I need the solution to an Integral of type f[z]*Log[z]/z

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Discussion Overview

The discussion revolves around finding the anti-derivative of a specific integral involving the function \( f[z] \log[z]/z \). The integral is defined with real values for both \( z \) and \( x \), constrained within the interval [0,1], and includes a condition for the square root to remain real. Participants explore potential forms of the solution, including combinations of logarithmic and polylogarithmic functions, as well as hypergeometric functions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the integral and expresses difficulty in finding the anti-derivative, noting that Mathematica does not provide a solution.
  • The same participant speculates that the solution may involve \( \log[z] \), \( \log[z]^2 \), and \( \text{PolyLog}[2,z] \), but cannot manipulate the integral into a solvable form.
  • Another participant clarifies that \( x \) and \( z \) are independent variables at this stage, although \( x \) will later be integrated from 0 to 1 in conjunction with other integrals.
  • A third participant provides specific solutions for the integral at the boundary conditions \( x = 0 \) and \( x = 1 \), presenting complex expressions for both cases.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method to solve the integral, and multiple approaches and potential forms for the solution are discussed without resolution.

Contextual Notes

The discussion includes assumptions about the independence of variables and the conditions under which the square root remains real, which may affect the validity of proposed solutions.

Hepth
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$$\int dz \frac{\sqrt{\frac{1}{4} (x+1)^2 (z-1)^2-x} \log (z)}{z}$$

All values are real. The domain for z and x are both [0,1], with also the constraint that the Sqrt is real ( which means z is really from ##[0, 1-\frac{2 \sqrt{x}}{x+1}]##. I'm just trying to get the anti-derivative, and not apply the limits yet. No complex values for x or z.

I've tried many ways, and I can't seem to get it. Mathematica doesn't give a solution, and I've included the assumptions.

I think it'll be a combination of Log[z], Log[z]^2, and PolyLog[2,z] though I can't get it into a form that I can do this. Its possible due to its nature that it could be solved in terms of HyperGeometric functions too.

Does anyone have any ideas? Is there a method to solving these?

Thanks!
-Hepth
 
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Hepth said:
$$\int dz \frac{\sqrt{\frac{1}{4} (x+1)^2 (z-1)^2-x} \log (z)}{z}$$

All values are real. The domain for z and x are both [0,1], with also the constraint that the Sqrt is real ( which means z is really from ##[0, 1-\frac{2 \sqrt{x}}{x+1}]##. I'm just trying to get the anti-derivative, and not apply the limits yet. No complex values for x or z.

I've tried many ways, and I can't seem to get it. Mathematica doesn't give a solution, and I've included the assumptions.

I think it'll be a combination of Log[z], Log[z]^2, and PolyLog[2,z] though I can't get it into a form that I can do this. Its possible due to its nature that it could be solved in terms of HyperGeometric functions too.

Does anyone have any ideas? Is there a method to solving these?

Thanks!
-Hepth
Is there any relationship between x and z?
 
No. They are both independent at this point. In the end, x will also be integrated from 0..1, but combined with some other integrals it'll be convergent.
 
You can see that at x = 0 the solution is
##\frac{z}{2}+\frac{\log^2(z)}{4}-\frac{1}{2} z \log (z)##,

and at x=1 the solution is $$
\frac{1}{\sqrt{4-2 z}}i \sqrt{\frac{2}{z}-1} \sqrt{z} \left(\sqrt{2} (\log (z)-1) \left(\sqrt{-(z-2) z}+2 \sin ^{-1}\left(\frac{\sqrt{z}}{\sqrt{2}}\right)\right)-4 \sqrt{z} \, 3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};\frac{z}{2}\right)\right)$$
 

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