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

In summary, the integral of $$\int dz \frac{\sqrt{\frac{1}{4} (x+1)^2 (z-1)^2-x} \log (z)}{z}$$ with real values for z and x in the domain [0,1], where the Sqrt is also real and z is from ##[0, 1-\frac{2 \sqrt{x}}{x+1}]##, cannot be solved using standard methods and may require HyperGeometric functions. The solutions for x = 0 and x = 1 are given by ##\frac{z}{2}+\frac{\log^2(z)}{4}-\frac{1}{2} z \log
  • #1
Hepth
Gold Member
464
40
$$\int dz \frac{\sqrt{\frac{1}{4} (x+1)^2 (z-1)^2-x} \log (z)}{z}$$

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
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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.
 
  • #4
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 $$
\frac{1}{\sqrt{4-2 z}}i \sqrt{\frac{2}{z}-1} \sqrt{z} \left(\sqrt{2} (\log (z)-1) \left(\sqrt{-(z-2) z}+2 \sin ^{-1}\left(\frac{\sqrt{z}}{\sqrt{2}}\right)\right)-4 \sqrt{z} \, 3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};\frac{z}{2}\right)\right)$$
 

1. What is an integral?

An integral is a mathematical operation that is used to find the area under a curve in a graph. It is often used to calculate quantities such as displacement, velocity, and acceleration in mathematics and physics.

2. What does f[z]*Log[z]/z mean?

f[z]*Log[z]/z is a type of function that combines a logarithm and a variable in the denominator. It is commonly used in calculus and can be solved through integration techniques.

3. How do you solve an integral of type f[z]*Log[z]/z?

To solve an integral of type f[z]*Log[z]/z, you can use the power rule, substitution, or integration by parts. The specific method used will depend on the form of the function and the level of complexity of the integral.

4. What is the purpose of solving an integral of type f[z]*Log[z]/z?

Solving an integral of type f[z]*Log[z]/z allows you to find the exact area under a curve and can be used to solve various problems in mathematics and physics. It also helps in finding the antiderivative of a function, which is useful in finding the original function from its derivative.

5. Can an integral of type f[z]*Log[z]/z have multiple solutions?

No, an integral of type f[z]*Log[z]/z should have one unique solution. However, the method used to solve it may vary, leading to different forms of the solution. It is important to check the validity of the solution by taking the derivative and comparing it to the original function.

Similar threads

Replies
2
Views
891
Replies
1
Views
2K
Replies
2
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
11
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
766
Replies
3
Views
655
Back
Top