Question

Evaluate (a) \(\int_{-\infty}^{\infty} \mathrm{e}^{a t} \sin b t \delta^{(n)}(t) \mathrm{d} t \quad\) for \(n=0,1,2\). (b) \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta^{(n)}\left(t^{3}+t^{2}+t\right) d t \quad\) for \(n=0,1\).

Short answer

(a) At n=0, the integral is 0; at n=1, the integral is b; at n=2, the integral is \(2ab\). (b) At n=0, the integral is \(\cos(0) + \sin(0) + \cos(-1)+ \sin(-1)\); at n=1, the integral is 0.

Step-by-step solution

Evaluate Part (a)

In Part (a), to simplify the problem, replace \(e^{at}\sin(bt)\) with a general function f(t). This can now be represented as \(\int_{-\infty}^{\infty} f(t) \delta^{(n)}(t) dt\). Recall that the integral of the product of a general function and the nth derivative of the delta function equals to the nth derivative of that general function at 0. Taking this into account, when n=0, the integral will be f(0), when n=1, it will be f'(0), and when n=2, it will be f''(0). View \(e^{at}\sin(bt)\) as f(t), then f'(t)=a\(e^{at}\sin(bt)\) + b\(e^{at}\cos(bt)\) and f''(t)=2ab\(e^{at}\cos(bt)\) - b^2\(e^{at}\sin(bt)\) + a^2\(\e^{at}\sin(bt)\)

Plug in n values for Part (a)

By plugging these values, at n=0, the integral is \(e^{a*0}\sin(b*0)=0\), at n=1, the integral is \(a*e^{a*0}\sin(b*0) + b*e^{a*0}\cos(b*0)=b\), at n=2, the integral is \(2ab*e^{a*0}\cos(b*0) - b^2*e^{a*0}\sin(b*0) + a^2*e^{a*0}\sin(b*0) = 2ab\).

Evaluate Part (b)

Similar approach to part (a), but this time, note that we have \(t^3+t^2+t\) inside the delta function, which means that the delta function will be non zero when \(t^3+t^2+t =0\). This means our a value in the property will be the roots of \(t^3+t^2+t =0\), instead of the conventional '0'. Then, find the roots of this equation (which are, 0 and -1) and calculate the sum of the values of \(\cos(t) + \sin(t)\) at these roots.

Plug in n Values for part (b)

At n=0, the integral is \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta(t^3+t^2+t) dt = \cos(0) + \sin(0) + \cos(-1)+ \sin(-1)\), and at n=1, the integral becomes \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta'(t^3+t^2+t) dt\). But recall that the delta function only has 1st derivative when n=1 and its value is infinitive only at t=a, and we only have two roots at t=0 and t=-1, thus the integral value is zero.

Key concepts

Integrals

An integral is a fundamental concept in calculus used to find areas under curves and accumulate quantities. In the context of the delta function, integrals leverage the sifting property of the delta function.
The delta function, written as \( \delta(t) \), is a mathematical construct defined to be zero everywhere except at \( t=0 \), where it is infinitely high such that its integral over the entire real line is 1. This unique property allows integrals involving a delta function to "pick out" the values of other functions at specific points. In this exercise, when evaluating the integral \( \int_{-\infty}^{\infty} f(t) \delta^{(n)}(t) \, dt \), the solution boils down to evaluating the nth derivative of \( f(t) \) at \( t = 0 \).
For example:

• When \( n = 0 \), the value is simply \( f(0) \).
• When \( n = 1 \), it requires the first derivative, \( f'(0) \).
• If \( n = 2 \), then the second derivative, \( f''(0) \), is used.

Understanding integrals with the delta function is not just theoretical but prevents errors in accurately determining results specific to particular points of an equation.

Derivatives

Derivatives measure the rate at which a function changes as its input changes. In simplest terms, if you have a function \( f(t) \), its derivative \( f'(t) \) tells you how \( f(t) \) behaves for very small changes in \( t \).
When dealing with expressions like \( e^{at}\sin(bt) \), calculating derivatives becomes crucial:

• First derivative: \( f'(t) = a \cdot e^{at}\sin(bt) + b \cdot e^{at}\cos(bt) \)
• Second derivative: \( f''(t) = 2ab \cdot e^{at}\cos(bt) + (a^2 - b^2) \cdot e^{at}\sin(bt) \)

Evaluating derivatives at specific points allows us to simplify the integration process when paired with delta functions.
For instance, the nth order derivative evaluated directly at a root of the delta function's argument helps in capturing precise values needed in solving our integrals.

Roots of Equations

Finding the roots of an equation means solving for the values of \( t \) that make the entire expression zero. Roots are crucial not just in algebra but especially in evaluating integrals with delta functions when they're within a complex expression like \( t^3 + t^2 + t \).
In the given exercise, the expression inside the delta function leads us to find the roots of the polynomial equation:

• The equation is \( t^3 + t^2 + t = 0 \).
• Roots for this are determined using factorization or synthetic division resulting in solutions, which are \( t = 0 \) and \( t = -1 \).

Once identified, these roots are used to evaluate the function \( \cos(t) + \sin(t) \) at those specific points.
Solving for roots provides a pivot to transform intricate calculus problems into simpler arithmetic checks, which is both clever and efficient.

Sin and Cos Functions

The sine and cosine functions are essential trigonometric functions characterized by their periodic oscillations. They are often encountered in calculus problems involving waves and rotations.
The functions are defined as:

• Sine, \( \sin(t) \), represents the height of a point on a unit circle for a given angle \( t \).
• Cosine, \( \cos(t) \), signifies the horizontal distance from the center of the circle to a point along the circumference.

In the context of the exercise, both functions appear within the integral with the delta function. Their roles are to be evaluated at specific root points determined earlier.
For instance, within the integral \( \int_{-\infty}^{\infty} (\cos t + \sin t) \delta(t^3 + t^2 + t) \, dt \), the values of sine and cosine at \( t = 0 \) and \( t = -1 \) directly affect the solution.
Proper evaluation of trigonometric attributes forms the backbone of accurately calculating integrals related to oscillatory behavior within physical and engineering contexts.