Using series to solve definite integrals

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SUMMARY

The discussion focuses on solving the definite integral of the function \(\int \log(x) \log(x+1) \, dx\) from 0 to 1 using series expansion. The series representation of \(\log(x+1)\) is given as \(-\sum(-1)^n \frac{x^n}{n}\). The antiderivative is explicitly stated as \(x - x(-1 + \log[x]) - \log[1 + x] + x(-1 + \log[x])\log[1 + x] + \log[x]\log[1 + x] + \text{PolyLog}[2, -x]\). The final result of the definite integral is \(2 - \frac{\pi^2}{12} - \log[4]\).

PREREQUISITES
  • Understanding of definite integrals and their properties
  • Familiarity with series expansions, particularly Taylor series
  • Knowledge of logarithmic functions and their properties
  • Experience with polylogarithmic functions, specifically PolyLog[2, -x]
NEXT STEPS
  • Study series expansions of logarithmic functions in detail
  • Learn about numerical integration techniques for complex functions
  • Explore the properties and applications of polylogarithmic functions
  • Investigate integration by parts and its effectiveness in solving integrals
USEFUL FOR

Mathematicians, students studying calculus, and anyone interested in advanced integration techniques and series analysis.

CDrappi
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(this is not homework)

Suppose I wanted to solve:

[tex]\int log(x) log(x+1) dx[/tex] from 0 to 1.

I would turn ln(x+1) into a series, namely, –∑(-1)^n * x^n / n

Any ideas? Besides substituting, pulling out the n's, and using intgration by parts?
 
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what? why not just do it numerically? using the series representation of a fn then integrating terms by term doesn't help you in the least since if the integrated series had a simple representation then so would the antiderivative of the original function.

for what it's worth here's the antiderivative:

x - x (-1 + Log[x]) - Log[1 + x] + x (-1 + Log[x]) Log[1 + x] +
Log[x] Log[1 + x] + PolyLog[2, -x]

where PolyLog[2, -x] is the second order polylog function in -x.

the definite integral turns out to be:

2 - pi^2/12 - Log[4]
 

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