Here is the integral for which I want a symbolic result:

Integrate[x^(z - 1)PolyLog[2, x]/(1 + x), {x, 0, 1}]

But the output is the same as the input without any error message. What I expected to get is a function of z, and without this result, I cannot continue my symbolic calculations.

Can anyone help to understand this issue?

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    $\begingroup$ All computer algebra systems, including Mathematica, are limited in their capabilities. Do you have any reason to think there is a closed form? Most integrals don't have one. Maybe the best you can do is numerical methods. $\endgroup$ – Mariusz Iwaniuk Jan 24 at 17:17
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    $\begingroup$ I find the integral can be performed if you assign a value to $z$. $\endgroup$ – David G. Stork Jan 24 at 18:18
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    $\begingroup$ Mathematica, and you too, needs to know more about parameter z ! $\endgroup$ – Ulrich Neumann Jan 24 at 18:28
  • $\begingroup$ Wow great!, I see and I will try right away...Thank you very much for your outstanding answer. I may come back with other questions soon. Kind regards! $\endgroup$ – Redamy Perez Ramos Jan 27 at 16:26

You can find a general solution for integer z with the help of FindSequenceFunction and integer sequence search engines.

int1[z_] := Integrate[x^(z - 1)*PolyLog[2, x]/(1 + x), {x, 0, 1}]

int2[n_] := Sum[(-1)^(k + n) HarmonicNumber[k]/k^2, {k, 1, n - 1}] + 1/24 (-1)^n (4 (-1)^n \[Pi]^2 LerchPhi[-1, 1, n] + 15 Zeta[3])

tab = Table[{z, int2[z] == int1[z]}, {z, 1, 10}] // Simplify

(*   {{1, True}, {2, True}, {3, True}, {4, True}, {5, True}, {6, True}, {7,
          True}, {8, True}, {9, True}, {10, True}}   *)
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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 $$\sum_{n=0}^\infty a_n H(b_n+\frac{z}{2}) \,,$$

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), {x, 0, 1}]

Plot[f[z], {z, 0, 2},
 GridLines -> Automatic, ImageSize -> Small, Frame -> True];

enter image description here

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[{nmax = 20, x0 = 0},
  s = Series[PolyLog[2, x], {x, x0, nmax}];
  t = Table[
    Integrate[x^(z - 1) SeriesCoefficient[s, n] (x - x0)^n/(1 + x),
     {x, 0, 1}, Assumptions -> Re[z] > 0], {n, 0, nmax}];
  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[{f[z], g}, {z, 0, 2},
 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_]] :> {b, a}];
{bn, an} = 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

h[z_, nmax_] := Sum[a[n] HarmonicNumber[b[n] + z/2], {n, 0, nmax}]

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[{f[z], h[z, 15]}, {z, 0, 2},
 GridLines -> Automatic, ImageSize -> Medium, Frame -> True]

enter image description here

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 $n_{max} = 150$ or greater.

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Not a solution but an extended comment.

For each integer $z$ the solution can be found analytically. These solutions consist of a Lerch transcendent, a Zeta function, and a rational number that I am unable to parametrize:

F[z_Integer?Nonnegative] := Integrate[(x^(z-1)*PolyLog[2,x])/(1+x), {x, 0, 1}]

Table[F[z] - (π^2/6*LerchPhi[-1,1,z] + 5/8*(-1)^z*Zeta[3]), {z, 0, 20}] // FullSimplify

(*    {Zeta[3], 0, -1, 5/8, -179/216, 1207/1728, -170603/216000, 155903/216000,
       -57395129/74088000, 433990957/592704000, -12276669439/16003008000,
       2361589283/3200601600, -3249595002553/4260000729600, 3157791670933/4260000729600,
       -7113784138562041/9359221602931200, 6958518361163881/9359221602931200,
       -35482726038639437/46796108014656000, 278917903336133521/374368864117248000,
       -1392213902396958791873/1839274229408039424000, 1372372969627754044673/1839274229408039424000,
       -9537086255033397455599307/12615581939509742409216000}    *)

Maybe at the Math StackExchange you'll have more success at getting a closed-form answer.

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