Re: integral of sin(x)/x

On 17 Nov 2005 06:44:18 -0800, stevendaryl3016@xxxxxxxxx (Daryl
McCullough) wrote:

>In article <dldgur$f6k$1@xxxxxxxxxxxxxxxxxx>, david says...
>>
>>can someone give me a quick explanation why
>>intergral from 0 to infinity of (sin(x)/x) dx
>>is pi/2?
>
>Here's an argument which might be begging the question,
>but let me give it anyway.
>
>Let f(x) be the discontinuous function defined by
>
> f(x) = 1, if |x| < 1
> = 0 otherwise
>
>Take the Fourier transform of f(x) to get g(k):
>
> g(k) = integral from x= -infinity to +infinity
> of f(x) exp(ikx) dx
>
>Using the definition of f(x), this simplifies to
>
> = integral from x=-1 to +1 of exp(ikx) dx
> = 2/(ik) (exp(ik) - exp(-ik))
> = 2 sin(k)/k
>
>Now, use the inverse transform:
>
> f(x) = 1/(2 pi) integral from k= -infinity to +infinity
> of g(k) exp(-ikx) dk
>
>At x=0, this becomes
>
> f(0) = 1/(2 pi) integral from k= -infinity to +infinity
> of 2 sin(k)/k dk
>
> = 2/pi integral from k=0 to infinity of sin(k)/k dk
>
>So,
>
> integral from k=0 to infinity of sin(k)/k dk
> = pi/2 f(0)
> = pi/2
>
>The reason that this derivation begs the question is that
>you actually have to know that the integral of sin(k)/k
>is pi/2 to prove that taking the Fourier transform of
>the Fourier transform gives the original function back.

As Grubb pointed out, that's not really a problem,
but it's not the only problem here. However it's
not hard to give a rigorous argument inspired by
the above - see my reply to his post.


************************

David C. Ullrich
.

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