Question

Using integration by parts, find the Laplace transform of the given function; \(n\) is a positive integer and \(a\) is a real constant. $$ t \sin a t $$

Short answer

#Question# Find the Laplace transform of the function \(t \sin at\). #Answer# The Laplace transform of the function \(t \sin at\) is \(-\frac{s}{a(s^2 + a^2)}\).

Step-by-step solution

Recall the Laplace Transform Definition

The Laplace transform of a function \(f(t)\) is defined as: $$ \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)dt $$ where \(s\) is a complex variable.

Identify Functions for Integration by Parts

In this case, we are given the function \(f(t) = t \sin at\). To apply integration by parts, we will need to choose functions \(u(t)\) and \(v'(t)\). We choose: $$ u(t) = t \quad\text{and}\quad v'(t) = \sin at $$

Find Derivatives and Integrals

Now, we will find the derivative of \(u(t)\) and the integral of \(v'(t)\): $$ u'(t) = \frac{du}{dt} = 1 $$ $$ v(t) = \int v'(t)dt = \int \sin at dt = -\frac{1}{a}\cos at \;+ C $$ Since we are integrating over a definite range, we can ignore the constant \(C\) for now.

Apply Integration by Parts Formula

Now we will apply the integration by parts formula for the Laplace transform: $$ \int_0^\infty e^{-st} t\sin at dt = \left[-e^{-st}t \cdot \frac{1}{a}\cos at\right]_0^\infty + \int_0^\infty e^{-st} \left(-\frac{1}{a}\cos at\right) dt $$

Evaluate the Integral

We can now evaluate the integral: $$ F(s) = \left[-e^{-st}t \cdot \frac{1}{a}\cos at\right]_0^\infty + \int_0^\infty e^{-st} \left(-\frac{1}{a}\cos at\right) dt = -\frac{1}{a}\int_0^\infty e^{-st}\cos(at) dt $$ Now we apply the Laplace transform formula for \(g(t) = \cos(at)\): $$ \mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2} $$ So, $$ F(s) = -\frac{1}{a} \cdot \frac{s}{s^2 + a^2} $$ Hence, the Laplace transform of the given function, \(t \sin at\), is: $$ \mathcal{L}\{t\sin at\} = -\frac{s}{a(s^2 + a^2)} $$

Key concepts

Integration by Parts

When faced with the daunting task of integrating the product of two functions, the method of integration by parts proves immensely useful. This technique is based on the product rule for differentiation and is formally stated as follows: for functions u(t) and v(t), the integral of u dv equals uv minus the integral of v du, or in mathematical terms,
\[\int u(t) \frac{dv(t)}{dt} dt = u(t)v(t) - \int v(t) \frac{du(t)}{dt} dt\].
The strategy involves choosing u(t) and dv(t) in such a manner that the resulting integral \int v du is simpler to evaluate than the original one. In our exercise, u(t) = t and dv(t) = \sin(at) dt are chosen, and through calculated differentiation and integration, we arrive at the functions u'(t) and v(t), which are used to transform and simplify the given integral into a more manageable form. This transformation is the key to solving integrals that are otherwise difficult to handle directly.

Definite Integrals

At the heart of many calculus problems lie definite integrals. These integrals are not merely operations of computation but represent the precise accumulation of quantities, such as areas under curves, over a specific interval. In contrast to indefinite integrals, which yield a family of functions plus a constant of integration, definite integrals have limits of integration and evaluate to a scalar number. Their notation is given by
\[\int_{a}^{b} f(t) dt\],
where a and b are the lower and upper limits, respectively. When computing the Laplace transform through definite integration from 0 to infinity, special consideration has to be given to the behavior of the function as t approaches infinity to ensure that the integral converges. Furthermore, while constants of integration C appear when finding indefinite integrals for intermediate steps, they do not affect the final result of a definite integral since they cancel out when the integral limits are applied.

Laplace Transform of Trigonometric Functions

The Laplace transform of trigonometric functions unveils a bridge between time-domain functions and their counterparts in the frequency domain. Trigonometric functions like \sin(at) and \cos(at) are essential in describing oscillatory systems, and their Laplace transforms reduce these infinite oscillations to algebraic forms, enabling easier analysis and solution to differential equations involving them. For instance,
\[\mathcal{L}\{\sin(at)\} = \frac{a}{s^2 + a^2}\]
and
\[\mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2}\].
These transforms play a pivotal role when we are working with the product of a polynomial function, such as t, and a trigonometric function, as seen in the exercise. Once the integration by parts is performed, we utilize the Laplace transform of \cos(at) to simplify the result into a final expression. Learning these transforms not only expedites the solving process but also enhances understanding of the behavior of electrical circuits, mechanical systems, and other applications within engineering and physics.