Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following mathematical equation was shown during the television show Fringe which aired on Friday, April 22nd. Any idea what it is?

Fringe formula

(edit by J.M.: for reference, this was Sam Weiss scribbling formulae in his notebook.)

share|improve this question
I haven't watched: was that Walternate writing? –  Guess who it is. Apr 28 '11 at 2:43
I appreciate this question. I think it's fun and lively - what is Fringe? –  mixedmath Apr 28 '11 at 4:14
@mix: This TV show. –  Guess who it is. 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 3

up vote 35 down vote accepted

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


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 $\lambda$:


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

share|improve this answer
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. –  Guess who it is. 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. –  Guess who it is. 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.

share|improve this answer
Actually it is $e^x + 1$, not $e^x -1$ (sure looked like that to me too!), so J.M's answer is more accurate. (And the symbol and $Re(s) \gt 1/2$ fit better for that I suppose). –  Aryabhata Apr 28 '11 at 2:38
Arturo, the DOI doesn't seem to work. Maybe link to JSTOR instead? –  Guess who it is. Apr 28 '11 at 2:38
@J.M., also available at umt.edu/math/reports/kayll/Karply-MMag-submit09.pdf –  lhf Apr 28 '11 at 2:44
@lhf: Thanks! I shall read it now... –  Guess who it is. Apr 28 '11 at 2:47
@J.M. I suspect the doi has not been created yet; the issue just came out in the past couple of weeks. –  Arturo Magidin Apr 28 '11 at 3:22

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

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

protected by Zev Chonoles Jul 18 at 7:41

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.