My Math Forum Is integral just trick?

 Real Analysis Real Analysis Math Forum

 June 3rd, 2013, 07:25 PM #1 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Is integral just trick? $\int_{\mathbb{R}}\frac{\exp{(-x^2)}\cos{(ax+b)}}{x^2+\pi ^2}dx$, where $a,b\in \mathbb{R}$
 June 3rd, 2013, 11:29 PM #2 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Is integral just trick? A simple trick: Let $f(x)=\frac{\exp{(-x^2)}}{x^2+\pi ^2}$. Note that $\cos(ax+b)=\frac{1}{2}(\exp(i(ax+b))+\exp(-i(ax+b)))$, so we have: $\int_{\mathbb{R}}f(x)\cos(ax+b)dx$ $=\frac{1}{2}\int_{\mathbb{R}}f(x)\exp(i(ax+b))dx+\ frac{1}{2}\int_{\mathbb{R}}f(x)\exp(-i(ax+b))dx$ $=\frac{1}{2}\exp(ib)\int_{\mathbb{R}}f(x)\exp{(iax )}dx+\frac{1}{2}\exp(-ib)\int_{\mathbb{R}}f(x)\exp(-iax)dx$ $=\frac{1}{2}\exp(ib)F^*(\omega)\|_{\omega=a}+\frac {1}{2}\exp(-ib)F(\omega)\|_{\omega=a}$, with notations: $F(\omega)=\int_{\mathbb{R}}f(x)\exp(-i\omega x)dx$ (the Fourier transform of f(x)) and $F^*(\omega)=\int_{\mathbb{R}}f(x)\exp(i\omega x)dx$ ('conjugate' Fourier transform of f(x)). Since $f(x)=f(-x)$, it is easy to get that: $F(\omega)=F^*(\omega)$. Thus, we have $\int_{\mathbb{R}}f(x)\cos(ax+b)dx=\frac{1}{2}\exp( ib)F^*(\omega)\|_{\omega=a}+\frac{1}{2}\exp(-ib)F(\omega)\|_{\omega=a}=F(a)cos{b}$ Now the problem is: how can we get the Fourier transform $F(\omega)$? The answer is convolution theorem! Let $h(x)=\exp(-x^2)$ and $g(x)=\frac{1}{x^2+\pi^2}$, we get $f(x)=h(x)g(x)$ (h(x), g(x) and f(x) are all absolutely integrable). The convolution theorem claims that: $F(\omega)=\int_{\mathbb{R}}h(x)g(x)\exp(-i\omega x)dx=\frac{1}{2\pi}\int_{\mathbb{R}}H(u)G(\omega-u)du$, where $H(\omega)$ and $G(\omega)$ denotes the Fourier transform of $h(x)$ and $g(x)$, respectively. $H(\omega)=\int_{\mathbb{R}}h(x)\exp(-i\omega x)dx=\sqrt{\pi}\exp(-\frac{\omega^2}{4})$. (Fourier transform of a Gaussian function is still Gaussian) $G(\omega)=\int_{\mathbb{R}}g(x)\exp(-i\omega x)dx=\exp(-\pi |\omega|)$. By applying the convolution theorem, we have $F(\omega)=\frac{1}{2\pi}\int_{\mathbb{R}}H(u)G(\om ega-u)du$ $=\frac{1}{2\pi}\int_{\mathbb{R}}\sqrt{\pi}\exp(-\frac{u^2}{4})\exp(-\pi |\omega-u|)du$ $=\frac{1}{2}\exp(\pi^2-\pi\omega)(1-\text{erf}(\pi-\frac{\omega}{2})+\exp(2\pi\omega)\text{erfc}(\pi+ \frac{\omega}{2}))$, with notations $\text{erf}(z)=\frac{2}{\sqrt\pi}\int_{-\infty}^{z}\exp(-t^2)dt$ and $\text{erfc}(z)=1-\text{erf}(z)$. Thus, the integral equals: $\int_{\mathbb{R}}f(x)\cos(ax+b)dx$ $=F(a)\cos b$ $=\frac{1}{2}\exp(\pi^2-\pi a)\cos b(1-\text{erf}(\pi-\frac{a}{2})+\exp(2\pi a)\text{erfc}(\pi+\frac{a}{2}))$. Moreover, if $a=b=0$, the integral will be degenerated to the form: $\int_{\mathbb{R}}\frac{\exp(-x^2)}{x^2+\pi^2}dx$, which shall be equal to $\exp(\pi^2)\text{erfc}(\pi)$. Of course, we can use Feynman's method to solve this 'degenerated' form and achieve the same conclusion.
 June 3rd, 2013, 11:52 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Is integral just trick? I have moved this to the university level analysis forum.
 June 4th, 2013, 12:04 AM #4 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Is integral just trick? By the way, If we consider the convolution operator $A$, and have: $[Ag](x)=\int f(y)g(x-y)dy$. If $f(x)$ is even and integrable absolutely, we can verify that the cosine functions $g(x)=\cos(ax)$ will meet the operator equation: $Ag-\lambda g=0$, $\lambda\in \mathbb{C}$ This conclusion indicates that $g(x)=\cos(ax)$ is the 'eigen' function of this operator. The corresponding eigenvalue is $\lambda=F(a)=\int f(x)\exp(-i\omega x)dx |_{\omega=a}$.
June 4th, 2013, 12:05 AM   #5
Senior Member

Joined: Oct 2010
From: Changchun, China

Posts: 492
Thanks: 14

Re: Is integral just trick?

Quote:
 Originally Posted by MarkFL I have moved this to the university level analysis forum.
Thanks!

 Tags integral, trick

### simple trick for convolution theorem

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post rain Algebra 8 July 30th, 2013 04:03 AM mathkid Algebra 3 February 25th, 2013 04:51 AM mathkid Calculus 1 September 22nd, 2012 12:42 PM ssjssujs Applied Math 1 May 16th, 2010 11:28 PM stainburg Calculus 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top