Evaluating the Improper Integral ( I int_0^infty frac{x^3 ln x}{1 - x^3} , dx ) Using Advanced Techniques
Consider the improper integral ( I int_0^infty frac{x^3 ln x}{1 - x^3} , dx ). This integral can be evaluated by applying complex techniques involving the gamma function and advanced integration methods.
1. Introduction
The goal of this article is to demonstrate how to evaluate the integral using an elegant combination of integration by parts and properties of the gamma function. We will go through the detailed process, including derivations and explanations of each step, to ensure a clear understanding.
2. Evaluating the Integral Using the Gamma Function
To start, we introduce the function ( J_a ) defined as:
[ J_a int_0^infty frac{x^a}{(1 - x^3)^2} , dx. ]
Note that we can express ( I ) in terms of ( J_a ) by adjusting the power of ( x ) and the logarithmic term:
[ I int_0^infty frac{x^3 ln x}{(1 - x^3)^2} , dx left. frac{d}{da} J_a right|_{a3}. ]
Next, we need to derive the form of ( J_a ).
2.1 Finding ( J_a )
Begin by making the substitution ( y frac{1}{1 - x^3} ), which leads to:
[ J_a int_0^infty frac{x^{-frac{1}{3}} left(1 - x^3right) y^2 left(1 - y^{frac{a-2}{3}}right)}{3 y^2} , dy frac{1}{3} int_0^infty y^{frac{2-a}{3}} left(1 - y^{frac{a-2}{3}}right) , dy. ]
This integral can be expressed using the Beta function:
[ J_a frac{1}{3} Bleft(frac{5-a}{3}, frac{a-2}{3}right) frac{1}{3} Gammaleft(frac{5-a}{3}right) Gammaleft(frac{a-2}{3}right). ]
Using the reflection property of the gamma function, we obtain:
[ J_a -frac{pi}{9} left(frac{a-2}{3}right) cscleft(frac{pi a}{3}right). ]
2.2 Differentiating ( J_a )
To find ( I ), we differentiate ( J_a ) with respect to ( a ) and evaluate at ( a 3 ):
[ J_a' -frac{pi}{9} cdot frac{1}{3} left(3 - 2 cotleft(frac{pi a}{3}right) - pi csc^2left(frac{pi a}{3}right)right). ]
Evaluating at ( a 3 ):
[ J_3' -frac{pi}{27} left(3 - 2 cotleft(frac{4pi}{3}right) - pi csc^2left(frac{4pi}{3}right)right). ]
Since ( cotleft(frac{4pi}{3}right) -frac{sqrt{3}}{3} ) and ( cscleft(frac{4pi}{3}right) -frac{2}{sqrt{3}} ), we get:
[ J_3' -frac{pi}{27} left(3 frac{2pi}{3sqrt{3}} - frac{2pi}{2sqrt{3}}right) frac{2pi}{81} left(3sqrt{3} - piright). ]
3. Confirming the Result
Combining all the intermediate results and simplifying, we conclude:
[ I int_0^infty frac{x^3 ln x}{1 - x^3} , dx boxed{frac{2pi}{81} 3sqrt{3} - pi}. ]
4. Conclusion
We have successfully evaluated the integral ( I ) using the gamma function and advanced integration techniques. This result highlights the power of these methods in solving challenging improper integrals.