Integral with stepfunction

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In summary, the conversation discusses evaluating a type of integral using iteration and calculating the last integral. However, the integral diverges due to the integrand being always equal to 1.
  • #1
Pietjuh
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For my statistical physics class I need to evaluate the following type of integral by iteration and I am not sure I'm doing it correctly

[tex]I_N = \int_0^{\infty}dx_1 \int_0^{\infty}dx_2\cdots \int_0^{\infty} \theta(a - \sum_{i=0}^N x_i) dx_N[/tex]

Since i must iterate this integral I'm going to try and calculate the last integral. I then wrote
[tex]a' = a - \sum_{i=0}^{N-1} x_i[/tex]
so,

[tex]I_N = \int_0^{\infty}dx_1 \int_0^{\infty}dx_2\cdots \int_0^{\infty} \theta(a' - x_N) dx_N = \int_0^{\infty}dx_1 \int_0^{\infty}dx_2\cdots \int_0^{a'} 1 dx_N = \int_0^{\infty}dx_1 \int_0^{\infty}dx_2\cdots \int_0^{\infty} a' dx_{N-1}[/tex]

But this integral diverges! Can someone tell me what I did wrong here?
 
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  • #2
You did not do anything wrong. The integral does indeed diverge because the integrand is always equal to 1, and the upper bound of integration does not go to 0 as N goes to infinity. This means that as N increases, the integral increases without limit.
 
  • #3



It looks like you are on the right track with your approach to evaluating this integral using iteration. However, it seems like there may be a mistake in your final step. Let's take a closer look at the integral you are trying to evaluate:

I_N = \int_0^{\infty}dx_1 \int_0^{\infty}dx_2\cdots \int_0^{\infty} \theta(a - \sum_{i=0}^N x_i) dx_N

In order to evaluate this integral by iteration, we can start by considering the last integral:

J_N = \int_0^{\infty} \theta(a - \sum_{i=0}^N x_i) dx_N

As you correctly pointed out, we can rewrite this as:

J_N = \int_0^{\infty} \theta(a' - x_N) dx_N

where a' = a - \sum_{i=0}^{N-1} x_i. However, in your next step, you seem to have made a mistake by replacing the upper limit of the integral with a'. This is incorrect because the upper limit should still be \infty. Instead, we can rewrite the integral as:

J_N = \int_0^{a'} 1 dx_N

This gives us a finite value for the integral, and we can now continue with the iteration process as you did by replacing x_N with a' in the previous integrals. This should give us a convergent result for the entire integral I_N.

In summary, it seems like the mistake was in replacing the upper limit of the integral with a' instead of keeping it as \infty. I hope this helps clarify the issue and allows you to successfully evaluate the integral.
 

1. What is an integral with step function?

An integral with step function is a type of mathematical integral that involves a piecewise-defined function called a step function. This means that the function is defined differently for different intervals, with a sudden "step" in between.

2. How is an integral with step function different from a regular integral?

The main difference is that a regular integral involves a continuous function, while an integral with step function involves a piecewise-defined function. This means that the integral with step function will have different values for different intervals, while a regular integral will have a single value.

3. What are some applications of integrals with step functions?

Integrals with step functions are commonly used in physics and engineering to model systems with sudden changes or discontinuities. They can also be used in economics to represent sudden changes in demand or supply.

4. How do you solve an integral with step function?

The process for solving an integral with step function is similar to solving a regular integral, but with additional steps to account for the different intervals. First, you must identify the intervals where the function is defined differently. Then, you can use the properties of integrals to solve each interval separately, and finally, combine the results to get the overall integral.

5. Are there any challenges or limitations when working with integrals with step functions?

One challenge with integrals with step functions is that they can be more complex and time-consuming to solve compared to regular integrals. Additionally, not all functions can be represented using step functions, so there may be limitations in certain applications.

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