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Chapter 4: Problem 17

(a) Show that the inverse Laplace transform of \(\hat{F}_{1}(s)=\log s\) is given by \(f_{1}(x)=-1 / x\) (b) Do the same for \(\hat{F}_{2}(s)=\log (s+1)\) to obtain \(f_{2}(x)=-e^{-x} / x\). (c) Find the inverse Laplace transform \(\hat{F}(s)=\log ((s+1) / s)\) to obtain \(f(x)=\left(1-e^{-x}\right) / x\), by subtracting the results of parts (a) and (b). (d) Show that we can get this result directly, by encircling both the \(s=0\) and \(s=-1\) branch points and using the polar representations \(s+1=\) \(r_{1} e^{i \theta_{1}}, s=r_{2} e^{i \theta_{2}},-\pi \leq \theta_{i}<\pi, i=1,2\)

Short Answer

Expert verified
The inverse Laplace transforms are: \( \hat{F}_{1}(s) = \log s \) yields \( f_{1}(x) = -\frac{1}{x} \); \( \hat{F}_{2}(s) = \log (s+1) \) yields \( f_{2}(x) = - \frac{e^{-x}}{x} \). \( \hat{F}(s) = \log \left( \frac{s+1}{s} \right) \) yields \( f(x) = \frac{1 - e^{-x}}{x} \) using parts (a) and (b) and also verified by integral representation.

Step by step solution

01

Given function and inverse Laplace transform formula

To find the inverse Laplace transform of \(\hat{F}_{1}(s)=\log s,\) recall that the inverse Laplace transform is \(\mathcal{L}^{-1}\{\hat{F}(s)\}(x)=f(x)\).
02

Use the known inverse Laplace data

For \(\hat{F}_{1}(s) = \log s \), the inverse Laplace transform is a known result: \(f_{1}(x) = - \frac{1}{x} \). Verify by noting that the Laplace transform of \(-\frac{1}{x} \) indeed gives \(\log s\).
03

Verify inverse Laplace transform for part (b)

Similar to part (a), for \( \hat{F}_{2}(s) = \log (s+1) \), the inverse Laplace transform yields \(f_{2}(x) = - \frac{e^{-x}}{x} \). This again follows from known transform pairs.
04

Find the inverse Laplace transform for part (c)

Given \(\hat{F}(s) = \log \left( \frac{s+1}{s} \right) \), rewrite the function as \(\hat{F}(s) = \log (s+1) - \log s\). Use linearity to find: \[ f(x) = \mathcal{L}^{-1} \{ \log (s+1)\}(x) - \mathcal{L}^{-1} \{ \log s \}(x). \] Applying parts (a) and (b), we get: \[ f(x) = - \frac{e^{-x}}{x} + \frac{1}{x} = \frac{1 - e^{-x}}{x}. \]
05

Verify the result with the integral for part (d)

Using contour integration, consider \(s+1 = r_{1} e^{i \theta_{1}}\) and \( s = r_{2} e^{i \theta_{2}} \), where \( - \pi \leq \theta_{i} < \pi \, i = 1, 2.\) Evaluate the integrals around branch cuts \( s = 0 \) and \( s = -1 \) confirming: \( f(x) = \frac{1 - e^{-x}}{x}.\)

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Key Concepts

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

Logarithmic Functions
Logarithmic functions play a crucial role in transformations within mathematics. A logarithm, represented as \(\text{log}(s)\), is the inverse function of exponentiation. In simpler terms, if \(\text{log}_b(a) = c\), then \((b^c = a)\). The natural logarithm, commonly denoted as \(\text{ln}(s)\), uses the base of the natural number \(e \approx 2.71828\).
Contour Integration
Contour integration is a method used in complex analysis for evaluating complex integral functions along certain paths in the complex plane. This technique helps solve integrals typically not resolvable using standard real-variable methods. Contour integration frequently involves applying the Cauchy integral theorem or residue theorem.
When dealing with inverse Laplace transforms, contour integration becomes pertinent. It considers the paths around singularities or branch points in the complex plane. This part of complex analysis allows us to solve integrals by taking advantage of the shapes and properties of these paths or contours.
During the exercise, specifically in part (d), contour integration was utilized around the branch points \(s = 0\) and \(s = -1\). By setting the polar representations \(s+1 = r_1 e^{i \theta_1}\) and \(s = r_2 e^{i \theta_2}\), the method evaluates the integral, confirming the derived solution for \(f(x) = \frac{1 - e^{-x}}{x} \).
Branch Points
Branch points are specific points in the complex plane where a multi-valued function (like a logarithmic function) switches branches, showing discontinuities or jumps in value. These points are often involved in the process of contour integration.
For the given exercise step (d), the branch points \(s = 0\) and \(s = -1\) play a significant role. By encircling these branch points and using their polar representations, the integral's evaluation confirms the inverse Laplace transformation results. Typically, dealing with these points requires identifying the nature and handling of multi-valued functions in the complex plane, such as logarithms or roots.
In simpler terms, branch points are where functions may 'split' into different values. It's essential to consider these branches carefully, ensuring an accurate mathematical evaluation, especially in complex analysis and transformations.

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Most popular questions from this chapter

Obtain the inverse Laplace transforms of the following functions, assuming \(\omega, \omega_{1}, \omega_{2}>0\) (a) \(\frac{s}{s^{2}+\omega^{2}}\) (b) \(\frac{1}{(s+\omega)^{2}}\) (c) \(\frac{1}{(s+\omega)^{n}}\) (d) \(\frac{s}{(s+\omega)^{n}}\) (e) \(\frac{1}{\left(s+\omega_{1}\right)\left(s+\omega_{2}\right)}\) (f) \(\frac{1}{s^{2}\left(s^{2}+\omega^{2}\right)}\) (g) \(\frac{1}{\left(s+\omega_{1}\right)^{2}+\omega_{2}^{2}}\) (h) \(\frac{1}{\left(s^{2}-\omega^{2}\right)^{2}}\)

Consider a rectangular contour \(C_{R}\) with corners at \((\pm R, 0)\) and \((\pm R, a)\). Show that $$ \oint_{C_{R}} e^{-z^{2}} d z=\int_{-R}^{R} e^{-x^{2}} d x-\int_{-R}^{R} e^{-(x+i a)^{2}} d x+J_{R}=0 $$ where $$ J_{R}=\int_{0}^{a} e^{-(R+i y)^{2}} i d y-\int_{0}^{a} e^{-(-R+i y)^{2}} i d y $$ Show \(\lim _{R \rightarrow \infty} J_{R}=0\), whereupon we have \(\int_{-\infty}^{\infty} e^{-(x+i a)^{2}} d x=\int_{-\infty}^{\infty} e^{-x^{2}} d x\) \(=\sqrt{\pi}\), and consequently, deduce that \(\int_{-\infty}^{\infty} e^{-x^{2}} \cos 2 a x d x=\sqrt{\pi} e^{-a^{2}}\)

Given the ODE $$ z y^{\prime \prime}+(2 r+1) y^{\prime}+z y=0 $$ look for a contour representation of the form \(y=\int_{C} e^{z \zeta} v(\zeta) d \zeta\). (a) Show that if \(C\) is a closed contour and \(v(\zeta)\) is single valued on this contour, then it follows that \(v(\zeta)=A\left(\zeta^{2}+1\right)^{r-1 / 2}\) (b) Show that if \(y=z^{-s} w\), then when \(s=r, w\) satisfies Bessel's equation \(z^{2} w^{\prime \prime}+z w^{\prime}+\left(z^{2}-r^{2}\right) w=0\), and a contour representation of the solution is given by $$ w=A z^{r} \oint_{C} e^{z \zeta}\left(\zeta^{2}+1\right)^{r-1 / 2} d \zeta $$ Note that if \(r=n+1 / 2\) for integer \(n\), then this representation yields the trivial solution. We take the branch cut to be inside the circle \(C\) when \((r-1 / 2) \neq\) integer.

The hypergeometric equation $$ z y^{\prime \prime}+(a-z) y^{\prime}-b y=0 $$ has a contour integral representation of the form \(y=\int_{C} e^{z \zeta} v(\zeta) d \zeta\). (a) Show that one solution is given by $$ y=\int_{0}^{1} e^{\tau 5} \zeta^{b-1}(1-\zeta)^{a-b-1} d \zeta $$ where \(\operatorname{Re} b>0\) and \(\operatorname{Re}(a-b)>0\) (b) Let \(b=1\), and \(a=2\); show that this solution is \(y=\left(e^{z}-1\right) /(z)\), and verify that it satisfies the equation. (c) Show that a second solution, \(y_{2}=v y_{1}\) (where the first solution is denoted as \(y_{1}\) ) obeys $$ z y_{1} v^{\prime \prime}+\left(2 z y_{1}^{\prime}+(a-z) y_{1}\right) v^{\prime}=0 $$ Integrate this equation to find \(v\), and thereby obtain a formal representation for \(y_{2}\). What can be said about the analytic behavior of \(y_{2}\), near \(z=0 ?\)

Consider a rectangular contour with corners \(\left(N+\frac{1}{2}\right)(\pm 1 \pm i)\) to evaluate $$ \frac{1}{2 \pi i} \oint_{C(N)} \frac{\pi \cot \pi z \operatorname{coth} \pi z}{z^{3}} d z $$ and in the limit as \(N \rightarrow \infty\), show that $$ \sum_{n=1}^{\infty} \frac{\operatorname{coth} n \pi}{n^{3}}=\frac{7}{180} \pi^{3} $$ Hint: note $$ \operatorname{Res}\left(\frac{\pi \cot \pi z \operatorname{coth} \pi z}{z^{3}} ; 0\right)=-\frac{7 \pi^{3}}{45} $$
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