Question

Evaluate (a) \(\int_{0}^{3}\left(1+e^{-t}\right) \delta(t-2) d t\) (b) \(\int_{-2}^{1}\left(1+e^{-t}\right) \delta(t-2) d t\) (c) \(\int_{-1}^{2}\left[\begin{array}{c}\cos 2 t \\ t e^{-t}\end{array}\right] \delta(t) d t\) (d) \(\int_{-3}^{2}\left(e^{2 t}+t\right)\left[\begin{array}{c}\delta(t+2) \\\ \delta(t-1) \\ \delta(t-3)\end{array}\right] d t\)

Short answer

Question: Evaluate the given integrals involving the Dirac Delta function. a) $\int_{0}^{3}\left(1+e^{-t}\right) \delta(t-2) dt$ b) $\int_{-2}^{1}\left(1+e^{-t}\right) \delta(t-2) dt$ c) $\int_{-1}^{2}\left[\begin{array}{c}\cos 2t \\ t e^{-t}\end{array}\right] \delta(t) dt$ d) $\int_{-3}^{2}\left(e^{2t}+t\right)\left[\begin{array}{c}\delta(t+2) \\ \delta(t-1) \\ \delta(t-3)\end{array}\right] dt$ Answer: a) $1 + e^{-2}$ b) $0$ c) $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ d) $\begin{bmatrix} \frac{1}{4}-2 \\ e+1 \\ 0 \end{bmatrix}$

Step-by-step solution

1. Evaluate the first integral (a)

Recall the sifting property of the Dirac Delta function. For the first integral: $$\int_{0}^{3}\left(1+e^{-t}\right) \delta(t-2) d t$$ Since the integration interval is \([0, 3]\), the Dirac Delta function is activated at \(t = 2\). Therefore, using the sifting property, we have: $$\int_{0}^{3}\left(1+e^{-t}\right) \delta(t-2) d t = 1 + e^{-2}$$

2. Evaluate the second integral (b)

In the second integral: $$\int_{-2}^{1}\left(1+e^{-t}\right) \delta(t-2) d t$$ Since the integration interval is \([-2, 1]\), the Dirac Delta function is not activated, as \(t = 2\) is not within this interval. Therefore, there is no contribution to the integral, and it evaluates to 0: $$\int_{-2}^{1}\left(1+e^{-t}\right) \delta(t-2) d t = 0$$

3. Evaluate the third integral (c)

In the third integral: $$\int_{-1}^{2}\left[\begin{array}{c}\cos 2 t \\\ t e^{-t}\end{array}\right] \delta(t) d t$$ The Dirac Delta function is activated at \(t = 0\), and we apply the sifting property to each component separately: $$\int_{-1}^{2}\cos 2t \cdot \delta(t) d t = \cos(2\cdot 0) = 1$$ $$\int_{-1}^{2} t e^{-t} \cdot \delta(t) d t = 0$$ Hence, the integral evaluates to: $$\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

4. Evaluate the fourth integral (d)

In the fourth integral: $$\int_{-3}^{2}\left(e^{2 t}+t\right)\left[\begin{array}{c}\delta(t+2) \\\ \delta(t-1) \\\ \delta(t-3)\end{array}\right] d t$$ We need to evaluate each component separately, applying the sifting property for each Dirac Delta function. The functions \(e^{2t}+t\) and \(t+2\), \(t-1\), \(t-3\) are continuous during the interval of integration. $$\int_{-3}^{2} (e^{2t}+t) \delta(t+2) d t = e^{2(-2)} + (-2) = \frac{1}{4}-2$$ $$\int_{-3}^{2} (e^{2t}+t) \delta(t-1) d t = e^{2(1)} + 1 = e+1$$ $$\int_{-3}^{2} (e^{2t}+t) \delta(t-3) d t = 0$$ Hence, the integral evaluates to: $$\begin{bmatrix} \frac{1}{4}-2 \\ e+1 \\ 0 \end{bmatrix}$$

Key concepts

Sifting Property

The sifting property, also known as the sampling property, is one of the cornerstone features of the Dirac Delta function, denoted as \( \delta(t) \). Imagine the Delta function as a filter that picks out the value of another function exactly at the point where the Delta function is 'activated'. Mathematically, when you integrate a function \( f(t) \) multiplied by \( \delta(t-a) \) over an interval that includes \( t = a \) the result is \( f(a) \) if \( a \) lies within the interval of integration.

For instance, given an integral \( \int_{a}^{b} f(t) \delta(t-c) dt \), if \( a \leq c \leq b \) then the result is simply \( f(c) \), just as we saw in Step 1 of the solution. However, if \( c \) is outside of the interval \( [a, b] \), as in Step 2, the integral evaluates to zero since the Delta function is not 'activated'. This remarkable property makes the Dirac Delta function an incredibly valuable tool for extracting information at specific points.

Definite Integral Evaluation

When evaluating definite integrals involving the Dirac Delta function, the process is guided by the limits of the integration interval. The definite integral \( \int_{a}^{b} f(t) \delta(t-c) dt \) will only be non-zero under the condition that \( c \) falls within the range \( a \leq c \leq b \). If this condition is met, the integration acts as a functional sifter, plucking out the value of \( f(t) \) at \( t = c \) and discarding everything else.

One common source of confusion is when multiple Delta functions are included, such as in Step 4 of the solution. It’s essential to evaluate each Delta function's contribution separately at its respective activation point. This step-wise approach simplifies the evaluation of complicated definite integrals, ensuring that each Delta activation is properly accounted for and the values of the function are correctly extracted.

Delta Function Activation

The term 'activation' in relation to the Dirac Delta function refers to the specific input value of \( t \) that causes the Delta function to effectively 'turn on' and exert its sifting effect. The Delta function is said to be 'activated' at \( t = c \) if \( \delta(t-c) \) is included within the integral. If the activation point falls within the integration range, as seen in Step 1 and Step 4, the Delta function performs its role, resulting in nonzero contributions.

The activation concept is not only theoretical but also has practical implications. For example, in signal processing, the Delta function is used to model impulses that are activated at specific times. Understanding when and how the Delta function activates is essential for correctly interpreting and solving integrals involving this sophisticated mathematical tool.