Question

Use the change of variables \(k=\ln t\) and \(i x=\omega-\alpha\) (where \(k\) and \(x\) are the common variables used in Fourier transform equations) to show that the Fourier transform changes into a Mellin transform, $$G(t)=\frac{1}{2 \pi i} \int_{-i \infty+\alpha}^{i \infty+\alpha} F(\omega) t^{-\omega} d \omega, \quad \text { where } \quad F(\omega)=\int_{0}^{\infty} G(t) t^{\omega-1} d t .$$

Short answer

The Fourier transform \(\int_{0}^{\infty} G(t) e^{-kx}\,dt\), becomes \(\int_{0}^{\infty} G(t) t^{-x}\,dt\) using \(k=ln(t)\), and then it becomes Mellin transform \(F(\omega)=\int_{0}^{\infty} G(t) t^{\omega-1} \, dt\) using variable substitution \(x=\omega/i +\alpha\). The Mellin inversion of \(F(w)\), is \(G(t)=(1/(2 \pi i)) \int_{-i\infty+\alpha}^{+i\infty+\alpha} F(w) t^{-w} \, dw\). Thus, the Fourier transform changes into a Mellin transform with the prescribed change of variables.

Step-by-step solution

Understanding the variables

The Fourier transform is being represented with common variables \(k\) and \(x\). The change of variables is defined as \(k=ln(t)\) and \(ix=\omega-\alpha\). The objective is to show that through this change of variables, the Fourier transform transforms into a Mellin transform. The Mellin transform involves an integral over a complex contour.

Change the Variables in the Fourier Transform

To begin, let us first substitute \(k=ln(t)\) in the Fourier transform of \(F(k)\). We write \(F(k)\) as \(\int_{0}^{\infty} G(t) e^{-kx}\,dt\). When we substitute \(k=ln(t)\), we get \(\int_{0}^{\infty} G(t) e^{-ln(t)x}\,dt\). This simplifies to \(\int_{0}^{\infty} G(t) t^{-x}\,dt\).

Change the exponent into \(\omega\) by substitution

According to our change of variables \(ix=\omega-\alpha\), so \(x=\omega/i +\alpha\). Substitute \(x\) in \(\int_{0}^{\infty} G(t) t^{-x}\,dt\) with this value of \(x\). The transform now becomes \(F(\omega)=\int_{0}^{\infty} G(t) e^{-t^{(\omega/i +\alpha)}}\,dt\). Now, multiply numerator and denominator by \(t\), the integral becomes \(F(w) = \int_{0}^{∞} G(t) t^{\omega-1} \, dt\), which corresponds to the Mellin transform of \(F(w)\), given that \(F(w)=\int_{0}^{\infty} G(t) t^{\omega-1} \, dt\).

Represent \(G(t)\) in terms of Mellin Inversion Integral

Now, \(G(t)\) is the Mellin inversion of \(F(\omega)\), which means \(G(t)=(1/(2 \pi i)) \int_{-i\infty+\alpha}^{+i\infty+\alpha} F(w) t^{-w} \, dw\). This completes the demonstration that using the given change of variables, the Fourier transform changes into a Mellin transform.

Key concepts

Change of Variables

Understanding the 'change of variables' is crucial in transforming one type of integral into another. It's like translating a sentence from one language to another; we maintain the essence but change the form. In the context of mathematical transforms, a clever change of variables can reveal connections between seemingly different concepts.

In the exercise, we encounter a change of variables that alters the appearance of a Fourier transform into that of a Mellin transform. The variables are set as follows: for the Fourier transform, typically denoted with variables such as 'k' and 'x', we define new relationships by setting k = ln(t) and ix = \(\omega - \alpha\), where 't' and '\(\omega\)' become the new variables of interest.

When this substitution is used in the Fourier integral, it translates the exponential function into a power function. To put it simply, where you originally had an exponent involving 'k' and 'x', after substituting 'k' with 'ln(t)', this exponent turns into a power of 't' with 'x' as the exponent. Once we address the variable 'x' with '\(\omega/i + \alpha\)', it changes the form of the integral to one that characterizes the Mellin transform. This is an elegant illustration of how a strategic change of variables can link two different mathematical frameworks.

Complex Contour Integral

Delving into the realm of complex analysis, a 'complex contour integral' might sound intimidating, but at its heart, it's a way to integrate functions over a path in the complex plane. Imagine tracing a path through a landscape of complex numbers and adding up values of a function along this journey.

In the Fourier and Mellin transforms, we work with complex contour integrals because we're integrating functions of a complex variable, 'w', over a path that lies in the complex plane. This path is defined by the limits of integration: from '\( -i \infty + \alpha \)' to '\( i \infty + \alpha \)', a vertical line in the complex plane shifted by a real number '\(\alpha\)'.

Why does this matter?

These types of integrals are essential in the definition of the inverse Mellin transform, which allow us to recover our original function 'G(t)'. What we are effectively doing is summing the contributions of the function 'F(w)' along the vertical contour, weighted by the factor '\(t^{-w}\)'. It's a process that might seem abstract but enables us to traverse back and forth between two different function representations, thanks to the power of complex analysis.

Mellin Inversion Integral

The Mellin inversion integral is the reverse process of the Mellin transform. If we think of the Mellin transform as a way to encode a function into a new form, then the Mellin inversion integral decodes it back to its original form. It's akin to having a secret message that you encode into cipher text—the Mellin inversion is the key you use to decipher that message back into plain text.

In the exercise, we express 'G(t)' as an integral that involves 'F(w)', the Mellin transform of 'G(t)'. This representation is given by:\[ G(t) = \frac{1}{2\pi i} \int_{-i\infty+\alpha}^{i\infty+\alpha} F(\omega) t^{-\omega} d\omega \].

Picture this: you have a function 'F(w)' transformed into the 'frequency domain' by the Mellin transform, and you need to revert it into the original time function 'G(t)'. The Mellin inversion does this by integrating 'F(w)' over a complex contour, as mentioned previously, and scaling it with '\(t^{-\omega}\)', which adjusts the function for each 't' back into the time domain. This operation underscores the duality between the Mellin transform and its inversion, highlighting the symmetry between transforming and inverting functions within this mathematical framework.