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We know that $\int_{-\infty}^{\infty}e^{-itw}dt=2\pi\delta(w)$, but how to calculate the half integration $\int_{0}^{\infty}e^{-itw}dt$?

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The integral $\int_{-\infty}^\infty e^{-i\omega t}\,dt$ diverges as either a Lebesgue integral or improper Riemann integral. However, for every "suitable" (smooth with compact support) test function $f(\omega)$, we have

$$\begin{align} \lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\left(\int_{-L}^L e^{-i\omega t}\,dt\right)\,d\omega&=\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\,\frac{2\sin(\omega L)}{\omega}\,d\omega\\\\ &=2\pi f(0) \end{align}$$

Therefore, we can interpret the symbol "$\int_{-\infty}^\infty e^{-i\omega t}\,dt$" to be the regularization of the Dirac Delta

$$\lim_{L\to \infty}\frac{1}{2\pi}\int_{-L}^L e^{-i\omega t}\,dt\sim \delta(\omega)$$

which is interpreted to mean

$$\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\left(\int_{-L}^L e^{-i\omega t}\,dt\right)\,d\omega =2\pi f(0)$$

Now, applying the regularization $\lim_{L\to \infty}\int_{0}^L e^{-i\omega t}\,dt$ to a suitable test function reveals

$$\begin{align} \lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\left(\int_{0}^L e^{-i\omega t}\,dt\right)\,d\omega&=\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\,\left(\frac{\sin(\omega L)}{\omega}-i\,\frac{1-\cos(\omega L)}{\omega}\right)\,d\omega\\\\ &=\pi f(0)-i\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\,\frac{1-\cos(\omega L)}{\omega}\,d\omega \tag 1\\\\ &=\pi f(0)-i\text{PV}\left(\int_{-\infty}^\infty \frac{f(\omega)}{\omega}\,d\omega\right) \end{align}$$

where $\text{PV}$ denotes that Cauchy-Principal Value. Note that we used the Riemann-Lebesgue Lemma assuming that $\frac{f(\omega)}{\omega}$ is smooth and of compact support for $|\omega| <\epsilon$ for all $\epsilon >0$ to evaluate the integral on the right-hand side of $(1)$.

Therefore, we can write

$$\lim_{L\to \infty}\int_{0}^L e^{-i\omega t}\,dt\sim \pi \delta(\omega)+ \text{PV}\left(\frac{-i}{\omega}\right)$$


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  • $\begingroup$ Thank you so much! Your answer is really what I want to know! $\endgroup$ – zito Aug 26 '16 at 1:30
  • $\begingroup$ @zito You're welcome! My pleasure. -Mark $\endgroup$ – Mark Viola Aug 26 '16 at 2:03
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The integral $$ \int_{0}^{\infty}e^{-itw}dt $$ doesn't exist in the normal functional sense. So it's pretty much the same as in the case of $\displaystyle\int_{-\infty}^{\infty}e^{-itw}dt=2\pi\delta(w)$, where you already (maybe implicitly) applied the distributional interpretation.

However, where you can see $$ \int_{-\infty}^{\infty}e^{-itw}dt=2\pi\delta(w) $$ as the Fourier transform of the constant function $f(x)=1$, you can see $$ \int_{0}^{\infty}e^{-itw}dt $$ as the Fourier transform of the Heaviside function $H(x)=\begin{cases}1,x>0\\0,x<0\end{cases}$. As a result you'll get $$ \hat{H}(w)=\int_{0}^{\infty}e^{-itw}dt=\pi\delta(w)+\frac 1{i w} $$ for a detailed derivation and further references I recommend you for example this and of course this answer.

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  • $\begingroup$ Thank you so much! It is helpful to consider the integral as the Fourier transform of the Heaviside function. $\endgroup$ – zito Aug 26 '16 at 1:24
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, $\ds{\Theta\pars{t} = \int_{-\infty}^{\infty}{\expo{\ic\nu t} \over \nu - \ic 0^{+}} \,{\dd\nu \over 2\pi\ic}}$ is the Heaviside Step Function.

\begin{align} \color{#f00}{\int_{0}^{\infty}\expo{-\ic wt}\,\dd t} & = \int_{-\infty}^{\infty}\Theta\pars{t}\expo{-\ic wt}\,\dd t = \int_{-\infty}^{\infty}\bracks{\int_{-\infty}^{\infty} {\expo{\ic\nu t} \over \nu - \ic 0^{+}}\,{\dd\nu \over 2\pi\ic}} \expo{-\ic wt}\,\dd t \\[5mm] & = -\ic\int_{-\infty}^{\infty}{1 \over \nu - \ic 0^{+}} \bracks{\int_{-\infty}^{\infty} \expo{\ic\pars{\nu - w}t}\,{\dd t \over 2\pi}}\,\dd\nu = -\ic\int_{-\infty}^{\infty}{\delta\pars{\nu - w} \over \nu - \ic 0^{+}}\,\dd\nu \\[5mm] & = -\ic\,{1 \over w - \ic 0^{+}} = \color{#f00}{-\ic\,\mrm{P.V.}\pars{1 \over w} + \pi\delta\pars{w}} \end{align} The result must be understood 'under the integral sign'. Namely, $$ \int_{-\infty}^{\infty}\mrm{f}\pars{w}\int_{0}^{\infty}\expo{-\ic wt} \,\dd t\,\dd w = -\ic\,\mrm{P.V.}\int_{-\infty}^{\infty}{\mrm{f}\pars{w} \over w}\,\dd w + \pi\mrm{f}\pars{0} $$

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Hint: maybe you can show symmetry- think of euler's identity

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