Proving and Generalizing the Integral I_n

In this article, we explore the evaluation and generalization of the following integral:

[ I_n int_0^1 frac{ln^n x}{1-x} , mathrm{d}x ]

1. Evaluating Specific Case I_3

First, let's consider a specific case where ( n 3 ). We want to evaluate the integral:

[ I_3 int_0^1 frac{ln^3 x}{1-x} , mathrm{d}x ]

By noting the form of the denominator, we substitute the infinite geometric series formula which is valid in this region of integration and interchange the order of the sum and the integral:

[ I_3 sum_{n0}^{infty} (-1)^n int_0^1 x^n ln^3 x , mathrm{d}x ]

1.1. Evaluating the Inner Integral

For the inner integral, we apply integration by parts 3 times, followed by L'Hopital's rule to evaluate the resulting limits. This yields:

[ int_0^1 x^n ln^3 x , mathrm{d}x -frac{6}{(n 1)^4} ]

Therefore, we have:

[ I_3 6 sum_{n0}^{infty} frac{(-1)^{n 1}}{(n 1)^4} ]

Expanding this out term by term gives:

[ I_3 6 left( -frac{1}{1^4} - frac{1}{2^4} - frac{1}{3^4} - frac{1}{4^4} - ldots right) ]

This expression can be rewritten in terms of the Riemann zeta function as:

[ I_3 6 left( 2 sum_{n1}^{infty} frac{1}{(2n)^4} - zeta(4) right) ]

Since it is known that ( zeta(4) frac{pi^4}{90} ), we can conclude:

[ I_3 6 left( frac{1}{8} cdot frac{pi^4}{90} - frac{pi^4}{90} right) ]

Therefore:

[ I_3 -frac{7pi^4}{120} ]

2. Generalizing I_n

Now, we aim to derive a general formula for the integral ( I_n ). We introduce the generating function:

[ G(z) sum_{n0}^{infty} frac{I_n z^n}{n!} ]

2.1. The Generating Function Derivation

We have:

[ G(z) int_0^1 frac{1}{1-x} sum_{n0}^{infty} frac{(ln x)^n z^n}{n!} , mathrm{d}x ]

This simplifies to:

[ G(z) int_0^1 frac{e^{z ln x}}{1-x} , mathrm{d}x int_0^1 frac{x^z}{1-x} , mathrm{d}x ]

We split this integral into two parts:

[ G(z) frac{1}{2} int_0^1 frac{t^{z/2} - t^{1/2}}{1-t} , mathrm{d}t frac{1}{2} left[ int_0^1 frac{t^{z-1/2} - t^{z/2}}{1-t} , mathrm{d}t right] ]

These integrals are Euler's representations of the harmonic number, given by:

[ H_n int_0^1 frac{1-t^n}{1-t} , mathrm{d}t ]

Thus:

[ G(z) frac{1}{2} left[ H_{z/2} - H_{(z-1)/2} right] ]

Using the fact that the harmonic number relates to the Digamma function:

[ H_n psi(n 1) - gamma ]

We get:

[ G(z) frac{1}{2} left[ psi(z/2 1) - psi((z-1)/2 1) right] ]

2.2. Taylor Series Expansion

The Taylor series expansion of ( G(z) ) is:

[ G(z) sum_{n0}^{infty} frac{z^n}{n!} G^{(n)}(0) ]

This gives us:

[ I_n G^{(n)}(0) frac{1}{2^{n-1}} left[ psi^{(n)}(1) - psi^{(n)}left(frac{1}{2}right) right] ]

3. Specific Values of I_n

For the specific values, we calculate:

3.1. I_0

[ I_0 frac{1}{2} left[ psi(1) - psileft(frac{1}{2}right) right] ln 2 ]

3.2. I_1

[ I_1 frac{1}{2^2} left[ psi'(1) - psi'left(frac{1}{2}right) right] -frac{pi^2}{12} ]

3.3. I_2

[ I_2 frac{1}{2^3} left[ psi''(1) - psi''left(frac{1}{2}right) right] 1.8030ldots ]

3.4. I_3

[ I_3 frac{1}{2^4} left[ psi'''(1) - psi'''left(frac{1}{2}right) right] -frac{7pi^4}{120} ]