# Evaluating and integral including unit step function

• user3
In summary, the integral can be calculated by first finding the second derivative of the given function using distribution theory. This involves using unit step functions and Dirac deltas to represent the piecewise form of the derivative. The original method is recommended as it is more straightforward and accounts for any potential discontinuities.
user3
how can this integral be calculated:

∫[e^(−2mx) θ^2(x)+2θ(x)θ(−x)+e^(−2mx)θ^2(−x)]dx from -∞ to ∞

where θ(x) is the unit step function with its amplitude 0 everywhere before x=0 and θ(−x) is the unit step function with its amplitude 0 everywhere after x=0In Introduction to Quantum Mechanics by Griffiths there is this problem: ψ(x)=e^(−m|x|) (there are other constants in the ψ that I think are not necessary to write). we need to find ∫ψ d^2ψ/dx^2 from -∞ to ∞

In the book, what the author does is first get

dψ/dx = −me^(−mx) if x >0 and me^(mx) if x <0 =−mθ(x)e^(−mx)+mθ(−x)e^(mx)then he gets
d^2ψ/dx^2=−mδ(x)e^(−mx)+m^2 θ(x)e^(−mx) −mδ(−x)e^(mx)+m^2 θ(−x)emx=m∗(−2δ+me−m|x|)

where δ(x) is the dirac delta function (0 everywhere and infinite at x=0)

the integral we wanted is easy to get from there. But I wanted to try a new approach: get the second derivative in the piecewise form, then transform it to unit step functions and evaluate the integral.

I thought that if the author's method worked, then the other method must also work because it's logical. Is there something wrong with my reasoning ?

here's a cleaner statement of the question: http://math.stackexchange.com/quest...on-squared?noredirect=1#comment1427483_679623

user3 said:
I thought that if the author's method worked, then the other method must also work because it's logical.
It might seem logical, but I suspect that's because you don't understand distribution theory.

Is there something wrong with my reasoning?
You've already been given a partial answer on stackexchange. The term ##\theta(x)\theta(-x)## complicates matters more than you might think -- because of the lurking discontinuity at 0.

Stick with the original method, where the derivative is implicitly understood as a weak derivative.
See also distribution (the section on differentiation).

If you stick to such distributional methods, at least you don't miss anything, even though it might mean you must deal with Dirac deltas.

## 1. What is a unit step function?

A unit step function is a mathematical function that takes on two values: 0 and 1. It is denoted by the symbol u(t) and is defined as u(t) = 0 for t < 0 and u(t) = 1 for t ≥ 0. It is also known as the Heaviside function, named after the mathematician Oliver Heaviside.

## 2. How is a unit step function used in evaluating integrals?

Unit step functions are commonly used in evaluating integrals because they allow us to define piecewise functions. By breaking up the integral into different intervals and using the unit step function to define the function for each interval, we can simplify the integration process and solve for the integral more easily.

## 3. What is the importance of the unit step function in science and engineering?

The unit step function is important in science and engineering because it allows us to model and analyze systems that exhibit abrupt changes or discontinuities. It is used in fields such as electrical engineering, control theory, and signal processing to represent switch-on/switch-off behavior and to solve differential equations.

## 4. Can the unit step function be used for functions other than step functions?

Yes, the unit step function can be used to define and evaluate integrals for functions other than step functions. It can be used to represent any function that has a constant value over a specific interval, making it a versatile tool in mathematical analysis and modeling.

## 5. Are there any limitations to using the unit step function in evaluating integrals?

While the unit step function is a useful tool in evaluating integrals, it does have some limitations. It can only be used for functions that are piecewise continuous, meaning they have a finite number of discontinuities. It also cannot be used for functions that are not defined at t = 0, as the unit step function is only defined at t = 0.

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