Prove $\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$

Prove $$\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$$

Hardy uses this fact without proof in a monograph on different ways to evaluate $\int_0^{\infty}\frac{\sin(x)}{x} dx$.

• HINT: Use the Mittag-Leffler Theorem. – Mark Viola Oct 11 '16 at 4:48
• @Winther Oops... – StubbornAtom Oct 11 '16 at 10:08

We can use also this well known summation formula (a consequence of the residue theorem): $$\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}f\left(k\right)=-\sum\left\{ \textrm{residues of }\pi\csc\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\}$$ so if we take $f\left(z\right)=\frac{1}{x+z\pi}$ we get $$\sum_{k\in\mathbb{Z}}\frac{\left(-1\right)^{k}}{x+k\pi}=-\underset{z=-x/\pi}{\textrm{Res}}\left(\frac{\pi\csc\left(\pi z\right)}{x+z\pi}\right)=\csc\left(x\right)$$ as wanted.

• Ah, right-- I was familiar with this formula and should have thought to use it. Thanks! – Vik78 Oct 13 '16 at 16:05

Presumably the principal value of the two-sided infinite sum is what was intended in the question.

We'll solve this just using Euler's product formula for the sine function:

\begin{align} \sin x=x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2 \pi^2}\right). \end{align}

Compute the logarithmic derivative:

\begin{align} \cot x &= \frac1{x}+\sum_{n=1}^\infty \frac{-2x/n^2\pi^2}{1-\frac{x^2}{n^2 \pi^2}} \\&=\frac1{x}+\sum_{n=1}^\infty \frac{2x}{x^2-n^2\pi^2} \\&=\frac1{x}+\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big), \end{align}

so

$$\cot x-\frac1{x}=\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big).$$

It follows that

\begin{align} \cot\big(\frac{x}{2}\big)-\frac2{x}&=\sum_{n=1}^\infty\big(\frac{1}{x/2+n\pi}+\frac{1}{x/2-n\pi}\big) \\&=2\sum_{n=1}^\infty\big(\frac{1}{x+2n\pi}+\frac{1}{x-2n\pi}\big) \\&=2\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big). \end{align}

Subtracting now, we obtain

\begin{align} \cot\big(\frac{x}{2}\big)-\cot(x)-\frac1{x}&=2\cdot\!\!\!\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)-\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big) \\&=\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)-\sum_{\substack{n\ge 1\\n\text{ odd}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big) \\&=\sum_{n=1}^\infty (-1)^n \big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big). \end{align}

But

\begin{align} \cot\big(\frac{x}{2}\big)-\cot(x)&=\frac{\cos(x/2)}{\sin(x/2)}-\frac{\cos x}{\sin x} \\&=\frac{\cos(x/2)}{\sin(x/2)}\cdot\frac{2\sin(x/2)\cos(x/2)}{\sin x}-\frac{\cos x}{\sin x} \\&=\frac{2\cos^2(x/2)-\cos x}{\sin x} \\&=\frac1{\sin x} \\&= \csc x. \end{align}

So we've shown that

\begin{align} \csc x &= \frac1{x}+\sum_{n=1}^\infty (-1)^n \big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big), \end{align}

which is the principal value of $$\sum_{n=-\infty}^\infty (-1)^n \frac{1}{x+n\pi},$$

as desired.

In this answer, it is shown in detail that $$\sum_{k=-\infty}^\infty\frac{1}{z+k}=\pi\cot(\pi z)\tag{1}$$ $(1)$ is the sum for even and odd $k$. The sum for even $k$ would be $$\sum_{k=-\infty}^\infty\frac{1}{z+2k}=\frac\pi2\cot\left(\frac{\pi z}2\right)\tag{2}$$ The sum for even minus the sum for odd would be twice $(2)$ minus $(1)$ \begin{align} \sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k} &=\pi\cot\left(\frac{\pi z}2\right)-\pi\cot(\pi z)\\ &=\pi\frac{1+\cos(\pi z)}{\sin(\pi z)}-\pi\frac{\cos(\pi z)}{\sin(\pi z)}\\[9pt] &=\pi\csc(\pi z)\tag{3} \end{align} Therefore, $$\sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k\pi}=\csc(z)\tag{4}$$

• Hmm... What does this add to the solution I posted? Isn't it the same thing, but without the derivation for $\cot,$ and then leaving out details? – Mitchell Spector Oct 11 '16 at 20:04
• If you read my earlier answer, this provides a significantly different approach to deriving the non-alternating sum. Then the derivation of the alternating sum from the non-alternating sum seems a bit simpler here. – robjohn Oct 12 '16 at 0:49
• I don't mean to belabor this; it doesn't really matter. Yes, your earlier answer had a different derivation of the $\cot$ formula, using residues (which is nice since it's good to see completely different approaches to the same problem -- but that's not part of your answer here). But the key point, the computation of the alternating series by subtracting the sum for all $k$ from twice the sum for even $k,$ is exactly what I did; you just left out details that I included. (What differences exist are inconsequential: for example, you used $\pi \cot \pi z$ where I had $\cot z.)$ – Mitchell Spector Oct 12 '16 at 1:06

We begin by expanding the function $$\cos(xy)$$ in a Fourier series,

$$\cos(xy)=a_0/2+\sum_{k=1}^\infty a_k\cos(ky) \tag1$$

for $$x\in [-\pi/\pi]$$. The Fourier coefficients $$(1)$$ are given by

\begin{align} a_k&=\frac{2}{\pi}\int_0^\pi \cos(xy)\cos(ky)\,dy\\\\ &=\frac1\pi (-1)^k \sin(\pi x)\left(\frac{1}{x +k}+\frac{1}{x -k}\right)\tag2 \end{align}

Substituting $$(2)$$ into $$(1)$$, setting $$y=0$$, and dividing by $$\sin(\pi x)$$ reveals

\begin{align} \pi \csc(\pi x)&=\frac1y +\sum_{n=1}^\infty (-1)^k\left(\frac{1}{x -k}+\frac{1}{x +k}\right)\\\\ &=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x-k}\\\\ &=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x+k}\tag3 \end{align}

Finally, enforcing the substitution $$x\to x/\pi$$ and dividing by $$\pi$$ in $$(3)$$ yields the coveted result

$$\csc(x)=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x+k\pi}$$

• last sum should extend to $-\infty$ right ? – G Cab Jan 29 at 23:33
• @GCab Indeed. And thank you for catching the typographical error! Much appreciative. – Mark Viola Jan 29 at 23:42
• Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Jan 30 at 5:03