1^\infty \frac 2{(2\pi k)^2-t^2} dt\ $ we get :
$$\tag{4}S(0)=1+\sum_{k=1}^\infty\frac {\mathrm{atanh}\bigl(\frac 1{2\pi k}\bigr)}{\pi k}$$
and the result :
$$\tag{5}\boxed{\displaystyle\sum_{n=0}^\infty \left(\mathrm{Si}(n)-\frac{\pi}{2}\right)=-1-\frac{\pi}4-\sum_{k=1}^\infty\frac {\mathrm{atanh}\bigl(\frac 1{2\pi k}\bigr)}{\pi k}}$$$$\approx -1.8692011939218853347728379$$
(and I don't know why the $\frac {\pi}2$ term re-inserted from the case $n=0$ became a $\frac {\pi}4$ i.e. the awaited answer was $-S(0)-\frac{\pi}2$ !) 
Let's try to rewrite this result using the expansion of the $\mathrm{atanh}$ :
$$\mathrm{atanh(x)}=\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}$$
so that 
$$A=\sum_{k=1}^\infty\frac {\mathrm{atanh}\bigl(\frac 1{2\pi k}\bigr)}{\pi k}=\sum_{k=1}^\infty \sum_{n=0}^\infty \frac 1{\pi k(2\pi k)^{2n+1}(2n+1)}$$
$$=\sum_{n=0}^\infty \frac 2{(2n+1)(2\pi)^{2n+2}}\sum_{k=1}^\infty \frac 1{ k^{2n+2}}$$
$$=2\sum_{n=1}^\infty \frac {\zeta(2n)}{2n-1}a^{2n}\quad \text{with}\ \  a=\frac 1{2\pi} $$
$$\tag{6}\boxed{\displaystyle\sum_{n=0}^\infty \left(\mathrm{Si}(n)-\frac{\pi}{2}\right)=-1-\frac{\pi}4-2\sum_{n=1}^\infty \frac {\zeta(2n)}{(2n-1)(2\pi)^{2n}}}$$
and... we are back to the cotangent function again since it is a generating function for even $\zeta$ constants !
$$1-z\,\cot(z)=2\sum_{n=1}^\infty \zeta(2n)\left(\frac z{\pi}\right)^{2n}$$
Here we see directly that
$$A=\frac 12\int_0^{\frac 12} \frac {1-z\,\cot(z)}{z^2} dz$$ 
with the integral result :
$$\tag{7}\boxed{\displaystyle\sum_{n=0}^\infty \left(\mathrm{Si}(n)-\frac{\pi}{2}\right)=-1-\frac{\pi}4-\int_0^1 \frac 1{t^2}-\frac {\cot\left(\frac t2\right)}{2t} dt}$$
(this shows that there was probably a more direct way to make (3) converge but all journeys are interesting !)
