Find the Fourier transform off (t) ={ 1, |t| < 1;0, |t| > 1.Hence evaluate the integral ∫ ∞0 sin tt dt

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 for |t| < 1 and 0 for |t| > 1, the Fourier transform becomes:

F(ω) = ∫ e^(-iωt) dt, from -1 to 1

This integral can be solved by breaking it into real and imaginary parts:

F(ω) = ∫ cos(ωt) dt - i ∫ sin(ωt) dt, from -1 to 1

The first integral gives:

∫ cos(ωt) dt = [sin(ωt) / ω] from -1 to 1 = 2 sin(ω) / ω

The second integral gives:

∫ sin(ωt) dt = [-cos(ωt) / ω] from -1 to 1 = 0

So, the Fourier transform of f(t) is F(ω) = 2 sin(ω) / ω.

Now, to evaluate the integral ∫ sin(t) / t dt from 0 to ∞, we can use the Fourier transform result. The integral is equal to the value of the Fourier transform at ω = 1:

∫ sin(t) / t dt = F(1) = 2 sin(1) / 1 = 2 sin(1)

So, the value of the integral is approximately 1.682941969615793.