Question

By using the definition and properties of the Laplace transform, compute each of the following integrals: (a) \(\int_{0}^{\infty} t e^{-2 t} \cos t d t\) (b) \(\int_{0}^{\infty} t^{3} e^{-t} \sin t d t\) (c) \(\int_{0}^{\infty} x^{4} e^{-x} d x\) (d) \(\int_{0}^{\infty} x^{6} e^{-3 x} d x\)

Short answer

(a) \( \frac{3}{25} \), (b) \( \frac{3}{8} \), (c) 24, (d) \( \frac{240}{729} \)."

Step-by-step solution

Understanding the Laplace Transform

The Laplace transform of a function \( f(t) \) is given by \( \mathcal{L}\{f(t)\}(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \). To solve integrals like the ones in this exercise, we essentially evaluate these integrals by recognizing them as specific Laplace transforms or using properties of Laplace transforms to simplify the expressions.

Part (a): Applying the Laplace Transform to Compute the Integral

We need to compute \( \int_{0}^{\infty} t e^{-2t} \cos t \, dt \). Recognize that this involves the Laplace transform of \( t \cos t \), for which we can use the shifting property. The Laplace transform of \( t \cos(at) \) is given by \( \frac{s^2-a^2}{(s^2+a^2)^2} \). Here, the transform is shifted by \( e^{-2t} \), implying a shift by \( s = 2 \). Substituting \( a=1 \), we find:\[ \mathcal{L}\{t\cos t\}(s) = \frac{s^2 - 1}{(s^2 + 1)^2} \].Substituting \( s = 2 \), we find:\[ \int_{0}^{\infty} t e^{-2t} \cos t \, dt = \mathcal{L}\{t\cos t\}(2) = \frac{2^2 - 1}{(2^2 + 1)^2} = \frac{3}{25} \].

Part (b): Applying the Laplace Transform to Compute the Integral

We need to evaluate \( \int_{0}^{\infty} t^{3} e^{-t} \sin t \, dt \). This is recognized as the Laplace transform of \( t^3 \sin t \). The transform for \( t^n \sin(at) \) is \( \frac{n! \, a}{(s^2 + a^2)^{n+1}} \). For \( n=3, a=1, \) and \( s=1 \),\[ \int_{0}^{\infty} t^3 e^{-t} \sin t \, dt = \frac{3! \, 1}{(1^2 + 1^2)^4} = \frac{6}{16} = \frac{3}{8} \].

Part (c): Using the Gamma Function Property

The integral \( \int_{0}^{\infty} x^{4} e^{-x} \, dx \) can be recognized as the Gamma function: \( \Gamma(n+1) = n! \). This matches the form \( \Gamma(5) \) for \( n=4 \). Therefore,\[ \int_{0}^{\infty} x^{4} e^{-x} \, dx = \Gamma(5) = 4! = 24 \].

Part (d): Using the Gamma Function Property with a Transformation

The integral \( \int_{0}^{\infty} x^{6} e^{-3x} \, dx \) can be evaluated by transforming it to a standard Gamma form. Let \( u = 3x \), then \( du = 3dx \) so that \( dx = \frac{du}{3} \). Substitute into the integral,\[ \int_{0}^{\infty} \left( \frac{u}{3} \right)^6 e^{-u} \frac{du}{3} = \frac{1}{3^7} \int_{0}^{\infty} u^{6} e^{-u} \, du \].Recognizing the second part as \( \Gamma(7) \), we have,\[ \Gamma(7) = 6! = 720 \].Thus,\[ \int_{0}^{\infty} x^{6} e^{-3x} \, dx = \frac{720}{3^7} = \frac{720}{2187} = \frac{240}{729} \].

Key concepts

Gamma Function

The Gamma Function is an important concept in mathematics, especially in calculus and complex analysis. It is denoted by \( \Gamma(n) \) and serves as a continuous extension of the factorial function to real and complex numbers. For positive integers, it is defined as:

• \( \Gamma(n) = (n-1)! \)

This means that for non-negative integers, \( \Gamma(n+1) = n! \). This property is particularly useful in evaluating integrals that resemble the gamma function form.
The Gamma Function is typically given by the integral:

• \( \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} \, dx \)

where \( n > 0 \). In the textbook exercise above, we use the Gamma Function to solve integrals like \( \int_{0}^{\infty} x^4 e^{-x} \, dx \). Since this fits the gamma form with \( n=5 \), it equals \( 4! \), which is 24.

Cosine Integral

The Cosine Integral is a specific integral that involves a cosine function. In general, the Laplace Transform is especially useful in solving integrals where a cosine function is multiplied by other terms, like exponential decay.
When you have an integral like \( \int_{0}^{\infty} t e^{-2t} \cos t \, dt \), you can often use properties of transformations. The Laplace Transform for such an integral involves recognizing a standard form \( t \cos(at) \):

• \( \mathcal{L}\{t \cos(at)\}(s) = \frac{s^2 - a^2}{(s^2 + a^2)^2} \)

By making substitutions, we integrate between the constant shifts and real values to evaluate the result.

Sine Integral

Sine Integrals are similar to Cosine Integrals but involve the sine function. These are often handled using the properties of the Laplace Transform, which are ideally suited for functions comprising exponential terms and trigonometric terms.
For integrals like \( \int_{0}^{\infty} t^3 e^{-t} \sin t \, dt \), we recognize it as the Laplace Transform of \( t^n \sin(at) \):

• \( \mathcal{L}\{t^n \sin(at)\}(s) = \frac{n! \, a}{(s^2 + a^2)^{n+1}} \)

In this context, understanding these transforms helps to compute integrals without directly solving the integral manually.

Shifting Property

The Shifting Property is a valuable tool used with the Laplace Transform, which helps in shifting variables to simplify integrals. This property allows consideration of shifts in the exponential term within the Laplace framework without changing the actual function.

• For a Laplace Transform \( \mathcal{L}\{e^{at}f(t)\}(s) = F(s-a) \)

This property becomes relevant when solving integrals like those in the exercises you encounter, where a substitution shifts the coefficients appropriately. For example, if you have \( \,e^{-2t} \), you substitute \( s = s+2 \) in its Laplace Transform. This shifting simplifies the integrals significantly.