Find the Fourier transform off (t) ={ 1 − |t|, if |t| < 1;0, |t| > 1and hence deduce that∫ ∞0( sin tt)4dt = π3

The Fourier transform of a function f(t) is given by the integral:

F(ω) = ∫ f(t) e^(-iωt) dt, from -∞ to ∞

For the given function f(t) = 1 - |t| for |t| < 1 and 0 for |t| > 1, the Fourier transform becomes:

F(ω) = ∫ (1 - |t|) e^(-iωt) dt, from -1 to 1

This integral can be split into two parts, one from -1 to 0 and one from 0 to 1. The absolute value function |t| can be replaced by -t for the first part and t for the second part. The integral then becomes:

F(ω) = ∫ (1 + t) e^(-iωt) dt, from -1 to 0 + ∫ (1 - t) e^(-iωt) dt, from 0 to 1

These integrals can be solved by using the method of integration by parts. The result is:

F(ω) = [ (1 + t) (-i/ω) e^(-iωt) ] from -1 to 0 + [ (1 - t) (-i/ω) e^(-iωt) ] from 0 to 1 - ∫ (-i/ω) e^(-iωt) dt, from -1 to 0 - ∫ (-i/ω) e^(-iωt) dt, from 0 to 1

The first two terms evaluate to zero because e^(-iωt) goes to zero for t going to ±1. The remaining integrals can be solved easily, yielding:

F(ω) = 2/ω^2

The integral ∫ (sin(t)/t)^4 dt from 0 to ∞ is related to the Fourier transform of f(t) by Parseval's theorem, which states that the integral of the square of a function is equal to the integral of the square of its Fourier transform, both divided by 2π. Therefore, we have:

∫ (sin(t)/t)^4 dt = ∫ |F(ω)|^2 dω / (2π)

Substituting the result for F(ω), we get:

∫ (sin(t)/t)^4 dt = ∫ (2/ω^2)^2 dω / (2π) = ∫ 4/ω^4 dω / (2π) = [ -4/3ω^3 ] from 0 to ∞ / (2π) = π/3

Therefore, the integral ∫ (sin(t)/t)^4 dt from 0 to ∞ is equal to π/3.