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flouran
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How do you integrate (ln ln x)^n for any n?
nicksauce said:Well since mathematica isn't able to find a formula for n=2, I'm going to say "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?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.
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
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.
Yes, you can use substitution to integrate (ln ln x)^n. However, it may not always be the most efficient method.
The integral of (ln ln x)^n converges for all values of x > 1. For x = 1, the integral is divergent.
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.