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By directly evaluating the integral in the definition, show that the Fourier transform of U_{0}(x)=\left\{\begin{array}{clr} 1+\cos (x) & : & |x|<\pi \\ 0 & z & \text { otherwise }
\end{array}\right. is given by \tilde{U}_{0}(k)=\frac{2 \sin (\pi k)}{k\left(1-k^{2}\right)} The function u(z, y) satisfies the partial differential equation \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 \quad \text { in } \quad-\infty
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