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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?

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    @Jen, I didn't really object to the tag Argon was previously using, but I wasn't sure if we really needed very specific tags either...2012-08-19

4 Answers 4

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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 $(1)$: enter image description here

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: enter image description here


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.

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    @MhenniBenghorbal Thanks for the comment. I certainly should have the $-\frac{\pi}{2}$. It arises from singling out $n=0$ term.2012-08-23
15

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 !)

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    Thanks @Sasha! Your answer was nicely exposed and powerful too!2012-08-23
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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 $[1,\infty)$.

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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.

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    @HaskellCurry: If the downvoter does not intend to leave a comment then they should not downvote! Don't you think so?2013-02-17