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 stainburg June 3rd, 2013 06:25 PM

Is integral just trick?

$\int_{\mathbb{R}}\frac{\exp{(-x^2)}\cos{(ax+b)}}{x^2+\pi ^2}dx$, where $a,b\in \mathbb{R}$

 stainburg June 3rd, 2013 10:29 PM

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.

 MarkFL June 3rd, 2013 10:52 PM

Re: Is integral just trick?

I have moved this to the university level analysis forum. :D

 stainburg June 3rd, 2013 11:04 PM

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 $\lambda=F(a)=\int f(x)\exp(-i\omega x)dx |_{\omega=a}$.

 stainburg June 3rd, 2013 11:05 PM

Re: Is integral just trick?

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

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