We can expand the PolyLog function as a Taylor series and integrate each term, at least for Re[z] > 0. The result can be written in form $$sum_^infty a_n H(b_n+frac) ,,$$

where $H$ is the HarmonicNumber function. Closed form expressions for $a_n$ and $b_n$ can be found.

We might start by plotting the integral for real, positive values of $z$.

f[z_] := NIntegrate[x^(z – 1) PolyLog[2, x]/(1 + x), ] Plot[f[z], , GridLines -> Automatic, ImageSize -> Small, Frame -> True];

Now we can expand the PolyLog function at $x_0 = 0$, evaluate the integral termwise and combine the results to a function $g(z)$ like this

With[, s = Series[PolyLog[2, x], ]; t = Table[ Integrate[x^(z – 1) SeriesCoefficient[s, n] (x – x0)^n/(1 + x), , Assumptions -> Re[z] > 0], ]; g = Simplify[Total@t] ];

Just to show we’re on the right track we could plot the $g(z)$ and $f(z)$, but I won’t show the plot here.

Plot[, , GridLines -> Automatic, ImageSize -> Small, Frame -> True]

The expression for $g(z)$ is a little messy. It is a series of terms like 7/288 HarmonicNumber[1 + z/2]. We can extract the $a_n$’s and the $b_n$’s like this

p = Cases[g, Times[a_, HarmonicNumber[b_]] :> ]; = Transpose[Sort[p /. z -> 0]];

Then find closed form expressions for $a_n$ and $b_n$ like this

Clear[a, b, n] a[0] = an[[1]]; a[n_] = FindSequenceFunction[Most@Rest@an, n] b[n_] = FindSequenceFunction[Most@Rest@bn, n] (* (1 + 2 n)/(2 n^2 (1 + n)^2) *) (* 1/2 (-1 + n) *)

We don’t use the first and last terms of the sequence because they don’t quite fit the pattern.

Now we are ready to write our final expression for the original integral as

Clear[h] h[z_, nmax_] := Sum[a[n] HarmonicNumber[b[n] + z/2], ]

Of course, we would need an infinite number of terms for $h(z)$ to represent the integral exactly. Numerically, we can get good agreement with a finite number of terms. For example,

Plot[, , GridLines -> Automatic, ImageSize -> Medium, Frame -> True]

We can increase the agreement by adding more terms, which is necessary for larger values of $z$. For instance, to get decent agreement at $z = 200$, we might use $n_ = 150$ or greater.