Using Fourier integral representation show that 

To show this, we'll use the Fourier integral representation of the given function:

\[ f(x) = \int_{-\infty}^{\infty} F(k) e^{ikx} dk \]

where \( F(k) \) is the Fourier transform of \( f(x) \). In this case, \( f(x) \) is given as:

\[ f(x) = \frac{\cos(\alpha x) + \alpha \sin(\alpha x)}{1 + \alpha^2} \]

We need to find \( F(k) \), the Fourier transform of \( f(x) \). The Fourier transform \( F(k) \) of \( f(x) \) is given by:

\[ F(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} dx \]

Let's find \( F(k) \):

\[ F(k) = \int_{-\infty}^{\infty} \frac{\cos(\alpha x) + \alpha \sin(\alpha x)}{1 + \alpha^2} e^{-ikx} dx \]

\[ = \frac{1}{1 + \alpha^2} \left( \int_{-\infty}^{\infty} \cos(\alpha x) e^{-ikx} dx + \alpha \int_{-\infty}^{\infty} \sin(\alpha x) e^{-ikx} dx \right) \]

\[ = \frac{1}{1 + \alpha^2} \left( \frac{1}{2} \left( \int_{-\infty}^{\infty} e^{i(\alpha - k)x} dx + \int_{-\infty}^{\infty} e^{i(-\alpha - k)x} dx \right) + \alpha \frac{1}{2i} \left( \int_{-\infty}^{\infty} e^{i(\alpha - k)x} dx - \int_{-\infty}^{\infty} e^{i(-\alpha - k)x} dx \right) \right) \]

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