Question

If F(s) is the Fourier transform of f(x), then F
[f(x)cos ax] =______​

Answer 1

Step-by-step explanation:

If F(s) is the Fourier transform of f(x), then F

[f(x)cos ax] =______

Answer 2

Answer:

If F(s) is the Fourier transform of f(x), then  $F[f(x)cos ax]=\dfrac{1}{2} \Big\{F(s+a)+ F(s-a)\Big\}$

Step-by-step explanation:

Given F(s) is the Fourier transform of f(x).

The Fourier transform of f(x) is

$F(s)=\dfrac{1}{\sqrt{2\pi } } \Big {\displaystyle\int\limits^{\infty} _{-\infty} {e^{isx} f(x)} \, dx$

For $F[f(x)cos ax]$,

$F[f(x)cos ax]=\dfrac{1}{\sqrt{2\pi } } \Big {\displaystyle\int\limits^{\infty} _{-\infty} {e^{isx} f(x)cos ax} \, dx$

According to Euler's rule, $cox=\dfrac{e^{ix} +e^{-ix}}{2}$

Applying the same for $cosax$,  

$F[f(x)cos ax]=\dfrac{1}{\sqrt{2\pi } } \Big {\displaystyle\int\limits^{\infty} _{-\infty} {e^{isx} f(x)\dfrac{e^{iax} +e^{-iax}}{2}} \, dx$

$=\dfrac{1}{2} \Bigg\{\dfrac{1}{\sqrt{2\pi } } \Big {\displaystyle\int\limits^{\infty} _{-\infty} {e^{isx} \big (e^{iax} +e^{-iax}}\big) f(x)\ dx\Bigg\}$

$=\dfrac{1}{2} \Bigg\{\dfrac{1}{\sqrt{2\pi } }\Bigg\{\Big {\displaystyle\int\limits^{\infty} _{-\infty} {e^{isx} . e^{iax} f(x)\ dx+ \Big {\displaystyle\int\limits^{\infty} _{-\infty} {e^{isx} .e^{-iax}} f(x)\ dx\Bigg\}\Bigg\}$

$=\dfrac{1}{2} \Bigg\{\dfrac{1}{\sqrt{2\pi } }\Big {\displaystyle\int\limits^{\infty} _{-\infty} {e^{i(s+a)x} f(x)\ dx+ {\dfrac{1}{\sqrt{2\pi } }\Big {\displaystyle\int\limits^{\infty} _{-\infty} {e^{i(s-a)x} }} f(x)\ dx\Bigg\}$

Using the Fourier transform again,

$=\dfrac{1}{2} \Big\{F(s+a)+ F(s-a)\Big\}$

Therefore, if F(s) is the Fourier transform of f(x), then $F[f(x)cos ax]=\dfrac{1}{2} \Big\{F(s+a)+ F(s-a)\Big\}$