# What is this mathematical equation from Fringe?

The following mathematical equation was shown during the television show Fringe which aired on Friday, April 22nd. Any idea what it is? (edit by J.M.: for reference, this was Sam Weiss scribbling formulae in his notebook.)

• I haven't watched: was that Walternate writing? – J. M. isn't a mathematician Apr 28 '11 at 2:43
• I appreciate this question. I think it's fun and lively - what is Fringe? – davidlowryduda Apr 28 '11 at 4:14
• @mix: This TV show. – J. M. isn't a mathematician Apr 28 '11 at 4:30
• @J.M. : No, it was that bowling guy. He was observing a vortex and taking notes. – Gunnar Þór Magnússon Apr 28 '11 at 11:30
• adds popular-math to followed tags... +1 – BBischof Apr 28 '11 at 14:13

## 3 Answers

The last formula seems to be an integral expression for the Dirichlet $\eta$ function:

$$\eta(s)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^s}\quad \Re(s) > 0$$

Using the relationship with the usual Riemann $\zeta$ function

$$\eta(s)=(1-2^{1-s})\zeta(s)$$

and this integral expression, you get that integral in the notebook:

$$\eta(s) = \frac1{\Gamma(s)} \int_0^{\infty}\frac{x^{s-1}}{e^x+1}\mathrm dx$$

There is also this closely related integral (which Arturo mentions in his answer).

The (first part of the) second line looks to be the chain of relations relating Riemann $\zeta$, Dirichlet $\eta$, and Dirichlet $\lambda$:

$$\frac{\zeta(s)}{2^s}=\frac{\lambda(s)}{2^s-1}=\frac{\eta(s)}{2^s-2}$$

In the second part, the expressions look to be the differentiation of Dirichlet $\eta$, but the screenshot is fuzzy around that region...

• Ok. Now that I look closely, it is a $e^x +1$, not a $e^x-1$. +1. – Aryabhata Apr 28 '11 at 2:37
• A confession: I was actually looking at the left hand side, and it sure looked like an $\eta$ to me... as for the equations on the second line, check out formula 3 in the MathWorld link I gave. – J. M. isn't a mathematician Apr 28 '11 at 2:41
• It is quite hard to read the second line, so I actually ignored those. Now that you point out, it does look like Eqn (3) :-) – Aryabhata Apr 28 '11 at 2:44
• Given the R(s) > 1/2 above that last integral, I wonder if the show was trying to show him solving (or attempting to solve) the Riemann Hypothesis... – Steven Stadnicki May 23 '11 at 19:09
• The succeeding scene had Sam writing a 0 after the =, so I suppose yes. – J. M. isn't a mathematician Jun 2 '11 at 6:13

By sheer coincidence, this equation occurs in P. Mark Kayll's paper Integrals don't have anything to do with discrete Math, do they? (Mathematics Magazine 84 no. 2, April 2011, pages 108-119, doi:10.4169/math.mag.84.2.108, which I was reading through today. It is equation (6) on page 111.

$\zeta(s)$ is the Riemann zeta function, $$\zeta(s) = \sum_{k=1}^{\infty}\frac{1}{k^s}.$$

$\Gamma(s)$ is the Gamma function, $$\Gamma(s) = \int_0^{\infty} t^{s-1}e^{-t}\,dt.$$

$R(s)\gt \frac{1}{2}$ says that the real part of $s$ is greater than $\frac{1}{2}$.

The final equation is just a recasting of the equality $$\zeta(x)\Gamma(x) = \int_0^{\infty}\frac{t^{x-1}}{e^t-1}\,dt$$ which holds whenever $\Gamma(x)$ is finite.

Nothing to get too excited about... unless you don't know what any of the symbols mean, which I suspect holds for a very large portion of the viewership of Fringe.

It's Riemann's zeta-function as a Mellin transform. See http://en.wikipedia.org/wiki/Riemann_zeta_function#Mellin_transform

• Oh, it's $+1$, not $-1$. And it's $\eta$, not $\zeta$... – lhf Apr 28 '11 at 2:47