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If $f(x)$ is a "very sharply peaked function" why is $$\int_{-\infty}^\infty e^{ikpx}f(x)\;dx,$$ where $k$ is a constant , a "very broad function of $p$"? Is there a way to visualize this? Thanks :)

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Here is my take on this, based on the scaling properties of the Fourier Transform.

Note that as $\epsilon\to0$, the graph of $f(\epsilon x)$ stretches out along the $x$-axis, thus becomes more "broad". At the same time, the graph of $f(x/\epsilon)/\epsilon$ squeezes in along the $x$-axis and stretches out along the $y$-axis, thus becoming more "sharply peaked"

Fix $k$ and let $$ F(p)=\int_{-\infty}^\infty e^{ikpx}f(x)\;\mathrm{d}x $$ Then $$ \begin{align} F(\epsilon p)&=\int_{-\infty}^\infty e^{ik\epsilon px}f(x)\;\mathrm{d}x\\ &=\int_{-\infty}^\infty e^{ikpx}f(x/\epsilon)/\epsilon\;\mathrm{d}x \end{align} $$ Thus, as $\epsilon\to0$, $f(x/\epsilon)/\epsilon$ becomes more "sharply peaked" and $F(\epsilon p)$ becomes more "broad" .

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Thanks for the explanation :) –  Pete M Aug 27 '11 at 8:12

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