Proving the Falsehood of a Complex Exponential Integral Identity

While exploring a complex identity involving exponential integrals, we attempt to prove a statement that turns out to be false. This journey through the realm of complex analysis and integral calculus provides valuable insights into the properties and behavior of these mathematical objects.

Introduction

The identity in question involves the Exponential Integral Enx, which is defined as the following integral:

Enx ∫∞1 (e-xt/tn) dt

However, due to the complexity of this form, we will use a more simplified notation for simplicity:

Enx En(x)

The Statement to Prove

The statement we wish to prove is:

(frac{i}{2n} left( E_{frac{n-1}{n}}(i) - E_{frac{n-1}{n}}(-i) right) Gamma left( frac{n 1}{n} right) sin left( frac{pi}{2n} right) int_{0}^{1} sin(x^n) dx )

Step-by-Step Proof

The first step is to expand the Exponential Integrals, which will provide more insight into the problem:

( E_{frac{n-1}{n}}(i) int_{1}^{infty} frac{e^{it}}{t^{frac{n-1}{n}}} dt )

( E_{frac{n-1}{n}}(-i) int_{1}^{infty} frac{e^{-it}}{t^{frac{n-1}{n}}} dt )

Substituting these into our original equation, we get:

( frac{i}{2n} left( int_{1}^{infty} frac{e^{it}}{t^{frac{n-1}{n}}} dt - int_{1}^{infty} frac{e^{-it}}{t^{frac{n-1}{n}}} dt right) Gamma left( frac{n 1}{n} right) sin left( frac{pi}{2n} right) int_{0}^{1} sin(x^n) dx )

Using the Exponential Function Definition

Recall the definition of the sine function in terms of the exponential function:

( sin z frac{e^{iz} - e^{-iz}}{2i} )

Using this, we can simplify the summed integrals:

( int_{1}^{infty} frac{e^{it} - e^{-it}}{t^{frac{n-1}{n}}} dt int_{1}^{infty} frac{2i sin t}{t^{frac{n-1}{n}}} dt )

Substituting this back into our original equation:

( frac{i}{2n} int_{1}^{infty} frac{2i sin t}{t^{frac{n-1}{n}}} dt Gamma left( frac{n 1}{n} right) sin left( frac{pi}{2n} right) int_{0}^{1} sin(x^n) dx )

Simplifying further:

( frac{-1}{n} int_{1}^{infty} frac{sin t}{t^{frac{n-1}{n}}} dt Gamma left( frac{n 1}{n} right) sin left( frac{pi}{2n} right) int_{0}^{1} sin(x^n) dx )

Counterexample Analysis

To determine the validity of the statement, let's plug in n 1 and evaluate the integral:

( frac{-1}{1} int_{1}^{infty} frac{sin t}{t^{frac{1-1}{1}}} dt Gamma left( frac{1 1}{1} right) sin left( frac{pi}{2 cdot 1} right) int_{0}^{1} sin(x) dx )

Simplifying:

( -int_{1}^{infty} sin t dt cdot 1 cdot 1 int_{0}^{1} sin(x) dx )

( -int_{1}^{infty} sin t dt 1 - cos(1) )

( -int_{1}^{infty} sin t dt -cos(1) )

( int_{1}^{infty} sin t dt )

However, the integral ( int_{1}^{infty} sin t dt ) does not converge. This leads to a contradiction, thus proving the original statement is false:

( therefore frac{i}{2n} left( E_{frac{n-1}{n}}(i) - E_{frac{n-1}{n}}(-i) right) Gamma left( frac{n 1}{n} right) sin left( frac{pi}{2n} right) eq int_{0}^{1} sin(x^n) dx )