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Introduction to Linear Algebra with Mathematica

Preface


 

For a real number μ, a generalized Fourier transform is defined as
\[ \hat{q} (k) = \int_0^{\infty} q(r)\left[ \cos (kr) - \frac{\mu}{k}\,\sin (kr) \right] {\text d}r . \]
Its inverse transform is
\[ q (r) = \frac{2}{\pi} \int_0^{\infty} \hat{q} (k)\,\frac{k^2}{\mu^2 + k^2} \left[ \cos (kr) - \frac{\mu}{k}\,\sin (kr) \right] {\text d}k . \]
Example 8: Consider the exponential function \( f(r) = e^{-\alpha\,r} . \) Its generalized Fourier transform is
\[ \int_0^{\infty} \left[ \cos (kr) - \frac{\mu}{k}\,\sin (kr) \right] e^{-\alpha\,r} \, {\text d}r = \frac{\alpha - \mu}{\alpha^2 + k^2} , \qquad \Re\alpha > \Im k. \]
Integrate[Exp[-alpha*r]*(Cos[k*r] - mu/k*Sin[k*r]), {r, 0, Infinity}]
ConditionalExpression[(alpha - mu)/(alpha^2 + k^2), Re[alpha] > Im[k]]
Integrate[ Exp[-alpha*r^2]*(Cos[k*r] - mu/k*Sin[k*r]), {r, 0, Infinity}]
ConditionalExpression[( E^(-(k^2/(4 alpha))) Sqrt[\[Pi]] (k - mu Erfi[k/(2 Sqrt[alpha])]))/( 2 Sqrt[alpha] k), Re[alpha] >= 0]
Using previously determined formulas, we find
\begin{align*} \int_0^{\infty} \left[ \cos (kr) - \frac{\mu}{k}\,\sin (kr) \right] e^{-\alpha\,r^2} \, {\text d}r &= \int_0^{\infty} \cos (kr) \, e^{-\alpha\,r^2} \, {\text d}r - \frac{\mu}{k} \,\int_0^{\infty} \sin (kr) \, e^{-\alpha\,r^2} \, {\text d}r \\ &= \end{align*}