Understanding the Dirac Delta Function Integral

The problem asks us to evaluate the definite integral of the product of an exponential function, ${e^{-t}}$, and the Dirac delta function, \(\delta(2t - 2)\), over the entire real line from \(-\infty\) to \(+\infty\).

The key to solving this integral lies in understanding the properties of the Dirac delta function, \(\delta(t)\).

Key Properties of Dirac Delta Function

• Sifting Property: The most fundamental property is the sifting property, which states that for a continuous function \(f(t)\) and a point \(a\), the following holds:

  \(\mathop \smallint \limits_{ - \infty }^{ + \infty } f(t)\delta(t - a)dt = f(a)\)

• Scaling Property: When the argument of the delta function is scaled, like \(\delta(bt)\), it relates to \(\delta(t)\) as follows:

  \(\delta(bt) = \frac{1}{|b|}\delta(t)\)

  This can be extended to \(\delta(bt - c)\) by factoring out \(b\): \(\delta(b(t - c/b)) = \frac{1}{|b|}\delta(t - c/b)\).

Step-by-Step Solution

1. Rewrite the Delta Function Argument: We need to adjust the term \(\delta(2t - 2)\) to fit the standard form \(\delta(t - a)\). We can factor out the coefficient of \(t\):

   \(\delta(2t - 2) = \delta(2(t - 1))\)

2. Apply the Scaling Property: Using the scaling property \(\delta(bt) = \frac{1}{|b|}\delta(t)\), we have \(b=2\) and the argument is \((t-1)\). So:

   \(\delta(2(t - 1)) = \frac{1}{|2|}\delta(t - 1) = \frac{1}{2}\delta(t - 1)\)

3. Substitute into the Integral: Now, substitute this back into the original integral:

   \(\mathop \smallint \limits_{ - \infty }^{ + \infty } {e^{ - t}}\delta \left( {2t - 2} \right)dt = \mathop \smallint \limits_{ - \infty }^{ + \infty } {e^{ - t}} \left( \frac{1}{2}\delta(t - 1) \right) dt\)

4. Factor out the Constant: Constants can be pulled out of the integral:

   \(= \frac{1}{2} \mathop \smallint \limits_{ - \infty }^{ + \infty } {e^{ - t}}\delta(t - 1) dt\)

5. Apply the Sifting Property: The integral now matches the sifting property format, where \(f(t) = e^{-t}\) and \(a = 1\). Applying the property \(\mathop \smallint \limits_{ - \infty }^{ + \infty } f(t)\delta(t - a)dt = f(a)\):

   \(= \frac{1}{2} f(1)\)

6. Evaluate the Function: Calculate the value of \(f(t) = e^{-t}\) at \(t = 1\):

   \(f(1) = e^{-1} = \frac{1}{e}\)

7. Final Result: Combine the constant factor with the function value:

   \(= \frac{1}{2} \times \frac{1}{e} = \frac{1}{2e}\)

Therefore, the value of the given integral is \(\frac{1}{2e}\).