Series expansion of an integral

In summary, The function f(r) is defined as the integral of g(t) from r to infinity. The speaker is interested in finding the series expansion of f(r) around r=0 without actually doing the integral. They suggest using a Laurent expansion, but are unsure how to do so. They also mention that the function g(t) is too complicated to provide, but can be series expanded around other values. The speaker only needs the coefficients for the terms involving 1/r^2 and the constant term in the expansion.
  • #1
praharmitra
311
1
Given a function g(t), define the function f(r) as follows

[tex] f(r) = \int_r^\infty g(t) dt [/tex]

I want to find the series expansion of f(r) around the point r = 0, without actually doing the integral. Is this possible?

Basically, can i use any particular series expansion of g to find the series expansion of f?
 
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  • #2
hi praharmitra! :wink:

you want to find every dnf(r)/drn ?

hint: what is df(r)/dr ? :smile:
 
  • #3
hey tinytim!

I'm afraid we cannot use the usual expression for taylor expansions. You see, the function g(t) that I have does not converge on r=0. I believe there is a 1/r^2 divergence at r=0. So, the first term in the taylor series expansion, which is f(0) is not defined, so we cannot use what u have said.

I was thinking, we could use laurent expansion, but am unsure how to use it.
 
  • #4
Can you give us what the function g is? Or is this function to complicated?
 
  • #5
micromass said:
Can you give us what the function g is? Or is this function to complicated?

I don't think I can. Mathematica gave me a 10 line expression for the function. Its not able to integrate it either. However, it can series expand it about the usual values (0, infinity, 1).

So, i was hoping we could figure out the series expansion of f using the series expansion of g.

I'll tell you exactly what I need:

I believe f will diverge about r = 0 as 1/r^2 (from experience on previous similar calculations all of which i could integrate). All I want actually is the coefficient of the 1/r^2 in the expansion and the constant term. I don't want the full expansion really.
 

What is a series expansion of an integral?

A series expansion of an integral is a mathematical technique used to approximate the value of an integral, which is a mathematical concept that represents the area under a curve. The series expansion breaks down the integral into smaller, simpler parts that can be more easily calculated, allowing for a more accurate approximation of the integral's value.

What is the purpose of using a series expansion to calculate an integral?

The purpose of using a series expansion to calculate an integral is to simplify the calculation process and obtain a more accurate result. Integrals can be difficult to solve directly, but by breaking them down into a series of simpler terms, the calculation becomes more manageable. Additionally, series expansions can be used to approximate integrals that do not have a closed-form solution.

What is the formula for a series expansion of an integral?

The formula for a series expansion of an integral is known as the Maclaurin series or Taylor series. It is given by:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
where f(x) is the function being integrated, a is the lower limit of integration, and x is the upper limit of integration.

What are some applications of series expansion of an integral?

Series expansion of an integral has many applications in mathematics, engineering, and physics. It is commonly used in the fields of calculus, differential equations, and power series. It can also be applied in areas such as signal processing, control systems, and probability theory.

How can I determine the accuracy of a series expansion of an integral?

The accuracy of a series expansion of an integral depends on the number of terms used in the expansion. Generally, the more terms included in the expansion, the more accurate the approximation will be. However, it is important to consider the convergence of the series, which refers to the limit of the terms as the number of terms approaches infinity. If the series converges, the approximation will become increasingly accurate. If it diverges, the approximation will become less accurate as more terms are added.

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