# 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?

• I get $-1.93354$. – Fabian Aug 18 '12 at 21:13
• Where did this come up? I doubt it has any closed form. – Eric Naslund Aug 18 '12 at 23:41
• @EricNaslund I came up with this sum as it seemed quite natural, because $$\operatorname{Si}(x)\sim\frac{\pi}{2}$$ – Argon Aug 19 '12 at 0:04
• Since $$\mathrm{Si}(x) - \frac{\pi}{2} \sim -\frac{\cos x}{x} + O\left(\frac{1}{x^2}\right),$$ the sum converges conditionally. I'm testing whether the technique for evaluating $$\sum_{n=1}^{\infty} \frac{\mathrm{Si}(2n\pi)}{n} = \pi(1 - \log\sqrt{2\pi}), \quad \sum_{n=1}^{\infty} \frac{\mathrm{Si}(n\pi)}{n} = \frac{\pi}{2}(1 - \log\pi)$$ can also be applied to here or not. – Sangchul Lee Aug 19 '12 at 2:04
• Something I've noticed: why do you keep making new tags for each special function? – J. M. is a poor mathematician Aug 19 '12 at 3:42

Numerical value

First, let's establish the numerical value of the sum. Since $$\operatorname{Si}(n) - \frac{\pi}{2} = -\frac{\cos(n)}{n} \left(1 + \mathfrak{o}(n^{-2})\right) - \frac{\sin(n)}{n^2} \left(1 + \mathfrak{o}(n^{-2})\right)$$ we shall evaluate: $$\sum_{n=0}^\infty \left(\operatorname{Si}(n) - \frac{\pi}{2}\right) = -\frac{\pi}{2} + \sum_{n=0}^\infty \left(\operatorname{Si}(n) - \frac{\pi}{2} + \frac{\cos(n)}{n}\right) - \sum_{n=1}^\infty \frac{\cos(n)}{n} \tag{1}$$ The middle sum now converges unconditionally. The conditionally convergent sum is easy to evaluate in closed form: $$-\sum_{n=1}^\infty \frac{\cos(n)}{n} = -\Re \left(\sum_{n=1}^\infty \frac{\mathrm{e}^{i n}}{n} \right) = \Re\left( \log\left(1-\mathrm{e}^i \right) \right) = \log\left(2 \sin \left(\frac{1}{2}\right)\right)$$ The following image displays the partial sums of $(1)$:

Thus the numerical value of the sum is approximately $-1.869201$.

Mellin-Barnes representation

Using the Mellin-Barnes representation for the summand: $$\operatorname{Si}(n) - \frac{\pi}{2} = -\frac{\sqrt{\pi}}{2} \cdot \frac{1}{2 \pi i} \int_{\mathcal{L}} \,\, \frac{\Gamma(s) \Gamma\left(s+\frac{1}{2}\right)}{\Gamma(1+s)\Gamma(1-s)} \left( \frac{n}{2} \right)^{-2s} \mathrm{d} s$$ where the contour $\mathcal{L}$ goes from $\mathrm{e}^{-i \theta} \infty$ to $\mathrm{e}^{i \theta} \infty$, with $\frac{\pi}{2} < \theta < \pi$, leaving all the poles of $\Gamma$-functions in the numerator to the left.

Thus: $$\sum_{n=0}^\infty \left( \operatorname{Si}(n) - \frac{\pi}{2}\right) = -\frac{\pi}{2} - \frac{\sqrt{\pi}}{2} \cdot \frac{1}{2 \pi i} \int_{\mathcal{L}} \,\, \frac{\Gamma(s) \Gamma\left(s+\frac{1}{2}\right)}{\Gamma(1+s)\Gamma(1-s)} 4^s \zeta(2s) \mathrm{d} s$$ Here is a numerical confirmation:

The summand also admits the following integral representation $$\operatorname{Si}(n) - \frac{\pi}{2} = -\int_0^1 J_0\left(\frac{n}{x}\right) \frac{\mathrm{d} x}{\sqrt{1-x^2}}$$ where $J_0(x)$ stands for the Bessel function of the first kind, but I was not able to put it to a good use to answer of this question.

• Neat trick to use Mellin-Barnes! I was trying to do Laplace, but that wasn't very useful here... – J. M. is a poor mathematician Aug 19 '12 at 5:20
• On the right hand side of the equation above the numerical confirmation, you should not have $-\frac{\pi}{2}$. – Mhenni Benghorbal Aug 23 '12 at 10:35
• @MhenniBenghorbal Thanks for the comment. I certainly should have the $-\frac{\pi}{2}$. It arises from singling out $n=0$ term. – Sasha Aug 23 '12 at 19:37

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 !)

• Fantastic result! – Argon Aug 22 '12 at 0:17
• Glad you liked it @Argon, it was a nice question too ! Still no 'closed form' but some fast convergent series like this variant of (6) : $$\displaystyle\sum_{n=0}^\infty \left(\mathrm{Si}(n)-\frac{\pi}{2}\right)=-1-\frac{\pi}4-2\sum_{n=1}^\infty \frac {\zeta(2n)-1}{(2n-1)(2\pi)^{2n}}-\frac 1{\pi}\mathrm{atanh}\left(\frac 1{2\pi}\right)$$ delivering $26$ digits with only $10$ terms of the series. – Raymond Manzoni Aug 22 '12 at 8:03
• @RaymondManzoni +1 Very nice result! – Sasha Aug 23 '12 at 3:45
• Thanks @Sasha! Your answer was nicely exposed and powerful too! – Raymond Manzoni Aug 23 '12 at 8:55

There is this. If $$S(x) = \sum_{n=0}^\infty \left(\mathrm{Si}(n x)-\frac{\pi}{2}\right)$$ then differentiate term-by-term and sum to get $$S'(x) = \frac{1}{2x} \cot\frac{x}{2}$$ However, we cannot just write $$S(1) = -\int_1^\infty \frac{1}{2x} \cot\frac{x}{2}\,dx$$ for our answer, because that integrand has lots of poles in the interval $[1,\infty)$.

One can find the following series,

$$\displaystyle\sum_{n=0}^\infty \left(\mathrm{Si}(n)-\frac{\pi}{2}\right) = -1 - \frac{\pi}{4} - \sum _{k=0}^{\infty}{\frac { \left( -1 \right) ^{k}{\it B} \left( 2\,k+2 \right) }{ \left( 2\,k+1 \right) \Gamma \left( 2\,k+3 \right) }} \approx -1.869201195,$$

where $B(n)$ are the Bernoulli numbers.

• Looks like $\text{(6)}$ from @RaymondManzoni's answer. How did you derive this formula? Also, I think that the approximation should be negative. – Argon Aug 23 '12 at 14:22
• @Argon:It looks like, but they are not the same. – Mhenni Benghorbal Aug 23 '12 at 21:46
• @downvoter: What's the downvote for? – Mhenni Benghorbal Feb 16 '13 at 19:05
• @Mhenni: I’m not the one who downvoted you, but I need to say a few things. I understand that you would like to know the reason for the downvote; of course, downvoters are encouraged to state their reasons for downvoting so that improvements can be implemented. However, if nobody states their reason, then there’s no need to get so worked up. I’m very sure that your last comment isn’t going to elicit any response from the downvoter. – Haskell Curry Feb 16 '13 at 23:18
• @Mhenni: Please consult this discussion thread on Meta. – Haskell Curry Feb 16 '13 at 23:34