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Chapter 5: Problem 8

Develop an asymptotic series for $$ \int_{0}^{\infty} e^{-x v}\left(1+v^{2}\right)^{-2} d v . $$ Take \(x\) to be real and positive. $$ \text { ANS. } \frac{1}{x}-\frac{2 !}{x^{3}}+\frac{4 !}{x^{5}}-\cdots+\frac{(-1)^{n}(2 n) !}{x^{2 n+1}} \text { . } $$

Short Answer

Expert verified
The asymptotic series is \(\frac{1}{x} - \frac{2!}{x^3} + \frac{4!}{x^5} - \cdots + \frac{(-1)^n (2n)!}{x^{2n+1}}\).

Step by step solution

01

Simplify the Integrand

The given integrand is \[ \int_{0}^{\infty} e^{-x v}\left(1+v^{2}\right)^{-2} d v. \]We need to consider expanding it as an asymptotic series. The term \((1 + v^2)^{-2}\) can be expanded into a series if we express this as \((1+z)^{-n}\), where \(z = v^2\) and \(n = 2\).
02

Use Binomial Series Expansion

Expand \((1 + v^2)^{-2}\) using the binomial series:\[ (1 + z)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} z^k. \]For \((1 + v^2)^{-2}\) or \((1 + z)^{-2}\), this becomes:\[ \sum_{k=0}^{\infty} \binom{1}{k} (-1)^k (v^2)^k = \sum_{k=0}^{\infty} (-1)^k (v^2)^k. \]
03

Substitute the Series into the Integral

Substitute the expanded form back into the integral:\[ \int_{0}^{\infty} e^{-xv} \sum_{k=0}^{\infty} (-1)^k (v^2)^k dv. \]We can switch the order of integration and summation:\[ \sum_{k=0}^{\infty} (-1)^k \int_{0}^{\infty} v^{2k} e^{-xv} dv. \]
04

Solve the Inner Integral

Focus on solving the integral:\[ \int_{0}^{\infty} v^{2k} e^{-xv} dv. \]Use the substitution \( u = xv \), thus \( dv = du/x \), so:\[ \int_{0}^{\infty} \frac{u^{2k}}{x^{2k}} e^{-u} \frac{du}{x}. \]This simplifies to:\[ \frac{1}{x^{2k+1}} \int_{0}^{\infty} u^{2k} e^{-u} du. \]
05

Solve the Gamma Function Part

The integral \(\int_{0}^{\infty} u^{2k} e^{-u} du\) is the gamma function \(\Gamma(2k+1)\).Using the property \(\Gamma(n) = (n-1)!\), we evaluate:\[ \Gamma(2k+1) = (2k)! \]Thus, our integral becomes:\[ \frac{(2k)!}{x^{2k+1}}. \]
06

Construct the Asymptotic Series

Plug this back into the series:\[ \sum_{k=0}^{\infty} \frac{(-1)^k (2k)!}{x^{2k+1}}. \]Therefore, the asymptotic series for the integral is:\[ \frac{1}{x} - \frac{2!}{x^3} + \frac{4!}{x^5} - \cdots + \frac{(-1)^n (2n)!}{x^{2n+1}}, \] which matches the given answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Series Expansion
The **binomial series expansion** is an extremely useful tool when dealing with expressions of the form \((1 + z)^{-n}\), especially in situations requiring series and asymptotic approximations.
It allows us to express powers of binomials as a potentially infinite series. For our expression \((1 + v^2)^{-2}\), it helps to simplify by expanding it as follows:
  • Writing the expression as \((1 + z)^{-2}\) where \(z = v^2\).
  • The binomial series expansion formula is given as:\[(1 + z)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} z^k.\]
  • In our specific case, the series becomes:\[\sum_{k=0}^{\infty} (-1)^k (v^2)^k.\]
This transformation simplifies the problem significantly, making the integration and subsequent steps more manageable.
Gamma Function
The **gamma function** is a key concept in higher mathematics, particularly in calculus and complex analysis. It extends the factorial function to complex and real number arguments.
Mathematically, it's represented and evaluated as:
  • For a positive integer \( n \), \( \Gamma(n) = (n-1)! \) is widely used.
  • In the context of our problem, the integral \( \int_{0}^{\infty} u^{2k} e^{-u} du \) simplifies to \( \Gamma(2k+1) \).
  • This is directly derived from the definition of the gamma function and its integral representation:\[\Gamma(a) = \int_{0}^{\infty} x^{a-1} e^{-x} dx.\]
The use of the gamma function here allows us to express complex integrals in factorial terms, simplifying the integration process considerably.
Integral Calculus
**Integral calculus** is fundamental in mathematics, focusing on the accumulation of quantities and the areas under and between curves. In solving our integral \[\int_{0}^{\infty} v^{2k} e^{-xv} dv,\]we need to understand the transformation and evaluation techniques:
  • Using substitution, set \( u = xv \) to simplify the integral. Then, \( dv = du/x \).
  • This changes the limits and form to:\[\int_{0}^{\infty} \frac{u^{2k}}{x^{2k+1}} e^{-u} du = \frac{1}{x^{2k+1}} \int_{0}^{\infty} u^{2k} e^{-u} du.\]
  • The result of this integral directly associates with the gamma function: \( \Gamma(2k+1) \).
Therefore, integral calculus allows transitioning from complex integrals into simpler representations, essential for series expansions and further mathematical manipulations.
Series Expansion
**Series expansion** techniques involve expressing functions as sums of more straightforward terms, which can facilitate easier computation and approximation of functions.
In the constructed asymptotic series:
  • The primary task is to substitute the calculated terms into a coherent series, following the derived pattern.
  • This is achieved by plugging in the results from the Gamma function and previously calculated terms of the series.
  • The asymptotic series for the given integral ends up being:\[\frac{1}{x} - \frac{2!}{x^3} + \frac{4!}{x^5} - \cdots + \frac{(-1)^n (2n)!}{x^{2n+1}}.\]
This series provides an approximation that becomes more accurate for large \(x\). Understanding this concept allows students to expand their toolkit for tackling problems related to series and asymptotics.

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Most popular questions from this chapter

The behavior of a neutron losing energy by colliding elastically with nuclei of mass \(A\) is described by a parameter \(\xi_{1}\), $$\xi_{1}=1+\frac{(A-1)^{2}}{2 A} \ln \frac{A-1}{A+1}$$ An approximation, good for large \(A\), is $$\xi_{2}=\frac{2}{A+2 / 3}$$ Expand \(\xi_{1}\) and \(\xi_{2}\) in powers of \(A^{-1}\). Show that \(\xi_{2}\) agrees with \(\xi_{1}\) through \(\left(A^{-1}\right)^{2} .\) Find the difference in the coefficients of the \(\left(A^{-1}\right)^{3}\) term.

Show that the first Bernoulli polynomials are $$ \begin{array}{l} B_{0}(s)=1 \\ B_{1}(s)=s-\frac{1}{2} \\ B_{2}(s)=s^{2}-s+\frac{1}{6} . \end{array} $$ Note that \(B_{n}(0)=B_{n}\), the Bemoulli number.

Evaluate (a) \(\lim _{x \rightarrow 0}[\sin (\tan x)-\tan (\sin x)] x^{-7}\). (b) \(\lim _{x \rightarrow 0} x^{-n} j_{n}(x), n=3\), where \(j_{n}(x)\) is a spherical Bessel function (Section \(\left.11.7\right)\) defined by $$ j_{n}(x)=(-1)^{n} x^{n}\left(\frac{1}{x} \frac{d}{d x}\right)^{n}\left(\frac{\sin x}{x}\right). $$

Show that $$ \prod_{n=2}^{\infty}\left[1-\frac{2}{n(n+1)}\right]=\frac{1}{3} $$

Using binomial expansions, compare the three Doppler shift formulas: (a) \(v^{\prime}=v\left(1 \mp \frac{v}{c}\right)^{-1}\) moving source; (b) \(v^{\prime}=v\left(1 \pm \frac{v}{c}\right)\) moving observer; (c) \(v^{\prime}=v\left(1 \pm \frac{v}{c}\right)\left(1-\frac{v^{2}}{c^{2}}\right)^{-1 / 2}\) relativistic. Note. The relativistic formula agrees with the classical formulas if terms of ord \(v^{2} / c^{2}\) can be neglected.
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