Integral Transform

In this topic we have introduced two extremely powerful methods to solving differential equations i.e the Fourier and the Laplace transforms. Besides its practical use, the Fourier transform is also used in quantum mechanics, by providing the correspondence between the position and momentum representations of the Heisenberg commutation relations. An integral transform is very useful as it allows the transformation of a complicated problem into a simpler one. The transformation we will be studying in this chapter is mostly useful to solve differential equations and, to a lesser extent, integral equations. Here we will learn about Fourier transform integral, Laplace transform of integration, and their applications along with real-life applications.

Integral transform is a mathematical operation that produces a new function f(y) after integrating the product of an existing function F(x), and it has a kernel function K(x,y) between the suitable limits. The process is called transformation, and it is symbolically represented by the equation f(y) = ∫K(x, y)F(x)dx. Several transforms are named by the name of mathematicians who introduced them such as the Laplace transform the kernel is \[e^{-xy}\] and the limits of integration are zero and plus infinity. The Fourier transform the kernel is \[(2\prod)^{\frac{-1}{2}}e^{-ixy}\] and the limits are minuses and plus infinity.

What is the Laplace Transform?

A piecewise continuous function is a function that has a finite number of breaks and it does not continue up to infinity anywhere. Let assume the function f(t) is a piecewise continuous function, then f(t) is defined using the Laplace transform. The Laplace transformation of a function is represented by the equation L{f(t)} or F(s). It helps to solve the differential equations, where it reduces the differential equation into an algebraic problem.

Fourier Integral

The representation of a function given on a finite set of intervals by a Fourier series is very important. The representation of a function f given on the whole axis by a Fourier integral is given below:

\[f(x)=\int_{0}^{\infty}\left [ A(\lambda)cos\lambda x + B(\lambda)sin\lambda x\right]d\lambda\]

Where \[A(\lambda)=\frac{1}{\prod}\int_{-\infty}^{\infty}f(\xi)cos\lambda \xi d\xi\] 

\[B(\lambda)=\frac{1}{\prod}\int_{-\infty}^{\infty}f(\xi)sin\lambda \xi d\xi\]

Mellin Transform

The Mellin transform is mostly useful for various applications like solving Laplace’s equation in polar coordinates, as well as is used for estimating integrals.

Laplace Transform of Integral

If \[G(s)=\mathcal{L}\left\{g(t)\right\},then\mathcal{L}\left\{\int_{0}^{t}g(t)dt\right\}=\frac{G(s)}{s}\].

For the general integral, if

\[\left [ \int g(t)dt \right ]_{t=0}\]

when the value of the integral  t=0, then:

\[\mathcal{L}\left\{\int g(t)dt\right\}=\frac{G(s)}{s}+\frac{1}{s} \left [ \int g(t)dt \right ]_{t=0}\]

Let’s understand this with an example

Find the Laplace transform of \[\int_{0}^{t}sin\,at\,cos\,at\,dt\] 

Sol: We know the formula of \[sin\,2\alpha=2 sin\alpha\,cos\alpha\]

We the help of the above formula we can rearrange our integrand, and then integrate

\[sin\,at+cos\,at=\frac{1}{2}sin\,2at\]

So the Laplace Transform of the integral becomes:

\[\mathcal{L}\{\int_{0}^{t}sin\,at\,cos\,at\,dt\}=\frac{1}{2}\mathcal{L}\{\int_{0}^{t}sin\,2at\,dt\}\] 

\[=\frac{1}{2}\frac{2a}{s(s^{2}+4a^{2})}\]

\[=\frac{a}{s(s^{2}+4a^{2})}\]

Hence this is the required answer.

Fourier Transform Integral

If the complex function \[g\epsilon L^{2}(R)\] then the function is given by the Fourier integral, i.e.

\[f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{ikx}dk\]

Exists and \[f\epsilon L^{2}(R)\] Furthermore, we have the equality

\[\int_{-\infty}^{\infty}\left|f(x)\right|^{2}dx=\int_{-\infty}^{\infty}\left|g(k)\right|^{2}dx\],

The function g(k) is called Fourier transform of f(x) and it can be recovered from the

following inverse Fourier integral

\[g(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\]

Laplace Stieltjes Transform

When we have to investigate the growth and approximation of entire functions we use Laplace Stieltjes transform. It is convergent on the whole complex plane. For Laplace-Stieltjes transforms,

\[G(s)=\int_{0}^{+\infty}e^{-sx}d\alpha(x),\,s=\sigma +it\]

Where \alpha(x) is a bounded variation on any finite interval [0, Y]\[(0<Y<+\infty)\]  and σ and t are real variables. In fact, if \[\alpha(t)\] is absolutely continuous, then G(s) becomes the classical Laplace integral form:

\[G(s)=\int_{0}^{\infty}e^{-st}\varphi(t)dt\]

Fourier Sine Integral

The Fourier integral of an odd function in the interval \[(-\infty,\infty)\] is the sine integral of the function

\[f(x)=\frac{2}{\prod}\int_{0}^{\infty}B(\alpha)sin\alpha x\,d\alpha\]

Where \[B(\alpha)=\int_{0}^{\infty}f(x)sin\alpha x\,dx\]

Laplace Transform Formula

Laplace transform is the integral transform of the given derivative function with respect to real variable t.  It is also used to convert into a complex function with variable s. For t ≥ 0, let f(t) be given and assume the function satisfies certain conditions to be stated later on.

The Laplace transform of f(t), that is denoted by L{f(t)} or F(s) is defined by the Laplace transform formula:

\[F(s)=\int_{0}^{\infty}f(t).e^{-s.t}dt\]

whenever the improper integral converges.

Standard notation: Where the notation is clearly known, we use an uppercase letter to indicate the Laplace transform, for e.g, L(f; s) = F(s).

Sometimes we define Laplace transform as the one-sided Laplace transform.  Two-sided version Laplace transform is also available where the integral goes from \[(-\infty\,to\,\infty)\].

Integration of Fourier Series

Consider g(x) to be a \[2\prod\] periodic piecewise continuous function on the interval  \[\left[-\prod,\prod\right]\]. Then this function can be integrated term by term on this interval. The Fourier series for g(x) is given by

\[g(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}cosnx+b_{n}sinnx)\]

Consider the function

\[G(x)\int_{0}^{x}g(t)dt\sim\frac{A_{0}}{2}+\sum_{n=1}^{\infty}(A_{n}cosnx+B_{n}sinnx)\]

Where \[A_{n}=-\frac{b_{n}}{n},B_{n}=\frac{a_{n}}{n}\]

By setting x=0, we see that

\[G(0)=0=\frac{A_{0}}{2}+\sum_{n=1}^{\infty}A_{n}=\frac{A_{0}}{2}-\sum_{n=1}^{\infty}\frac{b_{n}}{n}\]

or \[\frac{A_{0}}{2}=\sum_{n=1}^{\infty}\frac{b_{n}}{n}\]

Therefore, the Fourier series expansion of the function G(x) is given  by 

\[G(x)\int_{0}^{x}g(t)dt=\int_{0}^{x}\frac{a_{0}}{2}dx+\sum_{n=1}^{\infty}(a_{n}cosnx+b_{n}sinnx)dx\]

\[=\frac{a_{0}x}{2}+\sum_{n=1}^{\infty}\frac{a_{n}sinnx+b_{n}(1-cosnx)}{n}\]

where the series on the right-hand side is obtained by the formal step-by-step integration of the Fourier series for g(x).

As the presence of the term depending on x, on the right-hand side, this is not clearly a Fourier series expansion of the integral of g(x). The result we can rearrange in the form of Fourier series expansion of the function

\[\Phi(x)=\int_{0}^{x}g(t)dt-\frac{a_{0}x}{2}\]

The Fourier series of the function (x)is given by the expression

\[\Phi(x)=\int_{0}^{x}g(t)dt-\frac{a_{0}x}{2}\]

\[=\frac{A_{0}}{2}+\sum_{n=1}^{\infty}{A_{n}cos\,nx+B_{n}sin\,nx}\]

where the Fourier coefficients are defined by using the following relationships:

\[\frac{A_{0}}{2}=\sum_{n=1}^{\infty}\frac{b_{n}}{n},A_{n}=-\frac{b_{n}}{n},B_{n}=\frac{a_{n}}{n}\]

Inverse Laplace Transform Integral

In the inverse Laplace transform, the transformation of  F(s) is given and we have to find what function we have initially. The inverse transform of the function F(s) is given as:

\[f(t)=L^{-1}\left\{{F(s)}\right\}\]   

Let’s understand the two Laplace transform.  Suppose the two functions are F(s) and G(s), the inverse Laplace transform is defined by:

\[L^{-1}\left\{aF(s)+bG(s)\right\}=aL^{-1}\left\{F(s)+bL^{-1}\{G(s)\right\}\]

Where a and b are constants.

In the above case, we can take the inverse transform for the individual transforms, and then add their constant values in their respective places, that perform the operation to get the result.

Laplace Transform of Integral Function

According to the theorem 

If \[L\left\{f(t)\right\}=F(s)\], then

\[L\left[\int_{0}^{t}f(u)du\right]=\frac{F(s)}{s}\]

Conclusion

Integral transforms are very useful in mathematical analysis. It is used to solve differential and integral equations, and to study special functions and compute integrals. Applications of integral transform fractional, including fractional methods.

Laplace transform is used to convert complex differential equations into a simpler form that has polynomials. It is used to convert derivatives into multiple domain variables and finally convert the polynomials back to the differential equation using Inverse Laplace transformation. When we apply Fourier transform to a partial differential equation it reduces the number of independent variables by one. We also use Fourier Transform in signal and image processing. It is also useful in cell phones, LTI systems & circuit analysis.