Question

Fourier cosine integral f(x) is​

Answer 1

Answer:

Fourier Integrals. - Application of Fourier series to nonperiodic function

Step-by-step explanation:

The function x(t) can be recovered by the inverse Fourier transform, i.e., ... It defines the Fourier transform (also the Fourier sine and cosine transforms) and develops the Fourier integral theorem

Answer 2

Answer:

Fourier cosine integral f(x) is​ $\hat{f}_c(\omega) = \int_{0}^{\infty} f(x) \cos{\omega x} \, dx.$

Step-by-step explanation:

Fourier Sine and Cosine Integrals:

The Fourier integral becomes simpler when the given function in either an even or odd function. The Fourier integral of f(x) is described as

$\left.f(x)=\int_{0}^{\infty} A(w) \cos w x+B(w) \sin w x\right) d w \\\\Where \mathrm{A}(\mathrm{w})=\frac{1}{\pi} \int_{-\infty}^{\infty} f(v) \cos w v d v \quad and \quad B(w)=\frac{1}{\pi} \int_{-\infty}^{\infty} f(v) \sin w v d v$

Case-1:

Suppose f(x) is an even function i.e.

f(-x)=f(x).

cos(wx) is also an even function and sinwx is an odd function. Therefore $f(x) \cos w x$ is also an even function and $f(x) \sin w x$ is an odd function.

$\ A(w)=\frac{1}{\pi} \int_{-\infty}^{\infty} f(v) \cos w v d v=\frac{2}{\pi} \int_{-\infty}^{\infty} f(v) \cos w v d v$$\ A(w)=\frac{1}{\pi} \int_{-\infty}^{\infty} f(v) \cos w v d v=\frac{2}{\pi} \int_{-\infty}^{\infty} f(v) \cos w v d v And B(w)=\frac{1}{\pi} \int_{-\infty}^{\infty} f(v) \sin w v d v=0 \\Therefore f(x)=\int_{0}^{\infty} A(w) \cos w x d w This is known as the Fourier Cosine integral representation.$