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Chapter 12: Problem 6

Evaluate (a) \(\int_{-\infty}^{\infty} \mathrm{e}^{a t} \sin b t \delta^{(n)}(t) \mathrm{d} t \quad\) for \(n=0,1,2\). (b) \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta^{(n)}\left(t^{3}+t^{2}+t\right) d t \quad\) for \(n=0,1\).

Short Answer

Expert verified
(a) At n=0, the integral is 0; at n=1, the integral is b; at n=2, the integral is \(2ab\). (b) At n=0, the integral is \(\cos(0) + \sin(0) + \cos(-1)+ \sin(-1)\); at n=1, the integral is 0.

Step by step solution

01

Evaluate Part (a)

In Part (a), to simplify the problem, replace \(e^{at}\sin(bt)\) with a general function f(t). This can now be represented as \(\int_{-\infty}^{\infty} f(t) \delta^{(n)}(t) dt\). Recall that the integral of the product of a general function and the nth derivative of the delta function equals to the nth derivative of that general function at 0. Taking this into account, when n=0, the integral will be f(0), when n=1, it will be f'(0), and when n=2, it will be f''(0). View \(e^{at}\sin(bt)\) as f(t), then f'(t)=a\(e^{at}\sin(bt)\) + b\(e^{at}\cos(bt)\) and f''(t)=2ab\(e^{at}\cos(bt)\) - b^2\(e^{at}\sin(bt)\) + a^2\(\e^{at}\sin(bt)\)
02

Plug in n values for Part (a)

By plugging these values, at n=0, the integral is \(e^{a*0}\sin(b*0)=0\), at n=1, the integral is \(a*e^{a*0}\sin(b*0) + b*e^{a*0}\cos(b*0)=b\), at n=2, the integral is \(2ab*e^{a*0}\cos(b*0) - b^2*e^{a*0}\sin(b*0) + a^2*e^{a*0}\sin(b*0) = 2ab\).
03

Evaluate Part (b)

Similar approach to part (a), but this time, note that we have \(t^3+t^2+t\) inside the delta function, which means that the delta function will be non zero when \(t^3+t^2+t =0\). This means our a value in the property will be the roots of \(t^3+t^2+t =0\), instead of the conventional '0'. Then, find the roots of this equation (which are, 0 and -1) and calculate the sum of the values of \(\cos(t) + \sin(t)\) at these roots.
04

Plug in n Values for part (b)

At n=0, the integral is \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta(t^3+t^2+t) dt = \cos(0) + \sin(0) + \cos(-1)+ \sin(-1)\), and at n=1, the integral becomes \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta'(t^3+t^2+t) dt\). But recall that the delta function only has 1st derivative when n=1 and its value is infinitive only at t=a, and we only have two roots at t=0 and t=-1, thus the integral value is zero.

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Most popular questions from this chapter

Which of the following is a distribution? (a) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi^{(n)}(0) \quad\left(\lambda_{n} \in \mathbb{R}\right)\) (b) \(T(\phi)=\sum_{n=1}^{m} \lambda_{n} \phi\left(x_{n}\right) \quad\left(\lambda_{n}, x_{n} \in \mathbb{R}\right)\). (c) \(T(\phi)=(\phi(0))^{2}\). (d) \(T(\phi)=\sup \phi\) (c) \(T(\phi)=\int_{-\infty}^{\infty}|\phi(x)| \mathrm{d} x\).

Show the identities $$ \frac{d}{d x}(\delta(f(x)))=f^{\prime}(x) \delta^{\prime}(f(x)) $$ and $$ \delta(f(x))+f(x) \delta^{\prime}(f(x))=0 $$ Hence show that \(\phi(x, y)=\delta\left(x^{2}-y^{2}\right)\) is a solution of the partial differcntial equation $$ x \frac{\partial \phi}{\partial x}+y \frac{\partial \phi}{\partial y}+2 \phi(x, y)=0 $$

Prove that $$ x^{m} \delta^{(n)}(x)=(-1)^{m} \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \quad \text { for } n \geq m $$ Hence show that the Fourier trarsform of the distrobution $$ \sqrt{2 \pi} \frac{k !}{(m+k) !} x^{m} \delta^{(m+k)}(-x) \quad(m, k \geq 0) $$ is \((-2 y)^{k}\)

We say a scquence of distributions \(T_{n}\) converges to a distribution \(T\), written \(T_{n} \rightarrow T\). if \(T_{n}(\phi) \rightarrow T(\phi)\) for all test functions \(\phi \in \mathcal{D}\) (this is sometimes called weak convergence). If a scquence of continuous functions \(f_{n}\) converges uniformly to a function \(f(x)\) on every compact subsct of \(\mathbb{R}\), show that the associated regular distributions \(T_{f_{n}} \rightarrow T_{f-}\) In the distributional sense, show that we have the following convergences. $$ \begin{aligned} f_{n}(x) &=\frac{n}{\pi\left(1+n^{2} x^{2}\right)} \rightarrow \delta(x) \\ g_{n}(x) &=\frac{n}{\sqrt{\pi}} \mathrm{e}^{-\alpha^{2} x^{2}} \rightarrow \delta(x) \end{aligned} $$

Show that the Fourier transform of the distribution $$ \delta_{0}+\delta_{d i}+\delta_{2 a}+\cdots+\delta_{(2 n-1)} $$ is a distribution with density $$ \frac{1}{\sqrt{2 \pi}} \frac{\sin (n a y)}{\sin \left(\frac{1}{2} a y\right)} \mathrm{e}^{-\left(n-\frac{1}{2}\right)+2 y} $$ Show that $$ \mathcal{F}^{-1}\left(f(y) e^{2 b y}\right)=\left(\mathcal{F}^{-1} f\right)(x+b) $$ Hcnce find the inverse Fourier transform of $$ g(y)=\frac{\text { sin } n a y}{\sin \left(\frac{1}{2} a y\right)} $$
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