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A257820
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Decimal expansion of the absolute value of the imaginary part of li(-1).
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3
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3, 4, 2, 2, 7, 3, 3, 3, 7, 8, 7, 7, 7, 3, 6, 2, 7, 8, 9, 5, 9, 2, 3, 7, 5, 0, 6, 1, 7, 9, 7, 7, 4, 2, 8, 0, 5, 4, 4, 4, 3, 9, 4, 4, 2, 8, 6, 6, 8, 7, 0, 7, 8, 2, 0, 2, 9, 2, 2, 5, 6, 0, 7, 8, 0, 3, 0, 8, 9, 0, 0, 9, 3, 3, 0, 9, 4, 5, 2, 8, 5, 7, 8, 4, 6, 7, 2, 7, 7, 4, 9, 1, 7, 4, 0, 1, 3, 2, 9, 1, 6, 9, 2, 7, 5
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OFFSET
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1,1
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COMMENTS
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The logarithmic integral function li(z) has a cut along the negative real axis which causes therein a discontinuity in the imaginary part of li(z). However, the absolute value of the imaginary part is continuous and its value is a well behaved function of any real argument, excepting z=+1. The above value corresponds to |imag(li(z))| at z=-1, the point where the corresponding real part (A257819) attains its maximum within the real interval (-infinity,+1).
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LINKS
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FORMULA
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Equals Pi*(1/2 + Sum[k=0..infinity]((-1)^k*Pi^(2*k)/(2*k+1)!/(2*k+1))).
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EXAMPLE
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3.422733378777362789592375061797742805444394428668707820292256...
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MAPLE
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MATHEMATICA
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RealDigits[Im[LogIntegral[-1]], 10, 120][[1]] (* Vaclav Kotesovec, May 11 2015 *)
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PROG
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(PARI) li(z) = {my(c=z+0.0*I); \\ If z is real, convert it to complex
if(imag(c)<0, return(-Pi*I-eint1(-log(c))),
return(+Pi*I-eint1(-log(c)))); }
a=imag(li(-1))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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