Li

_{2}(x) = å

_{n=1}[x

^{n}/ n

^{2}]

^{ }

Where the summation is to infinity (oo) and Euler was unable to evaluate Li

_{2}(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/ [x^{2}-1]^{1/2}
Integrating this, he got:

z/ i = log (x + [x

where i= [-1]

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

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

Then to evaluate Li

x

Landen then divided the preceding by x to obtain:

- Li

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

C = -2 å

He then had to evaluate the series above and did this by setting x = -1 in the expression for - Li

å

Landen’s computations indicated that the formula for - Li

0 < Θ < 2π

Note, however, that if this is substituted directly into the expression for - Li

z/ i = log (x + [x

^{2}-1]^{1/2}/ iwhere i= [-1]

^{1/2}Landen then set z = π/2 and x = 1 so that log ([-1]

^{1/2}) = log(i) = -π/2iSince a square root must have two values, he concluded: log (-1) = +/- π/ i

Then to evaluate Li

_{2}(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/iLanden then divided the preceding by x to obtain:

- Li

_{2}(1/x) = - π/i log x + ½ (log x)^{2}+ Li_{2}(x) + CTo find C, Landen took x = 1, and then using algebra (solving for C above) obtained:

C = -2 å

^{¥}_{n=1}(1/n^{2})He then had to evaluate the series above and did this by setting x = -1 in the expression for - Li

_{2}(1/x) . Using a (+) sign this time, he arrived at:å

^{¥}_{n=1}(1/n^{2}) = π^{2 }/ 6Landen’s computations indicated that the formula for - Li

_{2}(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 Li_{2}(x) and Li_{2}(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 - Li

_{2 }(1/x) the result is invalid. However, choosing the (+) sign instead makes it valid, yielding the Fourier series expansion of the 2nd Bernouli polynomial.
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