I seek to prove that $$\sum_{n=1}^\infty \frac{1}{\lambda_{n}^2} = \frac{1} {10},$$ by applying Cauchy Theorem to $$ f(z) = \left(\frac{z\tan(z)}{z-\tan(z)}+\frac{3}{z}\right) \frac{1}{z^2},$$ where $\lambda_n$ are positive solutions to $\tan(x) = x$.

How can I solve this?

  • $\begingroup$ There are tons of Cauchy Theorem. Please, be more specific. $\endgroup$ – the_candyman Nov 26 '15 at 20:59
  • $\begingroup$ where do u get $f(z)$ from? $\endgroup$ – tired Nov 26 '15 at 20:59
  • $\begingroup$ The exercise is in my professor's notes - the section covers residues and rouche's theorem so I'd assume it's cauchy's integral formula $\endgroup$ – Dr. John A Zoidberg Nov 26 '15 at 21:01
  • $\begingroup$ I haven't done the question, but $f(z)$ as stated clearly has poles at $z=0$ and the $\lambda_i$. It might help to compute the residue of $f$ at each of these points. $\endgroup$ – user3131035 Nov 26 '15 at 21:30
  • $\begingroup$ nice question btw. $\endgroup$ – tired Nov 26 '15 at 22:04

Let's see:

$f(z)$ has three types of poles $z_0=0$ , $z_{1,n}=n \pi/2$ and $z_{2,n}=\lambda_n$

It turns out that the poles at $z_{1,n}$ are removeable so the corresponding residue will yield zero (why is that? because $f(z)$ is exactly designed to do so). Now let us integrate $f(z)$ around a contour which consists of two lines just above and below the real axis (connected at infinity), with a sense of orientation which is clockwise. Due to the residue theorem we have

$$ \oint f(z)dz = -2\pi i [\text{res}(z_0)+\sum_{n=-\infty}^{\infty}\text{res}(z_{2,n})] $$

But on the other hand, because $tan(z)$ is a bounded function away from the real axis we can also close our contour of integration by adding big semicircles to the now disconnected lines. Because there will no now poles enclosed during this procedure we see that $\oint f(z)dz=0$ and therefore

$$ -\text{res}(z_0)=\sum_{n=-\infty}^{\infty}'\text{res}(z_{2,n}) \quad (1) $$

Here $\sum'$ means that we omit $0$ under the sum sign. The residue at $0$ is given by $\text{res}(z_0)=\frac{1}{5}$, which follows from a straightforward Laurent expansion.

The tricky part know is to calculate the residues at $\lambda_n$. First of all because the poles are simple we can use the

$$ \text{res}(\lambda_n)=\frac{g(\lambda_n)}{f'(\lambda_n)} \quad (2) $$

with $g(z)=\frac{z^2\tan(z)+3(z- \tan(z))}{z^3}$ and $f'(z)=1-\sec^2(z)=-\tan^2(z)$

using $\tan(\lambda_n)=\lambda_n$ we might rewrite (2) as $$ \text{res}(\lambda_n)=-\frac{\lambda_n^2 \lambda_n+ 3(0)}{\lambda_n^3\lambda_n^2}=-\frac{1}{\lambda_n^2} $$ Putting this into (1) we might observe that

$$ -\frac{1}{5}=-\sum_{n=-\infty}^{\infty}'\frac{1}{\lambda_n^2} $$

now we observe that the equation $\tan(x)=x$ is anti-symmetric under $x\rightarrow -x$ and all zeros come as pairs $\pm\lambda_n$. This allows us to rewrite

$$ \frac{1}{10}=\sum_{n=1}^{\infty}\frac{1}{\lambda_n^2} $$


  • $\begingroup$ You still need to address an issue I also struggled with: why do the semicircles you add make zero contribution in the limit? That isn't completely obvious because part of them is close to the real line. $\endgroup$ – user138530 Nov 26 '15 at 22:16
  • $\begingroup$ @ChristianRemling it doesn't really matter because i can put my lines paralell to the real axis as far away from it as needed. $\endgroup$ – tired Nov 26 '15 at 22:16
  • $\begingroup$ Maybe this picture helps, inspirehep.net/record/1095423/files/Contour_Boson.png i don't find anything better at the moment so u have to exchange $Im$ and $Re$ $\endgroup$ – tired Nov 26 '15 at 22:17
  • $\begingroup$ That doesn't help because if you move the lines away from $\mathbb R$, then the original contour now has parts connecting the two lines that are not short and could make non-zero contributions. $\endgroup$ – user138530 Nov 26 '15 at 22:17
  • 1
    $\begingroup$ I would use a path similar to the one in my answer. It is easier to show that the integral along that path vanishes as $k\to\infty$. $\endgroup$ – robjohn Nov 27 '15 at 17:58

Here is an approach that uses the function $\frac1{\tan(z)-z}$.

First, let's compute the residue of the function $\frac1{\tan(z)-z}$ at $\lambda_n$: $$ \lim_{z\to\lambda_n}\frac{z-\lambda_n}{\tan(z)-z}=\frac1{\sec^2(\lambda_n)-1}=\frac1{\tan^2(\lambda_n)}=\frac1{\lambda_n^2}\tag{1} $$ Note that along a contour whose real part is $k\pi$, we have $$ \tan(k\pi+ix)=i\tanh(x)\tag{2} $$ whose absolute value is less than $1$. Therefore, along the contour $\gamma$: $$ k\pi+i[-k,k]\cup\pi[k,-k]+ik\,\cup-k\pi+i[k,-k]\cup\pi[-k,k]-ik\tag{3} $$ we have $\frac1{\tan(z)-z}=-\frac1z+O\left(\frac1{z^2}\right)$. Thus, the integral along this contour tends to $$ \int_\gamma\frac{\mathrm{d}z}{\tan(z)-z}=-2\pi i\tag{4} $$ Near $z=0$, $$ \begin{align} \frac1{\tan(z)-z} &=\frac1{\frac{z^3}3+\frac{2z^5}{15}+O\left(z^7\right)}\\ &=\frac3{z^3}-\frac6{5z}+O(z)\tag{5} \end{align} $$ so the residue at $z=0$ is $-\frac65$.

Therefore, combining $(1)$, $(4)$, and $(5)$, we get $$ \overbrace{\ -2\pi i\ \vphantom{\sum_{n=1}^\infty}}^{\text{integral along $\gamma$}}=\overbrace{-\frac{12\pi i}5\vphantom{\sum_{n=1}^\infty}}^{\text{residue at $z=0$}}+\overbrace{4\pi i\sum_{n=1}^\infty\frac1{\lambda_n^2}}^{\substack{\text{residues at $\lambda_n$}\\\text{in both directions}}}\tag{6} $$ Solving $(6)$ yields $$ \sum_{n=1}^\infty\frac1{\lambda_n^2}=\frac1{10}\tag{7} $$

  • $\begingroup$ The contribution of $\gamma$ can be seen as the residue at infintiy right? $\endgroup$ – tired Nov 27 '15 at 18:23
  • $\begingroup$ @tired: I was never taught to consider residues at infinity, but I guess you could look at it that way. $\endgroup$ – robjohn Nov 27 '15 at 18:28

Let's integrate $f$ over a circle of radius $R\gg 1$. The poles of $f$ are at $\pm\lambda_n$ ($f$ is odd, in fact) and $0$. It's easy to see that the residues at $\pm \lambda_n$ equal $-1/\lambda_n^2$, so we obtain $-2\sum 1/\lambda_n^2$ from these. The residue at $z=0$ equals $1/5$; this follows from a calculation, using that $$ \tan z =z+z^3/3 + 2z^5/15 + O(z^7) . $$ Note that the first term in parentheses has residue $-3$ at $z=0$; this "explains" the second term.

So $$ \frac{1}{2\pi i}\int_{C_R} f(z)\, dz = - 2\sum_{\lambda_n<R} \frac{1}{\lambda_n^2} + \frac{1}{5} . $$ It remains to show that the integral goes to zero as $R\to\infty$. Here we send $R\to\infty$ through values that don't come too close to the $\lambda_n$.

As soon as $\textrm{Im}\, z$ is not too small, $\tan z$ becomes bounded, and thus the integrand is $\lesssim 1/R^2$. Close to the real line, we use what we just agreed on, that $|R-\lambda_n|\gtrsim 1$, where $\lambda_n$ now denotes the nearest $\lambda$. Since $\tan x$ has derivative $1+\lambda_n^2$ at $\lambda_n$, we will have that $|\tan z|\gtrsim \lambda_n^2$ for these $z$, and the integrand becomes $\lesssim 1/R$, which is good enough, since we're using this bound only on a part of bounded length of our circle.


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