Sunday, January 6, 2013

Evaluating the Dilogarithm Function

Among the many fascinating mathematical curiosities is the so-called dilogarithmic function. It was believed to be originally discussed by Leonard Euler in 1729 in the form:

Li2 (x) =  å n=1   [xn  / n2]                            

Where the summation is to infinity (oo) and Euler was unable to evaluate Li2 (1)   from the integral form. However, Landen, in a paper appearing in Philosophical Transactions (Landen, John: 1760,  A New Method of Computing The Sums of Certain Series, Vol. 51, Pt. 2, p. 553)showed the evaluation required the use of log (-1). Unaware of Euler’s work, Landen set out to determine log (-1) by first differentiating x =   sin (z) to obtain: dz/ i=   dx/ [x2 -1]1/2

Integrating this, he got:

z/ i = log (x + [x2 -1]1/2  / i

where i= [-1]1/2

Landen then set z = π/2 and x = 1 so that log ([-1]1/2) = log(i) = -π/2i

Since a square root must have two values, he concluded: log (-1) = +/-  π/ i

Then to evaluate Li2 (1) he commenced with the series form:

x-1 + x-2 /2 + x -3/ 3 + ……= log (1/ 1 – x) = log x + log (1/ 1- x) - π/i

Landen then divided the preceding by x to obtain:

- Li2 (1/x) = - π/i log x + ½ (log x)2 + Li2 (x) + C

To find C, Landen took x = 1, and then using algebra (solving for C above) obtained:

C = -2 å ¥ n=1  (1/n2)

He then had to evaluate the series above and did this by setting x = -1 in the expression for - Li2 (1/x) . Using a (+) sign this time, he arrived at:

å ¥ n=1  (1/n2) = π2 / 6

Landen’s computations indicated that the formula for - Li2 (1/x) was only correct for the case x > 1. His tactic of dividing by x, then integrating led ultimately to the recognition that the series Li2 (x) and Li2 (1/x) converged when x = exp(i Θ) for the interval:

0 < Θ < 2π

Note, however, that if this is substituted directly into the expression for - Li2 (1/x) the result is invalid. However, choosing the (+) sign instead makes it valid, yielding the Fourier series expansion of the 2nd Bernouli polynomial.

No comments: