Can You Integrate (ln ln x)^n Using a Taylor Series?

In summary, the conversation discusses methods for integrating (ln ln x)^n for any value of n. It is mentioned that Mathematica is unable to find a formula for n=2 and the only way to integrate it would be numerically. It is also noted that there is one specific value where the integral can be evaluated explicitly. One person suggests using a Taylor polynomial for (ln ln x) and integrating that instead.
  • #1
flouran
64
0
How do you integrate (ln ln x)^n for any n?
 
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  • #2
Well since mathematica isn't able to find a formula for n=2, I'm going to say "numerically".
 
  • #3
nicksauce said:
Well since mathematica isn't able to find a formula for n=2, I'm going to say "numerically".

What do you mean exactly by "numerically"? Do you mean that i should evaluate it as a definite integral?
 
  • #4
Well I just mean if you want to find a general closed for expression for the indefinite integral, you are out of luck. Therefore the only way I can conceive of doing an integral with this expression would be to do a definite integral numerically.
 
  • #5
there is one place where the integral can be evaluated explicitly that I know of.
[tex]\int[/tex]-log[-log[x]]dx=Euler Constant (.577...)
This follows from differentionation the gamma function in its product and integral forms and making a change of variables.
 
  • #6
nicksauce said:
Well I just mean if you want to find a general closed for expression for the indefinite integral, you are out of luck. Therefore the only way I can conceive of doing an integral with this expression would be to do a definite integral numerically.
Although this method would be painful, since I cannot express this integral in terms of elementary functions, could I represent (ln ln x)^(n) as a Taylor polynomial (what is the Taylor series for ln ln x, anyways?) and then integrate that and leave it as a Taylor Series?
 

1. What is the general formula for integrating (ln ln x)^n?

The general formula for integrating (ln ln x)^n is:
∫(ln ln x)^n dx = x(ln ln x)^n - n∫(ln x)(ln ln x)^(n-1)dx

2. How do I solve for the indefinite integral of (ln ln x)^n?

To solve for the indefinite integral of (ln ln x)^n, follow these steps:
1. Use the general formula: ∫(ln ln x)^n dx = x(ln ln x)^n - n∫(ln x)(ln ln x)^(n-1)dx
2. Use integration by parts to evaluate the second integral on the right side, with u = ln x and dv = (ln ln x)^(n-1)dx
3. Simplify the resulting expression and solve for the indefinite integral.

3. Can I use substitution to integrate (ln ln x)^n?

Yes, you can use substitution to integrate (ln ln x)^n. However, it may not always be the most efficient method.

4. Is there a specific range of values for which the integral of (ln ln x)^n converges?

The integral of (ln ln x)^n converges for all values of x > 1. For x = 1, the integral is divergent.

5. Can I use integration by parts multiple times to evaluate the integral of (ln ln x)^n?

Yes, you can use integration by parts multiple times to evaluate the integral of (ln ln x)^n. This may be necessary if the value of n is large.

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