Solving Integrals Using the Laplace Transform: A Comprehensive Guide

When dealing with complex integrals, one powerful method is the use of the Laplace transform. This article demonstrates how to solve the integral related to the given expression using this technique.

Introduction

Consider the integral:

$int_{0}^{infty} xsin{sqrt{x}}e^{-x/2}dx$

To solve this, we can follow the approach using the Laplace transform. Let's go through the steps in detail.

Step 1: Compute the Laplace Transform of $sin{sqrt{t}}$

We start with the Maclaurin series for sine:

$sin{sqrt{t}} sum_{n0}^{infty} frac{(-1)^n t^{(2n-1)/2}}{(2n-1)!}$

Applying the Laplace transform to this series term by term, we get:

$mathcal{L}{sin{sqrt{t}}} sum_{n0}^{infty} frac{(-1)^n}{(2n-1)!} cdot frac{Gammaleft(frac{2n-1}{2} - 1right)}{s^{frac{2n-1}{2} - 1}}$

Using the Gamma function properties, we have:

$Gammaleft(frac{2n-3}{2}right) frac{1 cdot 3 cdot ... cdot (2n-1)}{2^{n-1}} sqrt{pi} frac{(2n-1)!}{2^{n-1} cdot 2^n cdot n!} sqrt{pi} frac{(2n-1)!}{2^{2n-1} cdot n!} sqrt{pi}$

Substituting this back, we get:

$mathcal{L}{sin{sqrt{t}}} frac{sqrt{pi}}{2s^{3/2}} sum_{n0}^{infty} frac{left(-frac{1}{4s}right)^n}{n!} frac{sqrt{pi}}{2s^{3/2}} e^{-frac{1}{4s}}$

Step 2: Apply the “Multiplication by t” Rule

Next, we apply the “multiplication by t” rule, which gives us:

$mathcal{L}{tsin{sqrt{t}}} -frac{d}{ds} left(frac{sqrt{pi}}{2s^{3/2}} e^{-frac{1}{4s}}right)$

After differentiating and simplifying, we obtain:

$mathcal{L}{tsin{sqrt{t}}} frac{sqrt{pi}}{8s^{7/2}} (6s - 1) e^{-frac{1}{4s}}$

Step 3: Substituting s 1/2

To find the original integral, we substitute $s frac{1}{2}$:

$int_{0}^{infty} xsin{sqrt{x}}e^{-x/2}dx frac{sqrt{pi}}{8(1/2)^{7/2}} (6(1/2) - 1) e^{-frac{1}{4(1/2)}}$

After simplifying, we get:

$boxed{2sqrt{frac{pi}{2e}}}$