Integration of funtion multiplied by shifted heavyside step

• SpaceDomain
In summary, the conversation discusses the equality of two integrals involving a unit step function and its shifted version. The left side of the equation should be interpreted as a graph of the unit step function, which is symmetric about the vertical axis. The speaker initially made a mistake in their logic, but upon realizing that t is the independent variable and not the function being multiplied by the unit step, they clarify their understanding.
SpaceDomain
As I understand it, the following is true:

$$\int_{0}^{\infty}{u(t - \lambda) d\lambda} = \int_{0}^{t}{d\lambda}$$

But I do not understand why. It seems to me that the left side above should equal

$$\int_{\lambda}^{\infty}{d\lambda}$$

since

$$u(t - \lambda) = \left\{\begin{array}{cc}0,&\mbox{ if } t< \lambda \\ 1, & \mbox{ if } t> \lambda \end{array}\right.$$

I obviously don't understand this correctly. What am I not doing right?

I guess there is not a function multiplied by the unit step function here. But if there were, like:
$$\int_{0}^{\infty}{f( \lambda ) u(t - \lambda) d\lambda}$$

then it would be equal to

$$\int_{0}^{t}{f( \lambda) d\lambda}$$

Right?

Oh, I think I am seeing my mistake in logic here.

The unit step $$u(t - \lambda)$$ is a shift the unit step $$u(\lambda)$$ and then symmetric about the vertical axis.

I should be thinking of this as a graph of $$\lambda$$ vs. $$u(\lambda)$$.

I was getting thrown off by the variable t here because I am so used to it being the independent variable. Okay, okay. Never mind.

1. What is the definition of "integration of function multiplied by shifted heavyside step"?

Integration of function multiplied by shifted heavyside step is a mathematical operation that involves finding the area under the curve of a function multiplied by a shifted heavyside step function. The heavyside step function is a function that is equal to 0 for all values less than the shift and equal to 1 for all values greater than or equal to the shift.

2. What is the purpose of integrating a function multiplied by a shifted heavyside step?

The purpose of integrating a function multiplied by a shifted heavyside step is to calculate the total area under the curve of the function within a specific range, taking into account the shift in the heavyside step function. This can be useful in solving various mathematical and engineering problems.

3. How is the integration of function multiplied by shifted heavyside step calculated?

The integration of function multiplied by shifted heavyside step can be calculated by first determining the integral of the function, and then multiplying it by the value of the shifted heavyside step function for the given range. This results in a piecewise function that can be evaluated to find the total area under the curve.

4. What are some real-world applications of integrating a function multiplied by a shifted heavyside step?

One real-world application of integrating a function multiplied by a shifted heavyside step is in signal processing. The heavyside step function can represent an abrupt change in a signal, and by integrating the function multiplied by the heavyside step, the total energy of the signal can be calculated. This is useful in designing filters and analyzing the characteristics of signals.

5. Are there any limitations to integrating a function multiplied by a shifted heavyside step?

One limitation to integrating a function multiplied by a shifted heavyside step is that it can only be used for piecewise continuous functions. This means that the function must be continuous within each individual range, but can have discontinuities at the boundaries between ranges. Additionally, the heavyside step function must also be continuous at the shift value.

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