# Evaluating integrals with delta function

• nhoratiu
In summary: = \lim_{\lambda\to 0^+} \int_{-\infty}^\infty u(x)\cos(x)=-\frac{1}{2} \int_{-\infty}^\infty u(x)\sin(x)\,dx = \frac{1}{2} \cos(x)
nhoratiu
Hi there!
I have a problem with one of the questions given to us in the signals and systems course. If anyone could help me I would greatly appreciate it!

## Homework Statement

integral(from -infinity to +infinity) of u0(t) * cos(t) dt

u(t) is a step function.

## Homework Equations

Why is this equal to 1 and not to sin(t), for t>0 ?

## The Attempt at a Solution

I thought this is equal to sin(t) but the answer manual says it's equal to 1 because the whole thing transforms into a delta function. How does it do that?

You need to define $$u_0$$.

u0(t) is a step function. its "0" from negative infinity to t=0 and "1" from t=0 to positive infinity.

So, I guess u0 is the same as u. And, I also guess, "*" stands for the convolution?

hey, the * is for multiplication. Overall we have:

integral from negative infinity to positive infinity of ( u(t) times cos(t) dt)

the solution manual says it's equal to 1. I thought it was equal to sin(t).

I would, formally (which is, strictly speaking, illegal), do it like that:

$$\int_{-\infty}^{\infty}u(x)\cos(x)dx=\int_{-\infty}^{\infty}u(x)\sin'(x)dx=-\int_{-\infty}^{\infty}u'(x)\sin(x)dx=-\int_{-\infty}^{\infty}\delta(x)\sin(x)dx=-\sin(0)=0$$

But I do not know what your course is based on, therefore I can't help more. And probably my result is not helping you anyway.

When you integrate cos x, you get sin x, but the problem is evaluating the result at the upper limit since sin x doesn't converge in the limit as x→∞. To get around this problem, you can use the convergence factor e-λx and take the limit as λ→0+:

$$\int_{-\infty}^\infty u(x)\cos(x)\,dx = \lim_{\lambda\to 0^+}\int_{-\infty}^\infty u(x)\cos(x)e^{-\lambda x}\,dx$$

## 1. What is the delta function and how does it relate to integrals?

The delta function, denoted as δ(x), is a mathematical concept used to represent a very narrow, infinitely tall spike at a specific point in the real line. It is often used in physics and engineering to model point sources or impulses. When used in integrals, it acts as a weighting factor that concentrates the integral at a specific point.

## 2. How do you evaluate integrals involving the delta function?

To evaluate integrals with the delta function, you can use the following property: ∫f(x)δ(x-a)dx = f(a). This means that the integral of a function multiplied by the delta function at a specific point is equal to the value of the original function at that point. You can also use this property to evaluate integrals involving other functions, such as polynomials, by expanding them into a sum of delta functions.

## 3. Can the delta function be used to solve improper integrals?

Yes, the delta function can be used to solve improper integrals. By using the above property, you can evaluate the integral at the point where it becomes improper, often resulting in a finite value. However, it is important to note that this method should be used with caution and may not always work for all types of improper integrals.

## 4. How does the delta function affect the value of an integral?

The delta function acts as a weighting factor in integrals, meaning that it can significantly affect the value of the integral. When multiplied by a function, it concentrates the integral at a specific point, resulting in a large value. However, if the delta function is integrated by itself, it results in a value of 1.

## 5. Are there any limitations to using the delta function in integrals?

While the delta function is a powerful tool in evaluating integrals, it does have some limitations. It is not defined at zero and is only useful in integrals when multiplied by a function. Additionally, it can be challenging to use in certain scenarios, such as when dealing with multidimensional integrals.

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