# Integral with Dirac delta (me or wolfram mathematica?)

I tried to compute with Wolfram Mathematica the following integral $$I=\int_0^\pi\int_{-\infty}^\infty x e^{-jx\cos(\theta-\varphi)}f(\alpha\cos(\theta-\psi))\mathrm \, dx\mathrm \, d\theta$$ where $-\pi\leq\varphi,\psi\leq\pi$.

Assuming that the inner intergral is by definition the derivative of the Dirac delta function I get: $$I=2\pi j\int_0^\pi\delta'(\cos(\theta-\varphi))f(\alpha\cos(\theta-\psi)) \, \mathrm \, d\theta$$ Wolfram Mathematica tells me that $$I_\mathrm{wolfram}=0$$ Then I tried to do it by hand. If I use the definition of a delta function: $$\begin{eqnarray} I_\mathrm{me}=\delta^{'} (\cos(\theta-\varphi))&=&\left[ \sum_i \frac{\delta(\theta-\varphi-\frac{\pi}{2}-i\pi)}{|\sin(\frac{\pi}{2}+i\pi)|}\right]^{'}=\sum_i \delta^{'}\left(\theta-\left(\varphi+\frac{\pi}{2}+i\pi\right)\right) \end{eqnarray}$$ $$\begin{eqnarray} I_\mathrm{me}&=&2\pi j\sum_i\int_0^\pi\delta'\left(\theta-\left(\varphi+\frac{\pi}{2}+i\pi\right)\right)f(\alpha\cos(\theta-\psi)) \, \mathrm d\theta=\\ &=&-2\pi j\int_0^\pi\delta\left(\theta-\left(\varphi+\frac{\pi}{2}\right)\right)f'(\alpha\cos(\theta-\psi)) \, \mathrm d\theta=\\ &=&-2\pi j f'(\alpha\cos(\varphi+\frac{\pi}{2}-\psi))u(\pi -2 \varphi ) u(2 \varphi +\pi ) \end{eqnarray}$$ where $u(\cdot)$ is a Heaviside function. And it is actually not $0$. So where's the catch?

## 1 Answer

\begin{align} &{\rm I}\left(\varphi,\psi\right) \equiv \int_{0}^{\pi}{\rm d}\theta\ \ {\rm f}\left(\alpha\cos\left(\theta - \psi\right)\right) \int_{-\infty}^{\infty}{\rm d}x\ \ x\, {\rm e}^{-{\rm i}x\cos\left(\theta - \varphi\right)} \\[3mm]&= \int_{0}^{\pi}{\rm d}\theta\ \ {\rm f}\left(\alpha\cos\left(\theta - \psi\right)\right)\left\lbrack% {2\pi \over {\rm i}\sin\left(\theta - \psi\right)}\, {\partial \over \partial\theta}\ \overbrace{\quad% \int_{-\infty}^{\infty}{{\rm d}x \over 2\pi}\, {\rm e}^{-{\rm i}x\cos\left(\theta - \varphi\right)}\quad} ^{\delta\left(\cos\left(\theta - \varphi\right)\right)} \right\rbrack \\[3mm]&= 2\pi{\rm i}\int_{0}^{\pi}{\rm d}\theta\, \delta\left(\cos\left(\theta - \varphi\right)\right)\, {\partial \over \partial\theta}\left\lbrack% {{\rm f}\left(\alpha\cos\left(\theta - \psi\right)\right) \over \sin\left(\theta - \psi\right)} \right\rbrack = 2\pi{\rm i}\int_{0}^{\pi}\!\!\!{\rm d}\theta\, \delta\left(\cos\left(\theta - \varphi\right)\right)\, {\partial{\rm F}\left(\alpha,\theta - \psi\right) \over \partial\theta} \end{align}

$\displaystyle{% \mbox{where}\quad {\rm F}\left(\alpha,\phi\right) \equiv {{\rm f}\left(\alpha\cos\left(\phi\right)\right) \over \sin\left(\phi\right)}}$

\begin{align} -------&---------------------------------- \\[3mm] {\rm I}\left(\varphi,\psi\right) &= -2\pi{\rm i}\,{\partial \over \partial\psi} \int_{0}^{\pi}{\rm d}\theta\, {\rm F}\left(\alpha,\theta - \psi\right) \delta\left(\cos\left(\theta - \varphi\right)\right) \\[3mm]&= -2\pi{\rm i}\,{\partial \over \partial\psi} \int_{0}^{\pi}{\rm d}\theta\, {\rm F}\left(\alpha,\theta - \psi\right) \sum_{\ell = -\infty}^{\infty} {\delta\left(\theta - \theta_{\ell}\left(\varphi\right)\right) \over \left\vert -\sin\left(\theta_{\ell}\left(\varphi\right) - \varphi\right) \right\vert} \end{align} where $$\theta_{\ell}\left(\varphi\right) \equiv \varphi + \left(2\ell + 1\right)\pi/2. \quad\mbox{Notice that}\quad \left\vert% \begin{array}{rcl} \sin\left(\theta_{\ell}\left(\varphi\right) - \varphi\right) & = & \left(-1\right)^{\ell} \\ \sin\left(\theta_{\ell}\left(\varphi\right) - \psi\right) & = & \left(-1\right)^{\ell}\cos\left(\varphi - \psi\right) \\ \cos\left(\theta_{\ell}\left(\varphi\right) - \psi\right) & = & \left(-1\right)^{\ell + 1}\sin\left(\varphi - \psi\right) \end{array}\right.$$ Then \begin{align} {\rm I}\left(\varphi,\psi\right) &= -2\pi{\rm i}\sum_{\ell = -\infty}^{\infty} {\partial{\rm F}\left(\alpha,\theta_{\ell}\left(\varphi\right) - \psi\right) \over \partial\psi} \int_{0}^{\pi}{\rm d}\theta\, \delta\left(\theta - \theta_{\ell}\left(\varphi\right)\right) \\[3mm]&= -2\pi{\rm i}\!\!\!\!\!\!\!\!\!\! \sum_{% \ell = -\infty \atop {0\ <\ \theta_{\ell}\left(\varphi\right)\ <\ \pi\vphantom{\Huge A^{A}}}} ^{\infty}\!\!\!\!\!\!\!\!\!\! {\partial{\rm F}\left(\alpha,\theta_{\ell}\left(\varphi\right) - \psi\right) \over \partial\psi} = -2\pi{\rm i}\!\!\!\!\!\!\!\!\!\! \sum_{% \ell = -\infty \atop {\left\vert\ell + \varphi/\pi\right\vert\ <\ 1/2\vphantom{\Huge A}}} ^{\infty}\!\!\!\!\!\!\!\!\!\! {\partial{\rm F}\left(\alpha,\varphi - \psi + \ell\pi + \pi/2\right) \over \partial\psi} \end{align}

$$\left\lbrace% \begin{array}{rcl} \cos\left(\phi + \ell\pi + {\pi \over 2}\right) & = & -\sin\left(\ell\pi + \phi\right) = \left(-1\right)^{\ell + 1}\sin\left(\phi\right) \\ \sin\left(\phi + \ell\pi + {\pi \over 2}\right) & = & \phantom{-}\cos\left(\ell\pi + \phi\right) = \left(-1\right)^{\ell\phantom{+1}}\cos\left(\phi\right) \end{array}\right.$$

\begin{align} ---&-------------------------------------- \\ {\rm I}\left(\varphi,\psi\right) &= \color{#ff0000}{-2\pi{\rm i}\!\!\!\!\!\!\!\!\!\! \sum_{% \ell = -\infty \atop {\left\vert\ell + \varphi/\pi\right\vert\ <\ 1/2\vphantom{\Huge A}}} ^{\infty}\!\!\!\!\!\!\!\!\!\!\left(-1\right)^{\ell}\, {\partial \over \partial\psi}\left\lbrack% {{\rm f}\left(\left(-1\right)^{\ell + 1}\alpha \sin\left(\varphi - \psi\right)\right) \over \cos\left(\varphi - \psi\right)} \right\rbrack = {\rm I}_{1}\left(\phi,\psi\right) + {\rm I}_{2}\left(\phi,\psi\right)} \\[5mm] {\rm I}_{1}\left(\varphi,\psi\right) &= \color{#ff0000}{-2\pi{\rm i}\alpha\!\!\!\!\!\!\!\!\!\! \sum_{% \ell = -\infty \atop {\left\vert\ell + \varphi/\pi\right\vert\ <\ 1/2\vphantom{\Huge A}}} ^{\infty}\!\!\!\!\!\!\!\!\!\! {\rm f}'\left(\vphantom{\LARGE A}\left(-1\right)^{\ell + 1}\alpha \sin\left(\varphi - \psi\right)\right)\sec\left(\varphi - \psi\right)} \\[3mm] {\rm I}_{2}\left(\varphi,\psi\right) &= \color{#ff0000}{2\pi{\rm i}\!\!\!\!\!\!\!\!\!\! \sum_{% \ell = -\infty \atop {\left\vert\ell + \varphi/\pi\right\vert\ <\ 1/2\vphantom{\Huge A}}} ^{\infty}\!\!\!\!\!\!\!\!\!\!\left(-1\right)^{\ell} {\rm f}\left(\vphantom{\LARGE A}\left(-1\right)^{\ell + 1}\alpha \sin\left(\varphi - \psi\right)\right)\sec\left(\varphi - \psi\right) \tan\left(\varphi - \psi\right)} \end{align}

• There are some formatting issues in the 3 last expressions. – user93957 Oct 23 '13 at 12:45
• @Adobe I'll check later. I'm so busy right now. – Felix Marin Oct 24 '13 at 0:58