91Ó°ÊÓ

(a) Use principal value integrals to show that $$ \int_{0}^{\infty} \frac{\cos k x-\cos m x}{x^{2}} d x=\frac{-\pi}{2}(|k|-|m|), \quad k, m \text { real. } $$ Hint: note that the function \(f(z)=\left(e^{i k z}-e^{i m z}\right) / z^{2}\) has a simple pole at the origin. (b) Let \(k=2, m=0\) to deduce that $$ \int_{0}^{\infty} \frac{\sin ^{2} x}{x^{2}} d x=\frac{\pi}{2} $$

Show that the Fourier transform of the "Gaussian" \(f(x)=\exp \left(-\left(x-x_{0}\right)^{2} / a^{2}\right) x_{0}, a\) real, is also a Gaussian: $$ \hat{F}(k)=a \sqrt{\pi} e^{-(k a / 2)^{2}} e^{-i k x_{0}} $$

Obtain the Fourier transform of the following functions, and thereby show that the Fourier transforms of hyperbolic secant functions are also related to hyperbolic secant functions. (a) sech \(\left[a\left(x-x_{0}\right)\right] e^{i \omega x}, \quad a, x_{0}, \omega\) real (b) \(\operatorname{sech}^{2}\left[a\left(x-x_{0}\right)\right], \quad a, x_{0}\) real

Projection operators can be defined as follows. Consider a function \(F(z)\) $$ F(z)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta $$ where \(C\) is a contour, typically infinite (e.g. the real axis) or closed (e.g. a circle) and \(z\) lies off the contour. Then the "plus" and "minus" projections of \(F(z)\) at \(z=\zeta_{0}\) are defined by the following limit: $$ F^{\pm}\left(\zeta_{0}\right)=\lim _{z \rightarrow \zeta_{0}}\left[\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta\right] $$ where \(\zeta_{0}{ }^{\pm}\)are points just inside \((+)\)or outside (-) of a closed contour (i.e., \(\lim _{z \rightarrow 5_{0}+\text { denotes the limit from points } z \text { inside the contour } C \text { ) or to }}\) the left (t) or right (-) of an infinite contour. Note: the "+" region lies to the left of the contour; where we take the standard orientation for a contour, that is, the contour is taken with counterclockwise orientation. To simplify the analysis, we will assume that \(f(x)\) can be analytically extended in the neighborhood of the curve \(C\). (a) Show that $$ F^{\pm}\left(\zeta_{0}\right)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-\zeta_{0}} d \zeta \pm \frac{1}{2} f\left(\zeta_{0}\right) $$ where \(f_{C}\) is the principal value integral that omits the point \(\zeta=\zeta 0\). (b) Suppose that \(f(\zeta)=1 /\left(\zeta^{2}+1\right)\), and the contour \(C\) is the real axis (infinite), with orientation take from \(-\infty\) to \(\infty\); find \(F^{\pm}\left(\zeta_{0}\right)\). (c) Suppose that \(f(\zeta)=1 /\left(\zeta^{2}+a^{2}\right), a^{2}>1\), and the contour \(C\) is the unit circle centered at the origin with counterclockwise orientation; find \(F^{\pm}\left(\zeta_{0}\right)\)

Consider the integral $$ I_{R}=\frac{1}{2 \pi i} \oint_{C_{R}} \frac{d z}{\left(z^{2}-a^{2}\right)^{1 / 2}}, \quad a>0 $$ where \(C_{R}\) is a circle of radius R centered at the origin enclosing the points \(z=\pm a\). Take the principal value of the square root. (a) Evaluate the residue of the integrand at infinity and show that \(I_{R}=1\). (b) Evaluate the integral by defining the contour around the branch points and along the branch cuts between \(z=-a\) to \(z=a\), to find (see Section \(2.3\) ) that \(I_{R}=\frac{2}{\pi} \int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}\). Use the well-known indefinite integral \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} x / a+\) const to obtain the same result as in part (a).

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) \(\frac{1}{\left(s^{2}-\omega^{2}\right)^{2}}\)

Show explicitly that the Laplace transform of the second derivative of a function of \(x\) satisfies $$ \int_{0}^{\infty} f^{\prime \prime}(x) e^{-s x} d x=s^{2} \hat{F}(s)-s f(0)-f^{\prime}(0) $$