Difference between revisions of "Aufgaben:Problem 12"
(Created page with " ''Draft: 3 b)'' (wird noch alles schön dargestellt) $$ \vdash : ||\hat \chi_{[a,b]}||_2^2 = b - a $$ for $$ a, b \in \mathbb{R} $$. Proof. We show the identity witho...") |
|||
| Line 26: | Line 26: | ||
$$ ||\hat \chi_{[a,b]}(k)||_2^2 = \frac{(b-a)^2}{(b - a) \pi} \int_{\mathbb{R} } \frac{sin^2(\eta)}{\eta^2} d\eta $$ | $$ ||\hat \chi_{[a,b]}(k)||_2^2 = \frac{(b-a)^2}{(b - a) \pi} \int_{\mathbb{R} } \frac{sin^2(\eta)}{\eta^2} d\eta $$ | ||
| − | It remains to evaluate the integral $$ I(\alpha) = \int_{\mathbb{R} } \frac{sin^2(\alpha \eta)}{\eta^2} d\eta $$ at $$ \alpha = 1$$. | + | It remains to evaluate the integral $$ I(\alpha) = \int_{\mathbb{R} } \frac{\sin^2(\alpha \eta)}{\eta^2} d\eta $$ at $$ \alpha = 1$$. |
| − | $$ \frac{dI}{d\alpha} = 2 \int_{0}^{\infty} \frac{2 \eta sin(\alpha \eta) cos(\alpha \eta) }{\eta} $$ and with the identity $$ sin(x) cos(y) = \frac{1}{2}\Big(\sin (x-y) + \sin (x+y)\Big) $$ it follows that: | + | $$ \frac{dI}{d\alpha} = 2 \int_{0}^{\infty} \frac{2 \eta \sin(\alpha \eta) \cos(\alpha \eta) }{\eta^2} $$ and with the identity $$ \sin(x) \cos(y) = \frac{1}{2}\Big(\sin (x-y) + \sin (x+y)\Big) $$ it follows that: |
$$ \frac{dI}{d\alpha} = \pi $$. | $$ \frac{dI}{d\alpha} = \pi $$. | ||
Revision as of 10:31, 23 December 2014
Draft: 3 b) (wird noch alles schön dargestellt)
$$ \vdash : ||\hat \chi_{[a,b]}||_2^2 = b - a $$ for $$ a, b \in \mathbb{R} $$.
Proof.
We show the identity without making use of Plancherel's theorem.
We have: $$ \hat \chi_{[a,b]} = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \chi_{[a,b]}(x) e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} \int_{a}^b e^{-ikx} dx$$
Now, let: $$ x \equiv \frac{2(y-a)}{b-a} -1 $$
$$ \hat \chi_{[a,b]} = \frac{2}{(b-a)\sqrt{2\pi}} e^{-\frac{1}{2} ik(b + a)}\int_{-1}^{1} e^{-ik \frac{b-a}{2} y} dy = \frac{2}{\sqrt{2\pi}} e^{-\frac{1}{2} ik(b + a)} \frac{sin(\frac{b-a}{2}k)}{k} $$
And thus:
$$ ||\hat \chi_{[a,b]}(k)||_2^2 = \int_{\mathbb{R} } |\hat \chi_{[a,b]}(k)|^2 dk = \frac{2}{\pi} \int_{\mathbb{R} } \frac{sin^2(\frac{b-a}{2}k)}{k^2} dk $$
With the substitution $$ \eta \equiv \frac{b-a}{2}k $$ we get:
$$ ||\hat \chi_{[a,b]}(k)||_2^2 = \frac{(b-a)^2}{(b - a) \pi} \int_{\mathbb{R} } \frac{sin^2(\eta)}{\eta^2} d\eta $$
It remains to evaluate the integral $$ I(\alpha) = \int_{\mathbb{R} } \frac{\sin^2(\alpha \eta)}{\eta^2} d\eta $$ at $$ \alpha = 1$$.
$$ \frac{dI}{d\alpha} = 2 \int_{0}^{\infty} \frac{2 \eta \sin(\alpha \eta) \cos(\alpha \eta) }{\eta^2} $$ and with the identity $$ \sin(x) \cos(y) = \frac{1}{2}\Big(\sin (x-y) + \sin (x+y)\Big) $$ it follows that:
$$ \frac{dI}{d\alpha} = \pi $$.
Solving this equation is a thing we learned a long time ago:
$$ I(\alpha) = \pi \alpha + c $$ and since $$ I(0) = 0 $$ :
$$ I(\alpha) = \pi \alpha $$.
As $$ I(1) = \pi $$ the claim is proven. $$\square$$
