Prove. The following are equivalent forms of the Fourier integral:
Part 1. We shall first show that 2) is equivalent to 1). We can move the factor eiωt in 1) into the inner integral since it doesn’t involve the variable s. Thus 1) becomes
Now, using the general formula eiθ = cos θ + i sinθ, we replace the factor eiω(t-s) with its trigonometric equivalent to get
Now because f(t) is by hypothesis real, the imaginary part of 7) is zero. Thus 7) becomes
which is 2) above.
Part 2. We shall now show that 3) is equivalent to 2). First we note that because the integrand of 2) is an even function of ω, we need perform the ω-integration only between 0 and , provided we multiply the result by 2. Thus 2) is equivalent to
We now substitute 4) into 3) to obtain
Now using the general formula from trigonometry
cos (A - B) = cos A cos B + sin A sin B
we rewrite 10) as
which is 9) and equivalent to 2).