## A sum involving hyperbolic functions I

We had a user who habitually posted difficult sums and integrals. This annoyed many of the more experienced users because he showed little interest in anything other than seeing someone else write a complete solution. I didn’t mind – these were nifty and challenging.

One problem is an evaluation of a sum:

$$\sum_{n=1}^{\infty}\frac{\cosh(2nx)}{\cosh(4nx)-\cosh(2x)}=\frac1{4\sinh^2(x)}$$

We use the residue theorem; consider

$$\sum_{n=-\infty}^{\infty} \frac{\cosh{2 n x}}{\cosh{4 n x} – \cosh{2 x}} = 2 \sum_{n=1}^{\infty} \frac{\cosh{2 n x}}{\cosh{4 n x} – \cosh{2 x}} – \frac{1}{\cosh{2 x} – 1}$$

By the residue theorem, the sum on the LHS is zero. The reason is that

$$\mathrm{Res}_{z=1/2+i k} \frac{\pi \cot{\pi z} \cosh{2 x z}}{\cosh{4 x z}-\cosh{2 x}} = 0 \quad \forall k \in \mathbb{Z}$$

$$\therefore \sum_{n=1}^{\infty} \frac{\cosh{2 n x}}{\cosh{4 n x} – \cosh{2 x}} = \frac{1}{2}\frac{1}{\cosh{2 x} – 1} = \frac{1}{4 \sinh^2{x}}$$

• Hi Ron:

By happenstance, I found your blog while searching for ‘using contours to evaluate integrals with arctan’.

I wonder, am I the one that has irritated the “more experienced users”?. 🙂
If so, they need to pull the sticks out of their asses and quit taking life so serious. It’s a frickin’ math website. I’m not hacking into the Pentagon!!.
I did not realize they were all so anal about it. Chill out and relax. I have noticed that Sasha and others no longer respond. I reckon they can stick it then. I won’t bother anymore. Life will certainly go on without it. 🙂

Well, you’re not uppity like that. You look at it as a fun learning process the way I do. I always like your solutions. They are thorough and usually involve contours/residues.

Cheers and take care,
Cody

• rlgordonma@yahoo.com wrote:

Hey Cody,

Hmmmm…I doubt I had you in mind for any of that nonsense. You’re right – it’s just a math website. And for me (and you), it’s just a fun diversion. And, honestly, the people who understand residue theory are always the most pleasant to work with.

Thanks for the compliments. I look forward to many more fun interactions with you.

• Hi RG:

Maybe I’m being paranoid and presumptuous. But, I figure it’s a nice math site and that is what it’s for…to post problems, learn, and have fun. For some to get upset over that does not make much sense. “That’s like getting mad at a beagle for chasing a rabbit.:):)
It’s just that it sounded like me and I think I posted that hyperbolic series in question.
Well, enough paranoia. May I ask you something about contour integration?. Why is it one rarely sees integrals containing inverse trig functions evaluated via residues?. Is it because the branches of something like arctan make it tedious and difficult?. I posted a fun one yesterday on SE, and I was wondering if it could be done that way. Perhaps you saw it. It involved arctan(x^3) and an e term in the denominator.

Take care,
Cody

• rlgordonma@yahoo.com wrote:

Cody,

I got exasperated (quietly) when people continually complained when a few guys were posting these terrific integral/sum problems. It was obvious that these weren’t HW, and it was further obvious that they were putting the problems out there because they ran out of ideas themselves. I think it should have been clear that there were great answers and camaraderie being generated. So what’s the problem? The problem is some folks being confused about what they think Math.SE should be. Not I.

Inverse trig functions are nasty in the complex plane. I tried your problem but got stymied when I kept running against a singularity at $z=i$. They have branch points away from the real axis and just make life difficult in general.

• Thanks, Ron. What I do know of contour integration caused me to notice that singularity at z=i as well. Then, I didn’t know what to do with it. I thought maybe there was a clever way to deal with it, so I thought I would ask those more learned than myself in the area. Namely, you. 🙂

Take care and thanks,
Cody