Finding a characteristic function (integral help)

In summary, the conversation discusses finding the characteristic function of a given function using integration by parts. The attempt at a solution involves setting up the u and v terms, but there is confusion about how to handle the integral limits and the ∫ v du term.
  • #1
mjordan2nd
177
1

Homework Statement



I'm looking to find the characteristic function of

[tex]p(x)=\frac{1}{\pi}\frac{1}{1+x^2}[/tex][/B]

Homework Equations



The characteristic function is defined as

[tex]\int_{-\infty}^{\infty} e^{ikx}p(x)dx[/tex]

3. The Attempt at a Solution

I attempted to solve this using integration by parts. I get

[tex]u=\frac{1}{1+x^2}[/tex]
[tex]du = -\frac{2x}{(1+x^2)^2} dx[/tex]
[tex] v = \frac{1}{ik}e^{ikx}[/tex]
[tex] dv = e^{ikx}dx[/tex]

This gives me

[tex] \phi (k) = \frac{e^{ikx}}{ik(1+x^2)} |_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \frac{2xe^{ikx}}{ik(1+x^2)} dx [/tex]

I'm a little stuck here. My term on the left seems to diverge, and I'm not particularly sure about how to handle my term on the right. For the term on the right, setting [itex]u=1+x^2[/itex] seem to make both integral limits [itex]\infty[/itex], so that doesn't seem right.[/B]
 
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  • #2
You haven't written the correct ∫ v du term any way.
 

Related to Finding a characteristic function (integral help)

1. What is a characteristic function?

A characteristic function is a mathematical tool used to describe the probability distribution of a random variable. It is defined as the Fourier transform of the probability density function (PDF) of the random variable.

2. How do I find the characteristic function of a random variable?

To find the characteristic function of a random variable, you need to take the Fourier transform of its PDF. This can be done analytically or numerically, depending on the complexity of the PDF.

3. What is the importance of finding a characteristic function?

Finding a characteristic function allows us to describe the probability distribution of a random variable in a concise and useful way. It also allows us to easily calculate various statistical properties, such as moments and cumulants, which are important in statistical analysis.

4. Can a characteristic function be used to determine the distribution of a random variable?

Yes, the characteristic function uniquely determines the probability distribution of a random variable. This property is known as the inversion theorem and it allows us to use the characteristic function to determine the PDF, and therefore, the distribution of the random variable.

5. Are there any limitations to using a characteristic function?

While characteristic functions are a powerful tool in probability and statistics, they may not exist for all random variables. Additionally, they may be difficult to calculate for some complex distributions. In these cases, alternative methods, such as moment-generating functions, may be used instead.

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