Chapter 20: Problem 32
Find the function \(f(t)\) whose Laplace transform is $$ \bar{f}(s)=\frac{e^{-s}-1+s}{s^{2}} $$
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By integrating a suitable function around a large semicircle in the upper half plane and a small semicircle centred on the origin, determine the value of $$ I=\int_{0}^{\infty} \frac{(\ln x)^{2}}{1+x^{2}} d x $$ and deduce, as a by-product of your calculation, that $$ \int_{0}^{\infty} \frac{\ln x}{1+x^{2}} d x=0 $$
Prove that if \(f(z)\) has a simple pole at \(z_{0}\) then \(1 / f(z)\) has residue \(1 / f^{\prime}\left(z_{0}\right)\) there. Hence evaluate $$ \int_{-\pi}^{\pi} \frac{\sin \theta}{a-\sin \theta} d \theta $$ where \(a\) is real and \(>1\).
By applying the residue theorem around a wedge-shaped contour of angle \(2 \pi / n\), with one side along the real axis, prove that the integral $$ \int_{0}^{\infty} \frac{d x}{1+x^{n}} $$ where \(n\) is real and \(\geq 2\), has the value \((\pi / n) \operatorname{cosec}(\pi / n)\).
By considering the real part of $$ \int \frac{-i z^{n-1} d z}{1-a\left(z+z^{-1}\right)+a^{2}} $$ where \(z=\exp i \theta\) and \(n\) is a non-negative integer, evaluate $$ \int_{0}^{\pi} \frac{\cos n \theta}{1-2 a \cos \theta+a^{2}} d \theta $$ for \(a\) real and \(>1\)
The equation of an ellipse in plane polar coordinates \(r, \theta\), with one of its foci at the origin, is $$ \frac{l}{r}=1-\epsilon \cos \theta $$ where \(l\) is a length (that of the latus rectum) and \(\epsilon(0<\epsilon<1)\) is the eccentricity of the ellipse. Express the area of the ellipse as an integral around the unit circle in the complex plane, and show that the only singularity of the integrand inside the circle is a double pole at \(z_{0}=\epsilon^{-1}-\left(\epsilon^{-2}-1\right)^{1 / 2}\) By setting \(z=z_{0}+\xi\) and expanding the integrand in powers of \(\xi\), find the residue at \(z_{0}\) and hence show that the area is equal to \(\pi l^{2}\left(1-\epsilon^{2}\right)^{-3 / 2}\). (In terms of the semi-axes \(a\) and \(b\) of the ellipse, \(l=b^{2} / a\) and \(\left.\epsilon^{2}=\left(a^{2}-b^{2}\right) / a^{2} .\right)\)
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