Question

If f(x) represented by Fourier integral

 \(f(x)=\int_0^{\infty} [A(ω) cos~ω x + B(ω) sin~ω x]dω \)

then A(ω) is defined as

1. ​\(\frac{1}{\pi}\int_{-\infty}^{\infty} f(v) ~cos~\omega v ~dv\)
2. ​​\(​\frac{1}{\pi}\int_{-\infty}^{\infty} f(v) ~sin~\omega v ~dv\)
3. \(​\int_{-\infty}^{\infty} f(\omega) ~cos~\omega v ~dv\)
4. \(​\int_{-\infty}^{\infty} f(\omega) ~sin~\omega v ~dv\)

Answer 1

Correct Answer - Option 1 : ​\(\frac{1}{\pi}\int_{-\infty}^{\infty} f(v) ~cos~\omega v ~dv\)

Fourier series of f(x) of period 2T is given as:

\(f_T(x)=a_0+\sum_{n=1}^{∞} (a_n cos ~ω_n x+b_n sin~ ω_n x)\)

Putting the values of a₀, a_n and b_n we get,

\(f_T(x)=\frac{1}{2T}\int_{-T}^{T} f_T(v) dv + \frac{1}{T}\sum_{n=1}^{∞} [cos ~ω_nx\int_{-T}^{T} f_T(v) cos ~(ω_nv )dv+sin ~ω_nx\int_{-T}^{T} f_T(v) sin~(ω_nv) dv]\)

Δ ω = ωn+1 - ωn

= \(\frac{2\pi (n+1)}{2T}-\frac{2\pi n}{2T}\)

= \(\frac{\pi}{T}\)

\(\frac{1}{T}=\frac{\Delta \omega}{\pi}\)

∴ \(f_T(x)=\frac{1}{2T}\int_{-T}^{T} f_T(v) dv + \frac{1}{\pi}\sum_{n=1}^{∞} [cos ~(ω_nx) ~\Delta{\omega}\int_{-T}^{T} f_T(v) cos ~(ω_nv )dv+sin ~(ω_nx) \Delta\omega\int_{-T}^{T} f_T(v) sin~(ω_nv) dv]\)

When T → ∞ 

\(f(x)=\)

 \(\frac{1}{\pi}\int_{0}^{∞} [cos ~(ωx) \int_{-\infty}^{\infty} f(v) cos ~(ωv )dv]~d{\omega} +\frac{1}{\pi}[sin ~(ωx)\int_{-\infty}^{\infty} f(v) sin~(ωv) dv]~d{\omega}\)    ---(1)

In the question f(x) is given as:

 \(f(x)=\int_0^{\infty} [A(ω) cos~ω x + B(ω) sin~ω x]dω \)  ------(2)

By comparing (1) and (2), we get

\(A(\omega)= \frac{1}{\pi} \int_{-\infty}^{\infty} f(v) cos ~(ωv )dv \)

\(B(\omega)=\frac{1}{\pi}\int_{-\infty}^{\infty} f(v) sin~(ωv) dv\)