Evaluating the Laplace Transform and Inverse Transform of a Specific Integral

In this article, we will explore the evaluation of a specific integral It using the Laplace transform and its inverse. We will walk through each step, break down the process into comprehensible parts, and explain the underlying mathematical concepts.

Introduction

The integral in question is:

[I_{t} int_{0}^{infty} frac{x sin{xt}}{(x^{2} - 1)^{3}} , dx]

Our goal is to evaluate this integral using the Laplace transform method and subsequent inverse transform.

Transforming the Integral

We start by taking the Laplace transform of It:

[mathcal{L} [I_{t}] int_{0}^{infty} frac{x}{(x^{2} - 1)^{3}} int_{0}^{infty} sin{xt} e^{-st} , dt , dx]

By interchanging the order of integration, we obtain:

[mathcal{L} [I_{t}] int_{0}^{infty} frac{x}{(x^{2} - 1)^{3}} cdot frac{x}{x^{2} s^{2}} , dx]

This simplifies to:

[mathcal{L} [I_{t}] frac{1}{s^{2}} int_{0}^{infty} frac{x^{2}}{(x^{2} - 1)^{3}} , dx]

Decomposing the Integral

We now decompose the integral into partial fractions:

[mathcal{L} [I_{t}] frac{s^{2}}{(s^{2} - 1)^{3}} A - frac{s^{2}}{(s^{2} - 1)^{3}} B frac{s^{2}}{(s^{2} - 1)^{2}} C - frac{1}{s^{2} - 1} D]

Where:

[A int_{0}^{infty} frac{1}{x^{2} s^{2}} , dx frac{pi}{2s}] [B int_{0}^{infty} frac{1}{x^{2} - 1} , dx frac{pi}{2}] [C int_{0}^{infty} frac{1}{(x^{2} - 1)^{2}} , dx frac{pi}{4}] [D int_{0}^{infty} frac{1}{(x^{2} - 1)^{3}} , dx frac{3pi}{16}]

Simplifying and Evaluating the Transform

Substituting these values into the expression for (mathcal{L} [I_{t}]), we get:

[mathcal{L} [I_{t}] frac{pi}{16} cdot frac{1}{s 1} - frac{pi}{8} cdot frac{1}{s 1^{3}}]

This simplifies to:

[mathcal{L} [I_{t}] frac{pi}{16 (s 1)} - frac{pi}{8 (s 1)^{3}}]

Inverse Laplace Transformation

To find (I_{t}), we take the inverse Laplace transform:

[I_{t} frac{pi}{16} mathcal{L}^{-1}left[frac{1}{s 1}right] - frac{pi}{8} mathcal{L}^{-1}left[frac{1}{(s 1)^{3}}right]]

Which evaluates to:

[I_{t} frac{pi}{16} e^{-t} - frac{pi}{24} t^{2} e^{-t}]

As (t rightarrow 0), the integral simplifies to:

[I_{1} int_{0}^{infty} frac{x sin x}{(x^{2} - 1)^{3}} , dx frac{pi}{8e}]

Conclusion

Through the use of the Laplace transform and its inverse, we have successfully evaluated the integral (I_{1}). This process demonstrates the power and versatility of the Laplace transform method in solving complex integrals.