Proving the Integral Using Complex Analysis

In this article, we will discuss how to prove the following integral using complex analysis:

[int_0^{infty} logleft(frac{x^2}{x^2 pi^2}right) log(1 - e^{-2x}) dx frac{pi^2}{4}]

To prove this, we will closely follow the steps outlined in the original solution, ensuring each step is detailed and explained thoroughly.

Setup and Contour Integration

We start by considering the function:

[f(z) log z log(1 - e^{-2z})]

We integrate this function over a specific contour in the complex plane. The contour consists of four segments: γ1, γ2, γ4, and γ6. To simplify the problem, we show that the integrals over γ2, γ4, and γ6 vanish as ε approaches 0 and R approaches infinity. This leaves us with:

[int_0^{infty} log x log(1 - e^{-2x}) dx - n int_{infty}^{0} log x log(1 - e^{-2x}) dx 0]

Let's call the sum of the first two integrals I1 and the last one I2.

Evaluation of Integral I1

For I1, we have:

[I_1 int_0^{infty} logleft(frac{x}{sqrt{x^2 pi^2}}right) log(1 - e^{-2x}) dx - n int_0^{infty} arg x log(1 - e^{-2x}) dx]

Noting that the real part of the first term is very interesting, we will focus on that.

Evaluation of Integral I2

For I2, we have:

[I_2 -i int_{pi}^{0} log y i log(1 - e^{-2 i y}) dy -i int_{pi}^{0} left[ log y i frac{pi}{2} right] left[ log 2 - 2 cos 2 y arg 1 - e^{-2 i y} right] dy]

Noting that we are interested in the real part of this expression, which is:

[text{Re} I_2 frac{pi}{2} int_0^{pi} log 2 - 2 cos 2 y dy - int_0^{pi} log y arg 1 - e^{-2 i y} dy J_1 J_2]

Calculation of J1

First, let's evaluate J1:

[J_1 frac{pi}{2} int_0^{pi} log 2 - 2 cos 2 y dy frac{pi}{2} int_0^{pi} log 4 sin^2 y dy 0]

The last line follows from the well-known integral of log sin x.

Calculation of J2

Next, let's evaluate J2:

[1 - e^{-2 i y} 1 - cos 2 y i sin 2 y 2 sin^2 y 2i sin y cos y 2 i sin y sin y i cos y 2i sin y e^{-i y}]

Up to unimportant constants, the multiplication by 2 i sin y is a rotation by frac{pi}{2}. Thus, in terms of argument, it is equivalent to a multiplication by e^{i frac{pi}{2}}. Therefore:

[arg(1 - e^{-2 i y}) arg(e^{i frac{pi}{2}} - y) frac{pi}{2} - y]

So J2 becomes:

[J_2 int_0^{pi} log y frac{pi}{2} dy - int_0^{pi} y log y dy -frac{pi^2}{4}]

The last line follows from elementary integration techniques.

Putting Everything Together

Putting everything together, we get:

[text{Re } I_1 - text{Re } I_2]

[int_0^{infty} logleft(frac{x}{sqrt{x^2 pi^2}}right) log(1 - e^{-2x}) dx frac{pi^2}{4}]

Multiplying by 2 and remembering the original (1/2), we have:

[I frac{pi^2}{4}]

As desired.